exercises - department of statistics

Exercises

4.1 Let Y be a random variable with p(y) given in the table below.

y 2 3 4

p(y) I .4 .3 .2 .I

a Give the distribution function, F( y) . Be sure to specify the value of F(y) for ally, -oo <:

y < 00.

b Sketch the distribution function given in part (a).

4.2 A box contains five keys, only one of which will open a lock. Keys are randomly selected and tried, one at a time, until the lock is opened (keys that do not work are discarded before another is tried). Let Y be the number of the trial on which the lock is opened.

a Find the probability function for Y.

b Give the corresponding distribution function.

c What is P(Y < 3)? P(Y :S 3)? P(Y = 3)?

d If Y is a continuous random variable, we argued that, for all -oo < a < oo, P(Y =a) = 0. Do any of your answers in part (c) contradict this claim? Why?

4.3 A Bernoulli random variable is one that assumes only two values, 0 and I with p(l) = p and p(O) = I - p = q.

4.4

4.5

4.6

4.7

a Sketch the corresponding distribution function.

b Show that this distribution function has the properties given in Theorem 4.1.

Let Y be a binomial random variable with n = I and success probability p.

a Find the probability and distribution function for Y .

b Compare the distribution function from part (a) with that in Exercise 4.3(a). What do you conclude?

Suppose that Y is a random variable that takes on only integer values I , 2, ... and has distribution function F(y). Show that the probability function p(y) = P(Y = y) is given by

p(y) = {

F(I), y=l,

F(y) - F(y- 1) , y = 2, 3, ....

Consider a random variable with a geometric distribution (Section 3.5); that is,

p(y) = qv- 1 p, y = I, 2, 3, ... , 0 < p < I.

a Show that Y has distribution function F(y) such that F(i) = I - q;, i = 0, I , 2, .. ·and that, in general,

{ 0, y < 0,

F(y) = I_ q ;, i ::: y < i + I , fori = 0 , I, 2, ....

b Show that the preceding cumulative distribution function has the properties given in

Theorem 4.1.

Let Y be a binomial random variable with n = I 0 and p = .2 .

a Use Table I, Appendix 3, to obtain P(2 < Y < 5) and ?(2 ::: Y < 5). Are the probabilitieS

that Y falls in the intevals (2, 5) and [2. 5) equal? Why or why not?

Exercises 167

b Use Table 1, Appendix 3, to obtain P(2 < Y :S 5) and P(2 :S Y :S 5). Are these two probabilities equal? Why or why not?

c Earlier in this section, we argued that if Y is continuous and a < b, then P(a < Y < b)= P(a ::: Y <b). Does the result in part (a) contradict this claim? Why?

4.8 Suppose that Y has density function

f(y) = { ky(l- y),

0,

0 :S y :S I,

elsewhere.

a Find the value of k that makes f(y) a probability density function.

b Find ?(.4 ::: Y :S 1) .

c Find ?(.4 :S Y < 1).

d Find P(Y :S .41 Y :S .8).

e Find P(Y < .41 Y < .8).

4.9 A random variable Y has the following distribution function:

4.10

4.11

4.12

0, for y < 2,

1/8, for 2 :S y < 2.5,

3/16, for 2.5 ::: y < 4,

F(y) = P(Y :S y) = ~ l /2 for 4 ::: y < 5.5,

5/8, for 5.5 ::: y < 6,

11 / 16, for 6::: y < 7 ,

I , for y 2: 7.

a Is Y a continuous or discrete random variable? Why?

b What values of Y are assigned positive probabilities?

c Find the probability function for Y.

d What is the median, ¢ 5 , of Y?

Refer to the density function given in Exercise 4.8.

a Find the .95-quantile, ¢ 95 , such that P(Y :S ¢95) = .95.

b Find a value y0 so that P(Y < y0 ) = .95.

c Compare the values for ¢ 95 and y0 that you obtained in parts (a) and (b). Explain the relationship between these two values.

Suppose that Y possesses the density function

{ cy, 0 ::: y ::: 2,

f(y) = 0 , elsewhere.

a Find the value of c that makes f(y) a probability density function.

b Find F(y).

c Graph f(y) and F(y).

d Use F(y) to find P(l ::: Y::: 2).

e Use f(y) and geometry to find P(l ::: Y::: 2).

The length of time to failure (in hundreds of hours) for a transistor is a random variable Y with distribution function given by

{ 0,

F(y) = I- e_-"2, y < 0,

y 2: 0.

a Show that F(y) has the properties of a distribution function. b Find the .30-quantile, ¢ 30 , of Y. c Find f(y).

d Find the probability that the transistor operates for at least 200 hours. e Find P(Y > 100/ Y c:; 200).

4.13 A supplier of kerosene has a I 50-gallon tank that is filled at the beginning of each week. His weekly demand shows a relative frequency behavior that increases steadily up to I 00 gallons and then levels off between I 00 and 150 gallons. If Y denotes weekly demand in hundreds of gallons, the relative frequency of demand can be modeled by

4.14

a Find F(y ).

b Find P(O c:; Y c:; .5).

c Find P(.5 c:; Y c:; 1.2).

I y. 0 c:: y c:: l ,

f( y ) = I , I < y c:; 1.5,

0, elsewhere.

A gas station operates two pumps, each of which can pump up to I 0,000 gallons of gas in a month. The total amount of gas pumped at the station in a month is a random variable y (measured in 10,000 gallons) with a probability density function given by

IV, 0 < y < l ,

f( y ) = 2- y, I c:; y < 2,

0, elsewhere. a Graph f(y).

b Find F(y) and graph it.

c Find the probability that the station will pump between 8000 and 12,000 gallons in a particular month.

d Given that the station pumped more than I 0,000 gallons in a particular month, find the probability that the station pumped more than 15,000 gallons during the month.

4.15 As a measure of intelligence, mice are timed when going through a maze to reach a reward of food. The time (in seconds) required for any mouse is a random variable Y with a density function given by

y?:. b, f(y) = I :2.

0, elsewhere.

where b is the minimum possible time needed to traverse the maze.

a Show that f (y) has the properties of a density function. b Find F(y).

c Find P(Y > b +c) for a positive constant c.

d If c and dare both positive constants such that d > c, find P(Y > b + d/ Y > b +c).

4.16 Let Y possess a density function

f( y ) = { c(2- y ),

0.

0 c:: y c:: 2,

elsewhere.

4.17

4.18

L- ~ "'-' 1 "-'•·"'""' _., I V.J

a Find c.

b Find F(y).


d Use F(y) in part (b) to find P(l c:; Y c:; 2) .

e Use geometry and the graph for f( y ) to calculate P(l c:; Y c:; 2).

The length of time required by students to complete a one-hour exam is a random variable with a density function given by

a Find c.

b Find F(y ).


f(y) = { cy2 + y , 0,

0 c:: y c:: l,

elsewhere.

d Use F(y) in part (b) to find F( -I), F(O), and F(l).

e Find the probability that a randomly selected student will finish in less than half an hour.

f Given that a particular student needs at least 15 minutes to complete the exam, find the probability that she will require at least 30 minutes to finish.

Let Y have the density function given by

1.2, -I < y c:: 0,

f(y)= .2+cy, O < yc:; I,

0, elsewhere.

a Find c.

b Find F(y).


d Use F(y) in part (b) to find F(-1) , F(O), and F(l).

e Find P(O c:; Y c:; .5).

f Find P(Y > .5/Y > .1).

4.19 Let the distribution function of a random variable Y be

0 ,

y

F(y) = ~ 8 ' y2

a Find the density function of Y.

b FindP(lc:;Yc:;3) .

c Find P(Y?:. 1.5).

d FindP(Y?:_l/Yc:;3).

16'

I ,

y c:: 0,

0 < y < 2,

2 c:: y < 4,

y ?:. 4.

4.20

4.21

Exercises

If, as in Exercise 4.16, Y has density function

{ (1/2)(2- y),

f(y) = 0

find the mean and variance of Y.

If, as in Exercise 4.17, Y has density function

f(y) = { (3/2)y2 + y, 0,


0.:::: y.:::: 2,

elsewhere,

0 _:::: y _:::: I,

elsewhere,

4.22 If, as in Exercise 4.18, Y has density function

4 .23

4.24

4.25

4.26

4.27

4.28

1.2, -I < y .:::: 0,

f(y) = .2+(1.2)y, 0 < y _::::I,

0, elsewhere,


Prove Theorem 4.5.

If Y is a continuous random variable with density function f(y), use Theorem 4.5 to prove that a 2 = V(Y) = E(Y2 )- [E(YW.

If, as in Exercise 4. J 9, Y has distribution function

F(y) =


0, y

8'

l 16'

I ,

y.:::: 0,

0 < y < 2,

2.:::: y < 4,

y::: 4,

If Y is a continuous random variable with mean J-1- and variance a 2 and a and b are constants, use Theorem 4.5 to prove the following:

a E(aY+b)=aE(Y)+b=aJJ-+b.

b V(aY +b)= a2 V(Y) = a2a 2

For certain ore samples, the proportion Y of impurities per sample is a random variable with density function given in Exercise 4.21. The dollar value of each sample is W = 5- .SY. Find the mean and variance of W.

The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with density function

f(y) = { cy2(1- y)4,

0,

0 _:::: y _:::: I,

elsewhere.

a Find the value of c that makes f(y) a probability density fu~ction.

b Find E(Y).

4.29 The temperature Y at which a thermostatically controlled switch turns on has probability density

function given by

4.30

Find E(Y) and V(Y).

f(y) = { 1/ 2. 0,

59.:::: y .:::: 61,

elsewhere.

The proportion of time Y that an industrial robot is in operation during a 40-hour week is a

random variable with probability density function

f(y) = ' - - , {

2y 0 < y < I

0, elsewhere.

a Find E(Y) and V(Y).

b For the robot under study, the profit X for a week is given by X = 200Y - 60. Find E(X)

and V(X).

c Find an interval in which the profit should lie for at least 75% of the weeks that the robot

is in use.

4.31 Daily total solar radiation for a specified location in Florida in October has probability density

function given by

4.32

4.33

*4.34

f(y) = { (3/32)(y- 2)(6- y),

0,

2.:::: y.:::: 6,

elsewhere,

with measurements in hundreds of calories. Find the expected daily solar radiation for October.

