continuous probability distribution.pdf

8/17/2019 Continuous Probability Distribution.pdf

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Probability Distribution


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Continuous Probability Distributions

The probability of the random variable assuming a value

within some given interval from x1 to x2 is defined to be the

area under the graph of the probability density function

between x1 and x2.

f (x)

x

Uniform

x

1 x

2

x

f

(

x

) Normal

x

1

x

2

x

1 x

2

Exponential

x

f (x)

x

1 x

2


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Normal Probability Distributions

The normal probability distribution is the most important

distribution for describing a continuous random variable.

It is widely used in statistical inference.


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Normal Probability Density Function

2 2( ) /21( )2

x f x e

= mean

= standard deviation

= 3.14159

e = 2.71828

where:


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It’s a probability function, so no matter what the values of

and , must integrate to 1!

12

12)(

2

1

dxe

x

E(X)= =

Var(X)=2 =

Standard Deviation(X)=

dxe x

x

2)(

2

1

2

1

2)(

2

1

2 )2

1(

2

dxe x

x


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The distribution is symmetric; its skewness

measure is zero.


Characteristics

x


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The entire family of normal probability distributions

is defined by its mean and its standard deviation .


Characteristics

Standard Deviation

Mean x


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The highest point on the normal curve is at the mean,

which is also the median and mode.


Characteristics

x


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Characteristics

-10 0 20

The mean can be any numerical value: negative, zero,

or positive.

x


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Characteristics

= 15

= 25

The standard deviation determines the width of the

curve: larger values result in wider, flatter curves.

x


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Probabilities for the normal random variable are given

by areas under the curve. The total area under the curve

is 1 (.5 to the left of the mean and 0.5 to the right).


Characteristics

.5 .5

x


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Since the area under the curve represents probability, the probability of a normal random variable at one specific value is

zero . With a single value, one can’t find the area since the

area must be bound by two values. Thus,

P(x = 10) = 0 P(x = 3) = 0 P(x = 7.5) = 0

However, one can find the following probabilities:

P( 1 < x < 3) P(2.2 < x < 3.7) P(x > 3)


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Characteristics

of values of a normal random variable

are within of its mean.

68.26%

+/- 1 standard deviation

of values of a normal random variableare within of its mean.95.44%

+/- 2 standard deviations

of values of a normal random variable

are within of its mean.

99.72%

+/- 3 standard deviations


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Characteristics

x

– 3 – 1

– 2

+ 1

+ 2

+ 3

68.26%

95.44%

99.72%


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There may be thousands of normal distribution curves,

each with a different mean and a different standarddeviation. Since the shapes are different, the areas under the curves between any two points are also different.

To make life easier, all normal distributions can be

converted to a standard normal distribution. A standardnormal distribution has a mean of 0 and a standarddeviation of 1.

No matter what and are, the area between - and +is about 68.26%; the area between -2 and +2 is about95.44%; and the area between -3 and +3 is about99.72%. Almost all values fall within 3 standarddeviations.


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How good is rule for real data?

Check some example data:

The mean of the weight of the women = 127.8

The standard deviation (SD) = 15.5


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80 90 100 110 120 130 140 150 160

0

5

10

15

20

25

P e r c e n t

POUNDS

127.8 143.3112.3

68% of 120 = .68x120 = ~ 82 runners

In fact, 79 runners fall within 1-SD (15.5 lbs) of the mean.


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80 90 100 110 120 130 140 150 160

0

5

10

15

20

25

P e r c e n t

POUNDS

127.896.8

95% of 120 = .95 x 120 = ~ 114 runners

In fact, 115 runners fall within 2-SD’s of the mean.

158.8


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80 90 100 110 120 130 140 150 160

0

5

10

15

20

25

P e r c e n t

POUNDS

127.881.3

99.7% of 120 = .997 x 120 = 119.6 runners

In fact, all 120 runners fall within 3-SD’s of the mean.

174.3


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Example

Suppose SAT scores roughly follows a normal distribution in

the U.S. population of college-bound students (with range

restricted to 200-800), and the average math SAT is 500 with a

standard deviation of 50, then:

68% of students will have scores between 450 and 550

95% will be between 400 and 600

99.7% will be between 350 and 650


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Standard Normal Probability Distributions

The formula for the standardized normal probabilitydensity function is

22)(

2

1)

1

0(

2

1

2

1

2)1(

1)(

Z Z

ee Z p


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The Standard Normal Distribution Z)

All normal distributions can be converted into the standard

normal curve by subtracting the mean and dividing by the

standard deviation:

X

Z

Somebody calculated all the integrals for the standard

normal and put them in a table! So we never have tointegrate!

Even better, computers now do all the integration.


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0

z

The letter z is used to designate the standardnormal random variable.

Standard Normal Probability Distributions


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Applications of Standard Normal Distribution

Problem:

What’s the probability of getting a math SAT score of 575 or less,

=500 and =50?

5.150

500575

Z

i.e., A score of 575 is 1.5 standard deviations above the mean

5.1

2

1575

200

)

50

500(

2

1 22

2

1

2)50(

1)575( dz edxe X P

Z x

Yikes!

But to look up Z= 1.5 in standard normal chart (or enterinto SAS) no problem! = .9332


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Applications of Standard Normal Distribution

Problem:

Test scores of a special examination administered to all potential

employees of a firm are normally distributed with a mean of 500

points and a standard deviation of 100 points. What is the probability

that a score selected at random will be higher than 700?

P(x > 700) = ?

If we convert this normal variable, x, to a standard normal variable, z,

z = (x - µ) / σ = (700 – 500) / 100 = 2

-------------500----------700 x-scale

P(x > 700) = P(z > 2) ----------------0-----------2 z-scale


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Problem

If birth weights in a population are normally distributed with a

mean of 109 oz and a standard deviation of 13 oz,

a. What is the chance of obtaining a birth weight of 141 oz

or heavier when sampling birth records at random?

b. What is the chance of obtaining a birth weight of 120 orlighter ?


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Solution

a. What is the chance of obtaining a birth weight of 141 oz

or heavier when sampling birth records at random?

46.213

109141

Z

From the chart or SAS Z of 2.46 corresponds to a right tail (greater

than) area of: P(Z≥2.46) = 1-(.9931)= .0069 or .69 %


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Solution

b. What is the chance of obtaining a birth weight of 120

or lighter ?

From the chart or SAS Z of .85 corresponds to a left tail area of:

P(Z≤.85) = .8023= 80.23%

85.13

109120

Z


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Looking up probabilities in the

standard normal table

Area is 93.45%

Z=1.51

Z=1.51

What is the area to

the left of Z=1.51 in

a standard normal

curve?


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Are my data normally distributed?

1. Look at the histogram! Does it appear bell shaped?

2. Compute descriptive summary measures — are mean,median, and mode similar?

3. Do 2/3 of observations lie within 1 std dev of the mean? Do

95% of observations lie within 2 std dev of the mean?4. Look at a normal probability plot — is it approximatelylinear?

5. Run tests of normality (such as Kolmogorov-Smirnov). But,

be cautious, highly influenced by sample size!


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Normal approximation to the binomial

When you have a binomial distribution where n is large and p

is middle-of-the road (not too small, not too big, closer to .5),then the binomial starts to look like a normal distribution in

fact, this doesn’t even take a particularly large n

Recall: What is the probability of being a smoker among a

group of cases with lung cancer is .6, what’s the

probability that in a group of 8 cases you have less than 2

smokers