Weekly CPU time used by an accounting firm has probability density function (measured in

hours) given by

f(y) = { (3 / 64)y2(4- y ) ,

0,

0 .:::: y .:::: 4,

elsewhere.

a Find the expected value and variance of weekly CPU time.

b The CPU time costs the firm $200 per hour. Find the expected value and variance of the weekly cost for CPU time.

c Would you expect the weekly cost to exceed $600 very often? Why?

The pH of water samples from a specific lake is a random variable Y with probability density function given by

a Find E(Y) and V(Y).

f(y) = { (3 / 8)(7- y )2,

0,

5.:::: y.:::: 7,

elsewhere.

b Find an interval shorter than (5, 7) in which at least three-fourths of the pH measurements

must lie.

c Would you expect to see a pH measurement below 5.5 very often? Why?

Suppose that Y is a continuous random variable with density f (y) that is positive only if y ::: 0. If F(y) is the distribution function, show that

E(Y) = 1oo yf(y ) dy = 100

[1- F(y)] dy.

[Hint: If y > 0 , y = J0'" dt, and E(Y) = J000

yf(y) dy = J000 {!0' dtj f(y) dy. Exchange the

order of integration to obtain the desired result.)4

4. Exercises preceded by an asterisk are optional.

*4.35

*4.36

*4.37

4.4

DEFINITION 4.6

FIGURE 4.9 Density function

for Y

If Y is a continuous random variable such that E[(Y -a)2] < oo for all a, show that E[(Y -a)l ] is minimized when a= E(Y). [Hint: E[(Y- a)2

] = E(([Y- E(Y)] + [E(Y)- aJJ2).]

Is the result obtained in Exercise 4.35 also valid for discrete random variables? Why?

If Y is a continuous random variable with density function f(y) that is symmetric about 0

(that is , f(y) = f( - y) for ally) and E(Y) exists, show that E(Y) = 0. [Hint: E( Y) _ f~oo yf( y ) dy + f0"" yf( y ) dy. Make the change of variable w = -yin the first integral.] -

The Uniform Probability Distribution Suppose that a bus always arrives at a particular stop between 8:00 and 8:10A.M. and that the probability that the bus will arrive in any given subinterval of ti me is proportional only to the length of the subinterval. That is, the bus is as likely to arrive between 8:00 and 8:02 as it is to arrive between 8:06 and 8:08. Let Y denote the length of time a person must wait for the bus if that person arrived at the bus stop at exactly 8:00. If we carefully measured in minutes how long after 8:00 the bus arrived for several mornings, we could develop a relative frequency histogram for the data.

From the description just given, it should be clear that the relative frequency with which we observed a value of Y between 0 and 2 would be approximately the same as the relative frequency with which we observed a value of Y between 6 and 8. A reasonable model for the density function of Y is given in Figure 4.9. Because areas under curves represent probabilities for continuous random variables and A 1 = A2 (by inspection), it follows that P(O _::::: Y _::::: 2) = P(6 _::::: Y _::::: 8), as desireq.

The random variable Y just discussed is an example of a random variable that has a uniform distribution. The general form for the density function of a random variable with a uniform distribution is as follows.

If e) < e2, a random variable y is said to have a continuous uniform probability distribution on the interval (eJ ' e2) if and only if the density function of y is

f( y )

I AI

I 0 I

f(y) = I e2 ~ e1' 0,

A2

I I I I 2 3 4 5 6 7

e1.:::: y.:::: e2,

elsewhere.

8 9 10 y

DEFINITI ON 4.7

EXAMPL E 4.7

Solution

- - •· • • ~· ·· • • ,.....,...., .._.. ...., ,., ._, ,_, o..; l.O I U UI..IVII

In the bus problem we can take e1 = 0 and e2 = 10 because we are interested only in a particular ten-minute interval. The density function discussed in Example 4.2 is a uniform distribution with e1 = 0 and e2 = 1. Graphs of the distribution function and density function for the random variable in Example 4.2 are given in Figures 4.4

and 4.5, respectively.

The constants that determine the specific form of a density function are called

parameters of the density function .

The quantities e1 and e2 are parameters of the uniform density function and are clearly meaningful numerical values associated with the theoretical density function. Both the range and the probability that Y will fall in any given interval depend on the

values of e) and e2. Some continuous random variables in the physical, management, and biological

sciences have approximately uniform probability distributions. For example, suppose that the number of events, such as calls coming into a switchboard, that occur in the time interval (0, t) has a Poisson distribution. If it is known that exactly one such event has occurred in the interval (0, t), then the actual time of occurrence is distributed

uniformly over this interval.

Arrivals of customers at a checkout counter follow a Poisson distribution. It is known that, during a given 30-minute period, one customer arrived at the counter. Find the probability that the customer arrived during the last 5 minutes of the 30-minute

period.

As just mentioned, the actual time of arrival follows a uniform distribution over the

interval of (0, 30) . If Y denotes the arrival time, then

130 1 30 - 25 5 1

p (25 < y < 30) = - dy = = - = -. - - 25 30 30 30 6

The probability of the arrival occurring in any other 5-minute interval is also 1/6 . •

As we will see, the uniform distribution is very important for theoretical reasons. Simulation studies are valuable techniques for validating models in statistics. If we desire a set of observations on a random variable Y with distribution function F(y), we often can obtain the desired results by transforming a set of observations on a uniform random variable. For this reason most computer systems contain a random number generator that generates observed values for a random variable that has a

continuous uniform distribution .

176 Chapter 4 Continuous Variables and Their Probability Distributions

THEOREM 4.6

Proof

4.38

4 .39

4.40

4 .41

4 .42

4.43

4.44

If til < tiz and y is a random variable uniformly distributed on the interval (ti1, ti2), then

ti1 + tiz 1-L = E(Y) = -

2- and a 2 = V (Y) :::::: (tiz - tiJ)2

12 By Definition 4.5,

E(Y) = /_: yf(y) dy

= {Oz y (-1 ) dy lo, tiz - ti1

( 1 ) i]02

ti} - e z tiz-til 2 o, = ~

tiz + ti1 2

Note that the mean of a uniform random variable is sirnply the value midway between the two parameter values, ti1 and tiz. The derivation of the variance is left as an exercise.

Exercises

Suppose that Y has a uniform distribution over the interval (0, 1).

a Find F(y).

b Show that P(a :<:: Y :<:: a+ b), for a ;::: 0, b ;::: 0, and a + b ::; I depends only upon the value of b.

If a parachutist lands at a random point on a line between markers A and 8, find the probability that she is closer to A than to 8. Find the probability that her distance to A is more than three times her distance to 8.

Suppose that three parachutists operate independently as described in Exercise 4.39. What 1

the probability that exactly one of the three lands past the midpoint between A and B?

A random variable Y has a uniform distribution over the interval (81 · 0z>· Derive the variance

of Y. i' !he value tJ> ~ such thll

The median of the distribution of a continuous random variable Y ·rile interval (o ~) P(Y :<:: ¢ 5) = 0.5. What is the median of the uniform distribution oJl "

·II' A circle of radius r has area A = :rrr2 . If a random circle has a raJ' tributed on the interval (0, 1), what are the mean and varianceof thC

The change in depth of a river from one day to the next, measured (in is a random variable Y with the following density function:

-2:::: y::: 2 {

k , f(y) = 0,

elsewhere.

Exercises 177

a Determine the value of k.

b Obtain the distribution function for Y.

4.45 Upon studying low bids for shipping contracts, a microcomputer manufacturing company finds that intrastate contracts have low bids that are uniformly distributed between 20 and 25, in units of thousands of dollars. Find the probability that the low bid on the next intrastate shipping contract

a is below $22,000.

b is in excess of $24,000.

4.46 Refer to Exercise 4.45. Find the expected value of low bids on contracts of the type described there.

4.47 The failure of a circuit board interrupts work that utilizes a computing system until a new board is delivered . The delivery time, Y , is uniformly distributed on the interval one to five days. The cost of a board failure and interruption includes the fixed cost c0 of a new board and a cost that increases proportionally to Y2

. If C is the cost incurred, C = c0 + c 1 Y 2.

a Find the probability that the delivery time exceeds two days.

b In terms of c0 and c 1, find the expected cost associated with a single failed circuit board.

4.48 If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a , b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

a within 25 feet of the end of the line?

b within 25 feet of the beginning of the line?

c closer to the beginning of the line than to the end of the line?

4.49 A telephone call arrived at a switchboard at random within a one-minute interval. The switch board was fully busy for I 5 seconds into this one-minute period. What is the probability that the call arrived when the switchboard was not fully busy?

4.50 Beginning at I 2:00 midnight, a computer center is up for one hour and then down for two hours on a regular cycle. A person who is unaware of this schedule dials the center at a random time between 12:00 midnight and 5:00A.M. What is the probability that the center is up when the person's call comes in?

4.51 The cycle time for trucks hauling concrete to a highway construction site is uniformly distributed over the interval 50 to 70 minutes. What is the probability that the cycle time exceeds 65 minutes if it is known that the cycle time exceeds 55 minutes?

4.52 Refer to Exercise 4.5 I. Find the mean and variance of the cycle times for the trucks.

4.53 The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific eight-hour day, one defective circuit board was found.

a Find the probability that it was produced during the first hour of operation during that day.

b Find the probability that it was produced during the last hour of operation during that day.

c Given that no defective circuit boards were produced during the first four hours of operation, find the probability that the defective board was manufactured during the fifth hour.

4.54 In using the triangulation method to determine the range of an acoustic source, the test equipment must accurately measure the time at which the spherical wave front arrives at a receiving


sensor. According to Perruzzi and Hilliard ( 1984), measurement errors in these times can be modeled as possessing a uniform distribution from -0.05 to +0.05 tJ-S (microseconds).

a What is the probability that a particular arrival-time measurement will be accurate to within 0.01 /)-S?

b Find the mean and variance of the measurement errors.

4.55 Refer to Exercise 4.54. Suppose that measurement errors are uniformly distributed between -0.02 to +0.05 J..tS.

4.56

4.57

4 .5

DEFINITION 4.8

THEOREM 4.7

a What is the probability that a particular arrival-time measurement will be accurate to within 0.01 /)-S?

b Find the mean and variance of the measurement errors.

Refer to Example 4.7. Find the conditional probability that a customer arrives during the last 5 minutes of the 30-minute period if it is known that no one arrives during the first 10 minutes of the period.

According to Zimmels ( 1983), the sizes of particles used in sedimentation experiments often have a uniform distribution. In sedimentation involving mixtures of particles of various sizes, the larger particles hinder the movements of the smaller ones. Thus, it is important to study both the mean and the variance of particle sizes. Suppose that spherical particles have diameters that are uniformly distributed between .0 I and .05 centimeters. Find the mean and variance of the volumes of these particles. (Recall that the volume of a sphere is ( 4/ 3)rr r 3

.)

The Normal Probability Distribution The most widely used continuous probability distribution is the normal distribution, a distribution with the familiar bell shape that was discussed in c onnection with the empirical rule. The examples and exercises in this section illustrate some of the many random variables that have distributions that are closely approximated by a normal probability distribution.ln Chapter 7 we will present an argument that at least partially explains the common occurrence of normal distributions of data in nature. The normal

density function is as follows:

A random variable Y is said to have a normal probability distribution if and

only if, for a > 0 and -oo < f.L < oo, the density function of Y is

f(y) = _l_ e - (y-JJ.)2/ (2a2l, -oo < y < oo.

a$

Notice that the normal density function contains two parameters, f.L and a.

If Y is a normally distributed random variable with parameters f.L and a, then

E(Y)=f.L and V(Y)=a 2.

FIG U RE 4.10 The normal probability

density function

EXAMPLE 4.8

4. 5 The Normal Probability Distribution 179

j (y)

J.< y

The proof of this theorem will be deferred to Section 4.9, where we derive the moment-generating function of a normally distributed random variable. The results contained in Theorem 4.7 imply that the parameter f.L locates the center of the distribution and that a measures its spread. A graph of a normal density function is shown in Figure 4.1 0.

Areas under the normal density function corresponding to P(a .:<:: Y .:<:: b) require evaluation of the integral

- e - (y- JJ. )2 /(2a 2) 1

h 1

a a$ dy.

Unfortunately, a closed-form expression for this integral does not exist; hence, its evaluation requires the use of numerical integration techniques. Probabilities and quantiles for random variables with normal distributions are easily found using R and S-Plus. If Y has a normal distribution with mean f.L and standard deviation a, the R (or S-Pius) command pnorm(y0 ,JL,a) generates P(Y .:<:: y0) whereas qnorm (p, f.L, a) yields the pth quantile, the value of </Jp such that P(Y .:<:: ¢p) = p. Although there are infinitely many normal distributions (JL can take on any finite value, whereas a can assume any positive finite value), we need only one table-Table 4, Appendix 3-to compute areas under normal densities. Probabilities and quantiles associated with normally distributed random variables can also be found using the applet Normal Tail Areas and Quantiles accessible at www.thomsonedu.com/statistics/ wackerly. The only real benefit associated with using software to obtain probabilities and quantiles associated with normally distributed random variables is that the software provides answers that are correct to a greater number of decimal places.

The normal density function is symmetric around the value f.L, so areas need be tabulated on only one side of the mean. The tabulated areas are to the right of points z, where z is the distance from the mean, measured in standard deviations. This area is shaded in Figure 4.11.

Let Z denote a normal random variable with mean 0 and standard deviation I.

a Find P(Z > 2). b FindP(-2.:<::2.:<::2). c Find P(O .:<:: Z .:<:: 1.73).

F I G U R E. 4.11 Tabulated area

for the normal

density function

Solution

FIGURE 4.12 Desired area for

Example 4.8(b)

EXAMPLE 4.9

Solution

f(y)

JL JL + zcr y f-----z (T -----1

a Since JL = 0 and a = I, the value 2 is actually z = 2 standard deviations above the mean. Proceed down the first (z) column in Table 4, Appendix 3 and read the area opposite z = 2.0. This area, denoted by the symbol A(z), i~ A(2.0) = .0228. Thus, P(Z > 2) = .0228.

b Refer to Figure 4.12, where we have shaded the area of interest. In part (a) we determined that A 1 = A (2.0) = .0228. Because the density function is symmetric about the mean fL = 0, it follows that A2 = A 1 = .0228 and hence that

P(-2 :S Z :S 2) = 1- A1- A2 = 1- 2(.0228) = .9544.

c Because P(Z > 0) = A(O) = .5, we obtain that P(O .:::: Z .:::: 1.73) = . 5- A(1.73), where A(l.73) is obtained by proceeding down the z column in Table 4, Appendix 3, to the entry l. 7 and then across the top of the table to the column labeled .03 to read A(l.73) = .0418. Thus,

P(O :S Z :S I .73) = .5- .0418 = .4582.

0 y I

The achievement scores for a college entrance examination are normally distributed with mean 75 and standard deviation 10. What fraction of the scores lies between 80 and 90?

Recall that z is the distance from the mean of a normal distribution expressed in units of standard deviation. Thus,

z =Y-JL a

FIGURE 4.13

Required area for Example 4.9

0 .5 1.5

Then the desired fraction of the population is given by the area between

80-75 90-75 21= =.5 and 22 = =1.5.

>A 10

This area is shaded in Figure 4.13. You can see from Figure4.13 that A= A(.5)- A(l.5) = .3085-.0668 = .2417 . • We can always transform a normal random variable Y to a standard normal random

variable Z by using the relationship

Y-JL Z=--.

a Table 4, Appendix 3, can then be used to compute probabilities, as shown here . Z locates a point measured from the mean of a normal random variable, with the distance expressed in units of the standard deviation of the original normal random variable. Thus, the mean value of Z must be 0, and its standard deviation must equal 1. The proof that the standard normal random variable, Z, is normally distributed with mean 0 and standard deviation l is given in Chapter 6.

The applet Normal Probabilities, accessible at www.thomsonedu.com/statistics/ wackerly, illustrates the correspondence between normal probabilities on the original and transformed (z) scales. To answer the question posed in Example 4.9, locate the interval of interest, (80, 90), on the lower horizontal axis labeled Y. The corresponding z-scores are given on the upper horizontal axis, and it is clear that the shaded area gives P(80 < Y < 90) = P(0.5 < Z < 1.5) = 0.2417 (see Figure 4.14). A few of the exercises at the end of this section suggest that you use this applet to reinforce the calculations of probabilities associated with normally distributed ran

dom variables.

Exercises 4.58 Use Table 4, Appendix 3, to find the following probabilities for a standard normal random

variable Z:

a P(0 ::S Z ::S 1.2)

b P(- .9 ::S Z ::S 0)

c P( .3 ::S Z ::S 1.56)


F I G U R E 4.14 ?(80.0000 < Y < 90.0000) = P(0.50 < Z < 1.50) = 0.2417

Required area for 0.40

Example 4.9, using

both the original and transformed (z) scales 0.30

0.20

0.10

0.00 +-~~---------,------.--~~~~

-4.00

d P(-.2:::Z ::: .2)

e P(-1.56 ::: z::: -.2)

0.50 1.50 z

r--1 80.00 90.00

y

4.00

f Applet Exercise Use the applet Normal Probabilities to obtain P(O ::: Z ::: 1.2). Why are the values given on the two horizontal axes identical?

4.59 lf Z is a standard normal random variable, find the value zo such that

a P(Z > zo) = .5.

b P(Z < z0) = .8643.

c P(-zo < Z < zo) = .90.

d P(-zo < Z < zo) = .99.

4.60 A normally distributed random variable has density function

4.61

4.62

4.63

j(y) = _1_ e -(y-~<lz/ < 2"2l, a,J2ii

-00 < y < 00 .

Using the fundamental properties associated with any density function , argue that the parameter

a must be such that a > 0.

What is the median of a normally distributed random variable with mean f-L and standard

deviation a?

If Z is a standard normal random variable, what is

a P(Z2 < I)?

b P(Z2 < 3.84146)? . 1 fillS

A company that manufactures and bottles apple juice uses a machine that automaucal Y ne 16-ounce bottles. There is some variation, however, in the amounts of liquid dispensed 1nto tiiY bottles that are filled. The amount dispensed has been observed to be approximately norma distributed with mean 16 ounces and standard deviation I ounce.

Exercises 183

a Use Table 4, Appendix 3, to determine the proportion of bottles that will have more than 17 ounces dispensed into them.

b Applet Exercise Use the applet Normal Probabilities to obtain the answer to part (a).

4.64 The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $400 and standard deviation $20. If $450 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount?

a Answer the question, using Table 4, Appendix 3.

b Applet Exercise Use the applet Normal Probabilities to obtain the answer.

c Why are the labeled values different on the two horizontal axes?

4.65 In Exercise 4.64, how much shou ld be budgeted for weekly repairs and maintenance to provide that the probability the budgeted amount will be exceeded in a given week is only .1?

4.66 A machining operation produces bearings with diameters that are normally distributed with mean 3.0005 inches and standard deviation .0010 inch. Specifications require the bearing diameters to lie in the interval 3.000 ± .0020 inches. Those outside the interval are considered scrap and must be remachined . With the existing machine setting, what fraction of total production will be scrap?


b Applet Exercise Obtain the answer, using the applet Normal Probabilities.

4.67 In Exercise 4.66, what should the mean diameter be in order that the fraction of bearings scrapped be minimized?

4.68 The grade point averages (GPAs) of a large population of college students are approximately normally distributed with mean 2.4 and standard deviation .8 . What fraction of the students will possess a GPA in excess of 3.0?


b Applet Exercise Obtain the answer, using the applet Normal Tail Areas and Quantiles.

4.69 Refer to Exercise 4.68. If students possessing a GPA less than 1.9 are dropped from college, what percentage of the students will be dropped?

4.70 Refer to Exercise 4.68. Suppose that three students are randomly selected from the student body. What is the probability that all three will possess a GPA in excess of 3.0?

4.71

4.72

Wires manufactured for use in a computer system are specified to have resistances between .12 and .14 ohms. The actual measured resistances of the wires produced by company A have a normal probability distribution with mean .13 ohm and standard deviation .005 ohm.

a What is the probability that a randomly selected wire from company A's production will meet the specifications?

b If four of these wires are used in each computer system and all are selected from company A, what is the probability that all four in a randomly selected system will meet the specifications?

One method of arriv ing at economic forecasts is to use a consensus approach. A forecast is obtained from each of a large number of analysts; the average of these individual forecasts is the consensus forecast. Suppose that the individual 1996 January prime interest-rate forecasts of all economic analysts are approximately normally distributed with mean 7% and standard


4.73

4.74

4.75

4.76

4 .77

4.78

4.79

4.80

deviation 2.6%. If a single analyst is randomly selected from among this group, what is he probability that the analyst's forecast of the prime interest rate will t

a exceed II %?

b be less than 9%?

The width of bolts of fabric is normally distrib4ted with mean 950 mm (millimeters) and standard deviation I 0 mm.

a What is the probability that a randomly chosen bolt has a width ofbetween947 and 958 mm~

b What is the appropriate value for C such that a randomly chosen bolt has a width less than C with probability .8531?

Scores on an examination are assumed to be normally distributed with mean78 and variance36.

a What is the probability that a person taking the examination scores higher than 72?

b Suppose that students scoring in the top I 0% of this distribution are to receive an A grade. What is the minimum score a student must achieve to earn an A grade?

c What must be the cutoff point for passing the examination if the examiner wants only the top 28.1 % of all scores to be passing?

d Approximately what proportion of students have scores 5 or more points above the score that cuts off the lowest 25%?

e Applet Exercise Answer parts (a)-(d), using the applet Normal Tail Areas and Quanriles.

f If it is known that a student's score exceeds 72, what is the probability that his or her score exceeds 84?

A soft-drink machine can be regulated so that it discharges an average of J.L ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.3 ounce, give the setting for J.L so that 8-ounce cups will overflow only I% of the time.

The machine described in Exercise 4 .75 has standard deviation a that can be fixed at certain levels by carefully adjusting the machine. What is the largest value of a that will allow the actual amount dispensed to fall within I ounce of the mean with probability at least .95?

The SAT and ACT college entrance exams are taken by thousands of students each year. The mathematics portions of each of these exams produce scores that are approximately normally distributed. [n recent years , SAT mathematics exam scores have averaged 480 with standard deviation I 00. The average and standard deviation for ACT mathematics scores are 18 and 6.

respectively.

a An engineering school sets 550 as the minimum SAT math score for new students. What

percentage of students will score below 550 in a typical year?

b What score should the engineering school set as a comparable standard on the ACT

math test?

Show that the maximum value of the normal density with parameters J.L and a is I f(a j2ifl

and occurs when y = J.L.

Show that the normal density with parameters J.L and a has inflection points at the values iJ _(!

and J.L +a. (Recall that an inflection point is a point where the curve changes direction frortl concave up to concave down, or vice versa, and occurs when the second derivative changes sign. Such a change in sign may occur when the second derivative equals zero.)

Assume that Y is normally distributed with mean J.L and standard deviation a. After observinf a value of Y, a mathematician constructs a rectangle with length L = IYI and width W == 3I Y · Let A denote the area of the resulting rectangle. What is E(A)?

4.6 The Gamma Probabi I ity Distribution

DEFINITION 4 .9

FIGUR E 4.1 5 A skewed probabi lity

density function

Some random variables are always nonnegative and for various reasons yield distributions of data that are skewed (nonsymrnetric) to the right. That is, most of the area under the density function is located near the origin, and the density function drops gradually as y increases. A skewed probability density function is shown in

Figure 4.15. The lengths of time between malfunctions for aircraft engines possess a skewed

frequency distribution, as do the lengths of time between arrivals at a supermarket checkout queue (that is, the line at the checkout counter). Similarly, the lengths of time to complete a maintenance checkup for an automobile or aircraft engine possess a skewed frequency distribution. The populations associated with these random variables frequently possess density functions that are adequately modeled by a gamma

density function.

A random variable Y is said to have a gamma distribution with parameters

a > 0 and f3 > 0 if and only if the density function of Y is

{

"--Ya_-_1 e_-_Y l_f3 ' 0 < y < 00 ,

J(y) = {3ar(a) -

0, elsewhere,

where

r(a) = laoo ya-le- y dy.

The quantity r(a) is known as the gamma function. Direct integration will verify that f(l) = l. Integration by parts will verify that r(a) =(a- l)f(a- 1) for any a > I and that r(n) = (n - l)!, provided that n is an integer.

Graphs of gamma density functions for a = l, 2, and 4 and f3 = 1 are given in Figure 4.16. Notice in Figure 4.16 that the shape of the gamma density differs for the different values of a. For this reason, a is sometimes called the shape parameter associated with a gamma distribution. The parameter f3 is generally called the scale parameter because multiplying a gamma-distributed random variable by a positive constant (and thereby changing the scale on which the measurement is made) produces

f(y)

0 y


THEOREM 4.9

Proof

0 EFI N ITION 4.11

THEOREM 4.10

Proof

If Y is a chi-square random variable with v degrees of freedom, then

f.L = E(Y) = v and a 2 = V(Y) = 2v.

Apply Theorem 4.8 with a = vl2 and f3 = 2.

Tables that give probabilities associated with x 2 distributions are readily available in most statistics texts. Table 6, Appendix 3, gives percentage points associated with x2 distributions for many choices of v. Tables of the general gamma distribution are not so readily available. However, we will show in Exercise 6.46 that if Y has a gamma distribution with a= nl2 for some integer n, then 2Y I f3 has a x2 distribution with n degrees of freedom. Hence, for example, if Y has a gamma distribution with a = 1.5 = 312 and f3 = 4, then 2Y I f3 = 2Y 14 = Y 12 has a x2 distribution with 3 degrees of freedom. Thus, P (Y < 3.5) = P([Y 12] < 1.75) can be found by us ing readily available tables of the x2 distribution.

The gamma density function in which a = I is called the exponential density function.

A random variable Y is said to have an exponential distribution with parameter f3 > 0 if and only if the density function of Y is

11 -e-yf/3 0 < y < oo

f(y) = f3 , - , 0, elsewhere.

The exponential density function is often useful for modeling the length of life of electronic components. Suppose that the length of time a component already has operated does not affect its chance of operating for at least b additional time units. That is, the probability that the component will operate for more than a + b time units. given that it has already operated for at least a time units, is the same as the probability that a new component will operate for at least b time units if the new component is put into service at time 0. A fuse is an example of a component for which this assumption often is reasonable. We will see in the next example that the exponential distribution provides a model for the distribution of the lifetime of such a component.

If Y is an exponential random variable with parameter f3, then

f.L = E(Y) = f3 and a 2 = V(Y) = f3 2•

The proof follows directly from Theorem 4.8 with a = 1.

EXAMPLE 4.10 Suppose that Y has an exponential probability density function. Show that, if a 7 0

and b > 0,

P(Y >a+ biY > a)= P(Y > b).

Exercises 189

Solution From the definition of conditional probability, we have that

4.81

4.82

4.83

4.84

P(Y >a + biY >a) = P(Y >a+ b) P(Y > a)

because the intersection of the events (Y > a + b) and (Y > a) is the event (Y > a+ b). Now

P(Y > a+ b) = -e-yff3 dy = -e-yff3 = e-(a+b)/{3. 1oo 1 Joo a+b f3 a+b

Similarly,

100 )

P(Y >a) = a "f/ -y/{3 dy = e-a ff3,

and

e-<a+b)/{3 P(Y > a+ biY > a)= = e - bff3 = P(Y > b).

e-a/{3

This property of the exponential distribution is often called the memoryless property ~~ili~~~. •

You will recall from Chapter 3 that the geometric distribution, a discrete distribution, also had this memoryless property. An interesting relationship between the exponential and geometric distributions is given in Exercise 4.95.

Exercises

a If a> 0, r(a) is defined by r(a) = fo00 ya- le - Y dy, show that f(l) = l. *b If a > 1, integrate by parts to prove that r (a) = (a - l)f(a - 1 ).

Use the results obtained in Exercise 4.81 to prove that if n is a positive integer, then f(n) = (n- l)!. What are the numerical values of f(2), f(4) , and f(7)?

Applet Exercise Use the applet Comparison of Gamma Density Functions to obtain the results given in Figure 4. 16.

Applet Exercise Refer to Exercise 4.83. Use the applet Comparison of Gamma Density Functions to compare gamma density functions with (a = 4, f3 = 1), (a = 40, f3 = I), and (a= 80, f3 = 1).

a What do you observe about the shapes of these three density functions? Which are less skewed and more symmetric?

b What differences do you observe about the location of the centers of these density functions? c Give an explanation for what you observed in part (b).

4.85

4.86

Applet Exercise Use the applet Comparison of Gamma Density Functions to compare ga111 density functions with (a= l , (3 = 1), (a= l , (3 = 2), and (a= l , (3 = 4). 11la

a b c

What is another name for the density functions that you observed?

Do these densities have the same general shape?

The parameter (3 is a "scale" parameter. What do you observe about the "spread" of the three density functions ? se

Applet Exercise When we discussed the x2 distribution in this section, we presented (With justification to follow in Chapter 6) the. fact that if Y is gamma distributed with a == nj2 for some integer n, then 2Y j (3 has a x2 distribution. In particular, it was stated that when a :::

1_5 and (3 = 4, W = Y / 2 has a x 2 distribution with 3 degrees of freedom.

a

b Use the applet Gamma Probabilities and Quantiles to find P(Y < 3.5).

Use the applet Gamma Probabilities and Quantiles to find P(W < 1.75). [Hint: Recall that the x2

distribution with v degrees of freedom is just a gamma distribution with a == v;2 and (3 = 2.]

c Compare your answers to parts (a) and (b).

4.87 Applet Exercise Let Y and W have the distributions given in Exercise 4.86.

a Use the applet Gamma Probabilities and Quantiles to find the .05-quantile of the distribution of Y.

b

c

Use the applet Gamma Probabilities and Quanti/es to find the .05-quantile of the x ~ distribution with 3 degrees of freedom .

What is the relationship between the .05-quantile of the gamma (a = 1.5, (3 = 4) distribution and the .05-quantile of the x2 distribution with 3 degrees of freedom? Explain this relationship.

4.88 The magnitude of earthquakes recorded in a region of North America can be modeled as having an exponential distribution with mean 2.4, as measured on the Richter scale. Find the probability that an earthquake striking this region will

a exceed 3.0 on the Richter scale.

b fall between 2.0 and 3.0 on the Richter scale.

4.89 If Y has an exponential distribution and P(Y > 2) = .0821, what is

a (3 = E(Y)?

b P(Y ::; 1.7)?

4.90 Refer to Exercise 4.88. Of the next ten earthquakes to strike this region, what is the probability that at least one will exceed 5.0 on the Richter scale?

4.91 The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean I 00 cfs (cubic feet per second).

a Find the probability that the demand will exceed 200 cfs during the early afternoon on a randomly selected day.

b What water-pumping capacity should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day " only .01?

4.92 The length of time Y necessary to complete a key operatio11 in the construction of houses ha: an exponential distribution with mean I 0 hours. The formula C = I 00 + 40Y + 3Y2 relate.

4.93

4.94

the cost C of completing this operation to the square of the time to completion. Find the mean

and variance of C.

Historical evidence indicates that times between fatal accidents on scheduled American domestic passenger flights have an approximately exponential distribution. Assume that the mean time between accidents is 44 days.

a If one of the accidents occurred on July I of a randomly selected year in the study period, what is the probability that another accident occurred that same month?

b What is the variance of the times between accidents?

One-hour carbon monoxide concentrations in air samples from a large city have an approximately exponential distribution with mean 3.6 ppm (parts per million).

a Find the probability that the carbon monoxide concentration exceeds 9 ppm during a randomly selected one-hour period.

b A traffic-control strategy reduced the mean to 2.5 ppm. Now find the probability that the concentration exceeds 9 ppm.

4.95 Let Y be an exponentially distributed random variable with mean (3. Define a random variable X in the following way: X = k if k - 1 ::; Y < k fork = I , 2, . . ..

4.96

a Find P(X = k) for each k = I, 2, ....

b Show that your answer to part (a) can be written as

P(X = k) = (e - 11.B )k- 1 (I_ e- 11.8 ) , k =I , 2, . . .

and that X has a geometric distribution with p =(I- e- 11.8 ).

Suppose that a random variable Y has a probability density function given by

j(y) = { ky3e-yl2,

0,

y > 0,

elsewhere.

a Find the value of k that makes j(y) a density function.

b Does Y have a x2 distribution? If so, how many degrees of freedom?

c What are the mean and standard deviation of Y?

d Applet Exercise What is the probability that Y lies within 2 standard deviations of its mean?

4.97 A manufacturing plant uses a specific bulk product. The amount of product used in one day can be modeled by an exponential distribution with (3 = 4 (measurements in tons). Find the probability that the plant will use more than 4 tons on a given day.

4.98 Consider the plant of Exercise 4.97. How much of the bulk product should be stocked so that the plant's chance of running out of the product is only .05?

4.99 If A. > 0 and a is a positive integer, the relationship between incomplete gamma integrals and sums of Poisson probabilities is given by

I l oo r(a) A y"- 1e-y dy =I: A_Xe-1. x =O X! .

*4.1 00

If Y has a gamma distribution with a = 2 and ,B = I, find P(Y > I) by using the precedin equality and Table 3 of Appendix 3. g

a

Applet Exercise If Y has a gamma distribution with a = 2 and ,B = I, find P (Y > 1) b using the apple! Gamma Probabilities. y

b

Let Y be a gamma-distributed random variable where a is a positive integer and ,B == 1. Th result given in Exercise 4.99 implies that that if y > 0, e

a- 1 x -y

""~ = P(Y > y). L.., x! x=O

Suppose that X 1 is Poisson distributed with mean A1 and X2 is Poisson distributed with mean A2, where A2 > AJ.

a Show that P(X 1 = 0) > P(X2 = 0).

b Letk be any fixed positive integer. Show that P(X1 ::; k) = P(Y > A1) and P(X2 s k) == P(Y > A2) , where Y is has a gamma distribution with a= k + I and ,B = l.

c Let k be any fixed positive integer. Use the result derived in part (b) and the fact that A. 2 > ;.1

to show that P(X 1 S k) > P(X2 ::: k).

d Because the result in part (c) is valid for any k = I , 2, 3, ... and part (a) is also valid, we

have established that P(X1 S k) > P(X2 S k) for all k = 0, I , 2, . ... Interpret this result .

4.101 Applet Exercise Refer to Exercise 4.88. Suppose that the magnitude of earthquakes striking the region has a gamma distribution with a = .8 and ,B = 2.4.

4.102

4.103

4.104

4.105

a What is the mean magnitude of earthquakes striking the region?

b What is the probability that the magnitude an earthquake striking the region will exceed 3.0 on the Richter scale?

c Compare your answers to Exercise 4.88(a) . Which probability is larger? Explain.

d What is the probability that an earthquake striking the regions will fall between 2.0 and 3.0 on the Richter scale?

Applet Exercise Refer to Exercise 4.97. Suppose that the amount of product used in one day has a gamma distribution with a = 1.5 and ,B = 3.

a Find the probability that the plant will use more than 4 tons on a given day.

b How much of the bulk product should be stocked so that the plant's chance of runn ing out of the product is only .05?

Explosive devices used in mining operations produce nearly circular craters when detonated. The radii of these craters are exponentially distributed with mean 10 feet. Find the mean and variance of the areas produced by these explosive devices.

The lifetime (in hours) Y of an electronic component is a random variable with density fu nction given by

j(y) = ~~~0 e-rf l 00, 0,

y > 0,

elsewhere .

Three of these components operate independently in a piece of equipment. The equipment fa! Is if at least two of the components fail. Find the probability that the equipment will operate tor

at least 200 hours without failure.

Four-week summer rainfall totals in a section of the Midwest United States have approxi matelY

a gamma distribution with a = 1.6 and ,B = 2.0.

4.106

4.107

Exercises 193

a Find the mean and variance of the four-week rainfall totals .

b Applet Exercise What is the probability that the four-week rainfall total exceeds

4 inches?

The response times on an online computer terminal have approximately a gamma distribution

with mean four seconds and variance eight seconds2.

a Write the probability density function for the response times.

b Applet Exercise What is the probability that the response time on the terminal is less than

five seconds?

Refer to Exercise 4.106.

a Use Tchebysheff' s theorem to give an interval that contains at least 75% of the response

times. b Applet Exercise What is the actual probability of observing a response time in the interval

you obtained in part (a)?

4.1 08 Annual incomes for heads of household in a section of a city have approximately a gamma

distribution with a = 20 and ,B = 1000.

a Find the mean and variance of these incomes.

b Would you expect to find many incomes in excess of $30,000 in this section of the city?

c Applet Exercise What proportion of heads of households in this section of the city have

incomes in excess of $30,000?

4.1 09 The weekly amount of downtime Y (in hours) for an industrial machine has approximately a gamma distribution with a = 3 and ,B = 2. The loss L (in dollars) to the industrial operation as a result of this downtime is given by L = 30Y + 2Y2

. Find the expected value and variance

of L.

4.110 If Y has a probability density function given by

j(y) = { 4yle-2Y , 0,

obtain E(Y) and V(Y) by inspection.

y > 0,

elsewhere,

4.11 1 Suppose that Y has a gamma distribution with parameters a and ,B.

a If a is any positive or negative value such that a +a > 0, show that

E(Y" ) = ,B 'T(a +a) r(a)

b Why did your answer in part (a) require that a +a > 0?

c Show that, with a = I, the result in part (a) gives E (Y) = a,B. d Use the result in part (a) to give an expression for £(../Y). What do you need to assume

about a? e Use the result in part (a) to give an expression for £(1/ Y), E(l/ ../Y), and £(1/ Y

2). What

do you need to assume about a in each case?

4.11 2 Suppose that Y has a x 2 distribution with v degrees of freedom. Use the results in Exercise 4.111 in your answers to the following. These resu lts will be useful when we study the t and

F distributions in Chapter 7.

a Give an expression for E(Y") if v > -2a.

b Why did your answer in part (a) require that v > -2a?

c Use the result in part (a) to give an expression for £(../f). What do you need to assu about v? rne

d Use the result in part (a) to give an expression for E (I I Y), E (I I ../Y), and E (I 1 Y2). What do you need to assume about v in each case?

4.7 The Beta Probabi lity Distribution

DEFINITION 4.12

The beta density function is a two-parameter density function defined over the closed interval 0 ::: y ::: I. 1t is often used as a model for proportions, such as the proportion of impurities in a chemical product or the proportion of time that a machine is under repair.

A random variable Y is said to have a beta probability distribution with parameters a > 0 and fJ > 0 if and only if the density function of Y is

I a-1 (l _ )f!-1

y y 0< <1 f(y) = B(a, fJ) ' - y - '

0, elsewhere,

where

B(a, {3) = 1' ya-1 (I _ y)fl-1 dy = f(a)f({J) o f(a+fJ)'

The graphs of beta density functions assume widely differing shapes for various values of the two parameters a and {3. Some of these are shown in Figure 4.17. Some of the exercises at the end of this section ask you to use the applet Comparison of Belll Density Functions accessible at www.thomsonedu.com/statistics/wackerly to explore and compare the shapes of more beta densities.

Notice that defining y over the interval 0 ::: y ::: I does not restrict the use of the beta distribution. If c ::: y ::: d, then y* = (y - c)j(d - c) defines a new variable such that 0 ::: y* ::: l. Thus, the beta density function can be applied to a random variable defined on the interval c ::: y ::: d by translation and a change of scale.

The cumulative distribution function for the beta random variable is commonlY called the incomplete beta function and is denoted by

1y ta-1(1 _ t)fl - 1

F(y) = dt = ly(a, {3). o B(a, fJ)

A tabulation of fv(a, {3) is given in Tables of the Incomplete Beta Function (Pearsonj 1968). When a ~nd f3 are both positive integers, lv (a, fJ) is related to the binorma

4.7 The Beta Probability Distribution 195

f iGUR E 4.17 f(y)

Beta density functions

THEORE M 4.11

0 y

probability function. Integration by parts can be used to show that for 0 < y < I,

and a and f3 both integers,

F(y) = t dt = L . yi(l- y)" - i, 1Y a-1(1-t)fl - 1 "(n) 0 B(a, {3) i=a l

where n = a + fJ - I. Notice that the sum on the right-hand side of this expression is just the sum of probabilities associated with a binomial random variable with n = a + f3 - 1 and p = y. The binomial cumulative distribution function is presented in Table I, Appendix 3, for n = 5, 10, 15, 20, and 25 and p = .01, .05, .10, .20, .30, .40, .50, .60, .70, .80, .90, .95, and .99. The most efficient way to obtain binomial probabilities is to use statistical software such as R or S-Plus (see Chapter 3). An even easier way to find probabilities and quantiles associated with beta-distributed random variables is to use appropriate software directly. The Thomson website provides an applet, Beta Probabilities, that gives "upper-tail" probabilities [that is, P(Y > y0)] and quantiles associated with beta-distributed random variables. In addition, if Y is a beta-distributed random variable with parameters a and fJ, the R (or S-Pius) command pbeta (y0 , a, 1/ f3 l generates P(Y ::: Yo), whereas qbeta (p, a, 1/ fJ) yields the pth quanti le, the value of

¢, such that P(Y ::: ¢,) = p.

If Y is a beta-distributed random variable with parameters a > 0 and f3 > 0,

then a

1-L = E(Y) = a+ f3 a 2 = V(Y) = a{J and

. . _ -·- . ~ .. ~v·~~ ao •u "1e11 rruoaOIIJty UJstnbutions

Proof By definition,

EXAMPLE 4.11

Solution

E(Y) = 1: yf(y) dy

= rl , [y"-'o- y).B-'] d lo ) B(a, fJ) Y

= 1 fo'y"(l-y).B-ldy

B(a + 1, fJ)

- B(a, fJ) (because a > 0 implies that a + 1 > 0)

f'(a + fJ) f'(a + 1)f'(fJ) = X----f'(a)f'(fJ) f'(a + fJ + 1)

f'(a + fJ) af'(a)f'(fJ) - X -- f'(a)f'(fJ) (a+ fJ)f'(a + fJ) - (a+ fJ)

a

The derivation of the variance is left to the reader (see Exercise 4.130).

We will see in the next example that the beta density function can be integrated directly in the case when a and fJ are both integers.

A gasoline wholesale distributor has bulk storage tanks that hold fixed supplies and are filled every Monday. Of interest to the wholesaler is the proportion of this supply that is sold during the week. Over many weeks of observation, the distributor found that this proportion could be modeled by a beta distribution with a = 4 and f3 = 2. Find the probability that the wholesaler will sell at least 90% of her stock in a given week.

If Y denotes the proportion sold during the week, then

{

1(4 + 2) 3 I 0 1 f'(4)f'(2)y ( - y), .:::: y.:::: , f(y) =

0, and

elsewhere,

P(Y > .9) = ~co f(y) dy = ~ 1

20(y 3 - y 4 ) dy .9 .9

= 201 ~'I_ ~'I 1 = 20(004) = 08

It is not very likely that 90% of the stock will be sold in a given week.

Exercises 197

Exercises

4.113 Applet Exercise Use the applet Comparison of Beta Density Functions to obtain the results given in Figure 4.17.

4.114 Applet Exercise Refer to Exercise 4.113. Use the applet Comparison of Beta Density Functions to compare beta density functions with (a = I, fJ = I), (a = I, fJ = 2), and (a=2,fJ=I).

a What have we previously called the beta distribution with (a = I, fJ = I)?

b Which of these beta densities is symmetric?

c Which of these beta densities is skewed right?

d Which of these beta densities is skewed left?

*e In Chapter 6 we will see that if Y is beta distributed with parameters a and {J, then Y* = I - Y has a beta distribution with parameters a* = fJ and {J* =a. Does this explain the differences in the graphs of the beta densities?

4.115 Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta density functions with (a = 2, fJ = 2), (a = 3, fJ = 3), and (a = 9, fJ = 9).

4.116

4.117

a What are the means associated with random variables with each of these beta distributions?

b What is similar about these densities?

c How do these densities differ? In particular, what do you observe about the "spread" of these three density functions?

d Calculate the standard deviations associated with random variables with each of these beta densities. Do the values of these standard deviations explain what you observed in part (c)? Explain.

e Graph some more beta densities with a = {J. What do you conjecture about the shape of beta densities with a = {J?

Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta density functions with (a = 1.5 , fJ = 7), (a = 2.5, fJ = 7), and (a = 3.5, fJ = 7).

a Are these densities symmetric? Skewed left? Skewed right?

b What do you observe as the value of a gets closer to 7?

c Graph some more beta densities with a > I, fJ > I, and a < {J. What do you conjecture about the shape of beta densities when both a > I, fJ > I, and a < {J?

Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta density functions with (a = 9, fJ = 7), (a = I 0, fJ = 7), and (a = 12, fJ = 7).



c Graph some more beta densities with a > I, fJ > I, and a > {J. What do you conjecture about the shape of beta densities with a > fJ and both a > I and fJ > I?

4 .118 Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta density functions with (a= .3 , fJ = 4), (a= .3, fJ = 7), and (a= .3, fJ = 12).


b What do you observe as the value of fJ gets closer to 12?

4.119

*4.120

4.121

4.122

4.123

c Which of these beta distributions gives the highest probability of observing a va lue Ia rger

than 0.2?

d Graph some more beta densities with a < I and f3 > I. What do you conjecture about the shape of beta densities with a < I and f3 > I?

Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta d sity functions with (a= 4, f3 = 0.3), (a= 7, f3 = 0.3) , and (a= 12, f3 = 0.3). en.



c Which of these beta distributions gives the highest probabi lity of observing a value less than 0.8?

d Graph some more beta densities with a > I and f3 < l . What do you conjecture about the shape of beta densities with a > I and f3 < I?

In Chapter 6 we will see that if Y is beta distributed with parameters a and {3, then y• = 1 _ y has a beta distribution with parameters a• = f3 and {3* =a. Does this explain the differences and similarities in the graphs of the beta densities in Exercises 4.118 and 4.119?

Applet Exercise Use the applet Comparison of Beta Density Functions to compare beta density functions with (a = 0.5 , f3 = 0.7) , (a= 0.7, f3 = 0.7), and (a= 0.9, f3 = 0.7).

a What is the general shape of these densities?

b What do you observe as the value of a gets larger?

Applet Exercise Beta densities with a < I and f3 < I are difficult to display because of scaling/resolution problems.

a Use the applet Beta Probabilities and Quantiles to compute P(Y > 0.1) if Y has .a beta distribution with (a = 0.1 , f3 = 2).

b Use the applet Beta Probabilities and Quantiles to compute P(Y < 0.1) if Y has a beta distribution with (a = 0. 1, f3 = 2).

c Based on your answer to part (b), which values of Y are assigned high probabi lities if Y has a beta distribution with (a = 0.1, f3 = 2)?

d Use the applet Beta Probabilities and Quantiles to compute P(Y < 0.1) if Y has a beta distribution with (a = 0.1, f3 = 0.2).

e Use the applet Beta Probabilities and Quantiles to compute P(Y > 0.9) if Y has a beta distribution with (a = 0. 1, f3 = 0.2).

f Use the applet Beta Probabilities and Quantiles to compute P (0.1 < Y < 0.9) if Y has a beta distribution with (a = .l, f3 = 0.2).

. . h g Based on your answers to parts (d), (e), and (f), which values of Y are asstgned htg

probabilities if Y has a beta disttibution with (a = 0. 1, f3 = 0.2)?

The relative humidity Y , when measured at a location, has a probability density function given by

f(y) = { ky3(1- y)2,

0,

0 ::: y::: I ,

elsewhere.

a Find the value of k that makes f(y) a density function. . d' t value

b Applet Exercise Use the applet Beta Probabilities and Quantiles to find a hurru 1 Y that is exceeded only 5% of the time. ·

4.124

4.125

4.126

4.127

4.128

4.129

Exercises 199

The percentage of impurities per batch in a chemical product is a random variable Y with

density function

f(y) = { l2y2( 1 - y),

0,

0::: y::: l ,

elsewhere.

A batch with more than 40% impurities cannot be sold .

a Integrate the density directly to determine the probability that a randomly selected batch

cannot be sold because of excessive impurities. b Applet Exercise Use the applet Beta Probabilities a11d Quantiles to find the answer to

part (a).

Refer to Exercise 4.124. Find the mean and variance of the percentage of impurities in a

randomly selected batch of the chemical.

Suppose that a random variable Y has a probability density function given by

{

6y( l -y), O<y<l, f(y) = -. -

0, elsewhere.

a Find F(y).

b Graph F(y) and f(y).

c Find P(.5 :S Y :S .8).

Verify that if Y has a beta distribution with a = f3 = l , then Y has a uniform distribution over (0, l ). That is, the uniform distribution over the interval (0, l) is a special case of a beta

distribution.

The weekly repair cost Y for a machine has a probability density function given by

{

3(1- y)2 , 0 < y < I , f(y) = .

0, elsewhere,

with measurements in hundreds of dollars. How much money should be budgeted each week for repair costs so that the actual cost will exceed the budgeted amount only l 0% of the time?

During an eight-hour shift, the proportion of time Y that a sheet-metal stamping machine is down for maintenance or repairs has a beta distribution with a = I and f3 = 2. That is,

{

2(1-y), o:::y:::l, f(y) =

0, elsewhere.

The cost (in hundreds of dollars) of this downtime, due to lost production and cost of maintenance and repair, is given by C = 10 + 20Y + 4Y2 Find the mean and variance of C.

4.130 Prove that the variance of a beta-distributed random variable with parameters a and f3 is

4.131

, af3 a- = ----=-'----

(a + f3) 2 (a + f3 + l) ·

Errors in measuring the time of atTival of a wave front from an acoustic source sometimes have an approximate beta distribution. Suppose that these errors, measured in microseconds, have

approximately a beta distribution with a = I and f3 = 2.

a What is the probability that the measurement error in a randomly selected instance is less

than .5 J-iS? b Give the mean and standard deviation of the measurement errors.

-~"'" ouvu~ vaflaUie~ dna I heir 1-'robabi/ity Distributions

4.132 Proper blending of fine and coarse powders prior to copper sintering is essential for unifor . in the finished product. One way to check the homogeneity of the blend is to select many s illll)o

samples of the blended powders and measure the proportion of the total weight contribute;~! the fine powders in each. These measurements should be relatively stable if a homogeneo~ blend has been obtained.

a Suppose that the proportion of total weight contributed by the fine powders has a be distribution with a = f3 = 3. Find the mean and variance of the proportion of Weig: contributed by the fine powders.

1 b Repeat part (a) if a = f3 = 2.

c Repeat part (a) if a = f3 = l.

4.133 d Which of the cases-parts (a), (b), or (c)-yields the most homogeneous blending?

The proportion of time per day that all checkout counters in a supermarket are busy is a random variable Y with a density function given by

4.134

*4.135

{ cy

2 ( 1 - y) 4 , 0 < y < l, f(y) = - -

0, elsewhere.

a Find the value of c that makes f(y) a probability density function.

b Find E(Y) . (Use what you have learned about the beta-type distribution. Compare your answers to those obtained in Exercise 4.28.)

c Calculate the standard deviation of Y.

d Applet Exercise Use the applet Beta Probabilities and Quantiles to find P(Y > f.1. + 2a).

In the text of this section, we noted the relationship between the distribution function of.a beta-distributed random variable and sums of binomial probabilities. Specifically, if Y has a beta distribution with positive integer values for a and f3 and 0 < y < 1,

l y ta - l (l _ t) f3- l " ( ) F(y) = dt = L n /(1- y)" - i,

0 B(a. 8) i=a t

where n = a + f3 - 1.

a If Y has a beta distribution with a = 4 and f3 = 7, use the appropriate binomial tables 10 find P(Y .:5 .7) = F(.7).

b If Y has a beta distribution with a = 12 and f3 = 14, use the appropriate binomial tables to find P(Y .:5 .6) = F(.6).

c Applet Exercise Use the applet Beta Probabilities and Quantiles to find the probabilities in parts (a) and (b).

Suppose that Y1 and Y2 are binomial random variables with parameters (n, p1) and (n . P1),

respectively, where p 1 < p2 • (Note that the parameter n is the same for the two variables.)

a Use the binomial formula to deduce that P(Y1 = 0) > P(Y

2 = 0).

b Use the relationship between the beta distribution function and sums of binomial probabilities given in Exercise 4.134 to deduce that, if k is an integer between 1 and n - l,

k 1 tk (l _ t)n-k - 1 P(Y, .:5 k) = t;; (;)cp,i(l- p,)"-i = i, B(k + 1, n- k) dt.

4.8 Some General Comments 201

c lf k is an integer between 1 and 11- 1, the same argument used in part (b) yields that

k (11) j' t k(1- t)ll -k- 1 P(Y2 .:5 k) = L . (pz/ (1- P2)"-i = dt.

i = D t " 2 B(k + I, n - k)

Show that, if k is any integer between 1 and 11- I , P(Y1 .:5 k) > P(Y2 .:5 k). Interpret this result.

4.8 Some General Comments

Keep in mind that density functions are theoretical models for populations of real data that occur in random phenomena. How do we know which model to use? How much does it matter if we use the wrong density as our model for reality?

To answer the latter question first, we are unlikely ever to select a density function that provides a perfect representation of nature; but goodness of fit is not the criterion for assessing the adequacy of our model. The purpose of a probabilistic model is to provide the mechanism for making inferences about a population based on information contained in a sample. The probability of the observed sample (or a quantity proportional to it) is instrumental in making an inference about the population. It follows that a density function that provides a poor fit to the population frequency distribution could (but does not necessarily) yield incorrect probability statements and lead to erroneous inferences about the population. A good model is one that yields good inferences about the population of interest.

Selecting a reasonable model is sometimes a matter of acting on theoretical considerations. Often, for example, a situation in which the discrete Poisson random variable is appropriate is indicated by the random behavior of events in time. Knowing this, we can show that the length of time between any adjacent pair of events follows an exponential distribution. Similarly, if a and b are integers, a < b, then the length of time between the occurrences of the ath and bth events possesses a gamma distribution with a = b -a. We will later encounter a theorem (called the central limit theorem) that outlines some conditions that imply that a normal distribution would be a suitable approximation for the distribution of data.

A second way to select a model is to form a frequency histogram (Chapter I) for data drawn from the population and to choose a density function that would visually appear to give a similar frequency curve. For example, if a set of n = 100 sample measurements yielded a bell-shaped frequency distribution, we might conclude that the normal density function would adequately model the population frequency distribution.

Not all model selection is completely subjective. Statistical procedures are available to test a hypothesis that a population frequency distribution is of a particular type. We can also calculate a measure of goodness of fit for several distributions and select the best. Studies of many common inferential methods have been made to determine the magnitude of the errors of inference introduced by incorrect population models. It is comforting to know that many statistical methods of inference are insensitive to assumptions about the form of the underlying population frequency distribution.

I ' 1-:11"


4.136

The moments of U = Y - f.-t can be obtained from m(t) by differentiating rn (t) .

accordance with Theorem 3.12 or by expanding m(t) into a series. •

---The purpose of the preceding discussion of moments is twofold. First, momen can be used as numerical descriptive measures to describe the data that we obtain its an experiment. Second, they can be used in a theoretical sense to prove that a rando n variable possesses a particular probability distribution. It can be shown that if tw~ random variables Y and Z possess identical moment-generating functions, then y and Z possess identical probability distributions. This latter application of moments was mentioned in the discussion of moment-generating functions for discrete random variables in Section 3.9; it applies to continuous random variables as well.

For your convenience, the probability and density functions, means, variances, and moment-generating functions for some common random variables are given in Appendix 2 and inside the back cover of this text.

Exercises Suppose that the waiting time for the first customer to enter a retail shop after 9:00A.M. is a random variable Y with an exponential density function given by

f( y )= I G)e-rlo, 0,

y > 0,

elsewhere.

a Find the moment-generating function for Y.

b Use the answer from part (a) to find E(Y) and V(Y).

4.137 Show that the result given in Exercise 3.158 also holds for continuous random variables. That is, show that, if Y is a random variable with moment-generating function m(t) and U is given by U = aY + b, the moment-generating function of U is e1bm(at). If Y has mean f1 and variance a 2 , use the moment-generating function of U to derive the mean and variance of U.

4.138 Example 4.16 derives the moment-generating function for Y - J.L, where Y is normally dis·

tributed with mean J.L and variance a 2.

a Use the results in Example 4.16 and Exercise 4.137 to find the moment-generating function

for Y.

b Differentiate the moment-generating function found in part (a) to show that E(Y) == f1 and

V(Y) = a 2.

4.139 The moment-generating function of a normally distributed random varia~ I~, Y, with rne~ J.L and variance a 2 was shown in Exercise 4.138 to be m(t) = e~-' 1 +0 12>1 ·a·. Use the resu in Exercise 4.137 to derive the moment-generating function of X = -3Y + 4. What 15 the

distribution of X? Why?

4.140 Identify the distributions of the random variables with the following moment-generating

functions:

a m(t) =(I- 41) - 2 .

b m(t) = l/(1 - 3.2t). c m(t) = e- 5t+6t z

4.1 41

4.1 42

4.1 43

4.144

4.145

4.10 T chebysheff' s Theorem 207

If 81

< 82

, derive the moment-generating function of a random variable that has a uniform

distribution on the interval (81 , 82 ).

Refer to Exercises 4. 141 and 4.137. Suppose that Y is uniformly distributed on the interval

(0, I) and that a > 0 is a constant.

a Give the moment-generating function for Y.

b Derive the moment-generating function of W =a Y. What is the distribution of W? Why?

c Derive the moment-generating function of X = -aY. What is the distribution of X ? Why?

d If b is a fixed constant, derive the moment-generating function of V = aY +b. What is

the distribution of V ? Why?

The moment-generating function for the gamma random variable is derived in Example 4.13. Differentiate this moment-generating function to find the mean and variance of the gamma

distribution.

Consider a random variable Y with density function given by

f( y ) = ke_,.z /2,

a Find k. b Find the moment-generating function of Y.

c Find E(Y) and V(Y).

A random variable Y has the density function

f( y ) = { ~-~ ,

a Find E(e3YI2) .

b Find the moment-generating function for Y.

c Find V(Y) .

-00 < y < 00 .

y < o. elsewhere.

4.1 0 T chebysheff' s Theorem As was the case for discrete random variables, an interpretation of f.-t and CJ for continuous random variables is provided by the empirical rule and Tchebysheff's theorem. Even if the exact distributions are unknown for random variables of interest, knowledge of the associated means and standard deviations permits us to deduce meaningful bounds for the probabilities of events that are often of interest.

We stated and utilized Tchebysheff's theorem in Section 3.11. We now restate this theorem and give a proof applicable to a continuous random variable.

THEOREM 4.13 Tchebysheff's Theorem Let Y be a random variable with finite mean 1-L and

variance CJ 2 . Then, for any k > 0, 1

P(IY- ~-LI < kCJ) 2: 1- k2 or

I P(IY- 1-LI 2: kCJ) :::: k2 .

Proof We will give the proof for a continuous random variable. The proof for th di screte case proceeds similarly. Let f(y) denote the density function of / Then ·

V(Y) = a 2 = 1: (y -~J-) 2 j(y) dy

111-ku 1 11+ku = -ao (Y -~J-) 2/(y)dy + 11-ku (Y-~J-) 2 /(y)dy

+lao (y - P-) 2 f( y ) dy . 11+ku

The second integral is always greater than or equal to zero, and (y -11-) 2 :::: k2a 2

for all values of y between the limits of integration for the first and third integrals; that is, the regions of integration are in the tails of the density function and cover only values of y for which (y - ~J-) 2 ::=:: k2a 2. Replace the second integral by zero and substitute k

2a

2 for (y - ~J-) 2 in the first and third integrals to obtain

the inequality

111-ku l ao V(Y) = a

2 ::=:: k

2a

2 f(y) dy + k 2a 2 f(y) dy.

-ao 11+~ Then

[111-ku l +ao J a

2 :::: k

2a

2 f (y ) dy + f( y ) dy ,

-ao 11+ku or

a2

:::: ea2[P(Y :::: 11-- ka) + P(Y :::: 11- + ka)] = k 2a 2 P(/Y- 11-/ :::: ka) .

Dividing by k2a 2, we obtain

or, equivalently,

1 P(/Y -~J-/ :::: ka):::: k

2'

1 P(/Y -~J-/ < ka) :::: 1- k

2.

One real value of Tchebysheff's theorem is that it enables us to find bounds for probabilities that ordinarily would have to be obtained by tedious mathematical manipulations (integration or summation). Further, we often can obtain means and van~ ances of random variables (see Example 4.15) without specifying the distribution of the variable. In situations like these, Tchebysheff's theorem still provides meaningful bounds for probabilities of interest.

EX AMPLE 4.17 Suppose that experience has shown that the length of time Y (in minutes) required

to conduct a periodic maintenance check on a. dictating machine follows a gamma distribution with a = 3.1 and f3 = 2. A new maintenance worker takes 22.5 minutes to

Solution

check the machine. Does this length of time to perform a maintenance check disagree with prior experience?

The mean and variance for the length of maintenance check times (based on prior experience) are (from Theorem 4.8)

11- = a{J = (3.1)(2) = 6.2 and a 2 = af32 = (3.1)(22) = 12.4.

It follows that a = ..JI2.4 = 3.52. Notice that y = 22.5 minutes exceeds the mean 11- = 6.2 minutes by 16.3 minutes, or k = 16.3/ 3.52 = 4.63 standard deviations. Then from Tchebysheff's theorem,

1 P(/Y- 6.2/ :::: 16.3) = P(/Y- 11-/ :::: 4.63a) :::: (

4_63

)2 = .0466.

This probability is based on the assumption that the distribution of maintenance times has not changed from prior experience. Then, observing that P(Y ::=:: 22.5) is small, we must conclude either that our new maintenance worker has generated by chance a lengthy maintenance time that occurs with low probability or that the new worker is somewhat slower than preceding ones. Considering the low probability for P(Y ::=:: 22.5) , we favor the latter view. •

The exact probability, P(Y ::=:: 22.5), for Example 4.17 would require evaluation of the integral

lao yl·le -yf2

P(Y > 22.5) = 3 I dy . - 22.s 2 · r(3.1)

Although we could utilize tables given by Pearson ( 1965) to evaluate this integral, we cannot evaluate it directly. We could, of course useR or S-Plus or one of the provided applets to numerically evaluate this probability. Unless we use statistical software, similar integrals are difficult to evaluate for the beta density and for many other density functions . Tchebysheff's theorem often provides quick bounds for probabilities while circumventing laborious integration, utilization of software, or searches for appropriate tables.

Exerc ises 4.146 A manufacturer of tires wants to advertise a mileage interval that excludes no more than 10%

of the mileage on tires he sells. All he knows is that, for a large number of tires tested, the mean mileage was 25 ,000 miles, and the standard deviation was 4000 miles. What interval would you suggest?

4.147

4.148

A machine used to fill cereal boxes dispenses, on the average, J-t ounces per box. The manufacturer wants the actual ounces dispensed Y to be within I ounce of J-t at least 75% of the time. What is the largest value of a , the standard deviation of Y , that can be tolerated if the manufacturer's objectives are to be met?

Find P(IY - ~-tl :::: 2a) for Exercise 4.16. Compare with the corresponding probabilistic statements given by Tchebysheff 's theorem and the empirical rule.

4.149

4.150

4.151

4.152

4.153

4.154

Find P(IY- JLI ~ 2a) for the uniform random variable. Compare with the correspond· probabilistic statements given by Tchebysheff's theorem and the empirical rule. lng

Find P(l Y - JLI ~ 2a) for the exponential random variable. Compare with the correspond· probabilistic statements given by Tchebysheff's theorem and the empirical rule. lng

Refer to Exercise 4.92. Would you expect C to exceed 2000 very often?

Refer to Exercise 4.1 09. Find an interval that will contain L for at least 89% of the weeks th the machine is in use. at

Refer to Exercise 4.129. Find an interval for which the probability that C will lie within it 1·s . at

least .75.

Suppose that Y is a x2 distributed random variable with v = 7 degrees of freedom.

a What are the mean and variance of Y?

b Is it likely that Y will take on a value of23 or more?

c Applet Exercise Use the applet Gamma Probabilities and Quantiles to find P(Y > 23).

4.11 Expectations of Discontinuous Functions and Mixed Probability Distributions (Optional)

EXAMPLE 4.18

Problems in probability and statistics sometimes involve functions that are partly continuous and partly discrete, in one of two ways. First, we may be interested in the properties, perhaps the expectation, of a random variable g(Y) that is a discontinuous function of a discrete or continuous random variable Y. Second, the random variable of interest itself may have a distribution function that is continuous over some intervals and such that some isolated points have positive probabilities.

We illustrate these ideas with the following examples.

A retailer for a petroleum product sells a random amount Y each day. Suppose that Y, measured in thousands of gallons, has the probability density function

f(y) = { (3/8)y2, 0,

0 :::: y :::: 2,

elsewhere.

The retailer's profit turns out to be $100 for each lOOO gallons sold ( 10¢ per gallon) if Y _:::: l and $40 extra per 1000 gallons (an extra 4¢ per gallon) if Y > 1. Find the retailer's expected profit for any given day.

Solution Let g(Y) denote the retailer's daily profit. Then

{ IOOY,

g(Y) = 140Y, O_::::Y_::::l ,

1 < y :::: 2.

EXAMPLE 4.19

Solution

We want to find expected profit; by Theorem 4.4, the expectation is

E[g(Y)] =I: g(y)f(y) dy

= fo1

lOOy [ (~) l ] dy+ ~2

140y [ (~) i ] dy

300 ]I 420 ]2

= (8)(4) l 0 + (8)(4) l I

= 300

(1) + 420 (15) = 206.25.

32 32

Thus, the retailer can expect a profit of $206.25 on the daily sale of this particular

product. •

Suppose that Y denotes the amount paid out per policy in one year by an insurance company that provides automobile insurance. For many policies, Y = 0 because the insured individuals are not involved in accidents. For insured individuals who do have accidents, the amount paid by the company might be modeled with one of the density functions that we have previously studied. A random variable Y that has some of its probability at discrete points (0 in this example) and the remainder spread over intervals is said to have a mixed distribution. Let F(y) denote a distribution function of a random variable Y that has a mixed distribution. For all practical purposes, any

mixed distribution function F(y) can be written uniquely as

F(y) = C l Fl (y) + C2F2(y),

where F1 (y) is a step distribution function, F2 (y) is a continuous distribution function,

c1 is the accumulated probability of all discrete points, and c2 = 1 - c 1 is the accu

mulated probability of all continuous portions. The following example gives an illustration of a mixed distribution.

Let Y denote the length of life (in hundreds of hours) of electronic components. These components frequently fail immediately upon insertion into a system. It has been observed that the probability of immediate failure is 1/ 4. If a component does not fail immediately, the distribution for its length of life has the exponential density

function

{

e-Y, y > 0,

f(y) = O, elsewhere.

Find the distribution function for Y and evaluate P(Y > 10).

There is only one discrete point, y = 0, and this point has probability l / 4. Hence, c1 = 1/4 and c

2 = 3/4. It follows that Y is a mixture of the distributions of two

.


FIGURE 4.18 Distribution function

F (y) for Example 4.19

DEFINITION 4.15

F(y)

l/4

0 y

random variables, X 1 and Xz, where X 1 has probability I at point 0 and X2

has the given exponential density. That is,

and

Now

and, hence,

{

0, Fl (y) = I,

y < 0,

y::: 0,

{ 0,

Fz(y) = J; e- xdx = l-e-Y, y < 0,

y ::: 0.

F(y) = (I/4)F1 (y) + (3/4)Fz(y) ,

P(Y > 10) = 1 - P(Y S 10) = I - F(IO)

= 1- [(1/4) + (3/4)( 1 - e- 10)]

= (3/4)(1- (I - e- 10 )] = (3/4)e- 10 .

A graph of F(y) is given in Figure 4.18. I

An easy method for finding expectations of random variables with mixed distn· butions is given in Definition 4.15.

Let Y have the mixed distribution function

F(y) = c1 F 1 (y) + czFz(y)

and suppose that X 1 is a discrete random variable with distribution function F1 (y) and that X 2 is a continuous random variable with distribution function Fz(y). Let g(Y) denote a function of Y. Then

E[g(Y)] = c 1 E[g(X 1)] + czE[g(Xz) ].

---EXAMPLE 4.20

Solution

Find the mean and variance of the random variable defined in Example 4.19.

With all definitions as in Example 4.19, it follows that

E(X 1) = 0 and E(X2) = 100

ye-y dy = 1.

Therefore,

f.k = E(Y) = (l / 4)E(X 1) + (3/4)E(X2) = 3/4.

Also,

E(X~) = 0 and E(X~) = 100

/e-y dy = 2.

Therefore,

E(Y 2) = (l/4)E(XT) + (3/4)E(X~) = (1/4)(0) + (3/4)(2) = 3/2.

Then

V(Y) = E(Y 2)- {.k

2 = (3/2)- (3/4) 2 = 15/16. •

Exercises

*4.155 A builder of houses needs to order some supplies that have a waiting time Y for delivery, with a continuous uniform distribution over the interval from I to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at $100 for any waiting time up to 2 days. After 2 days, however, the cost of the delay is $100 plus $20 per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is $100 + $20( 1.5) = $130. Find the expected value of the builder's cost due to waiting for supplies.

*4.156 The duration Y of long-distance telephone calls (in minutes) monitored by a station is a random variable with the properties that

P(Y = 3) = .2 and P(Y = 6) =.I.

Otherwise, Y has a continuous density function given by

f(y) = { (l / 4)y e-YI2 , y > 0,

0, elsewhere.

The discrete points at 3 and 6 are due to the fact that the length of the caJI is announced to the caller in three-minute intervals and the caJler must pay for three minutes even if he talks less than three minutes. Find the expected duration of a randomly selected long-distance call.

*4.15 7 The life length Y of a component used in a complex electronic system is known to have an exponential density with a mean of I 00 hours. The component is replaced at failure or at age 200 hours, whichever comes first.

a Find the distribution function for X , the length of time the component is in use.

b Find E(X).

exercises - department of statistics

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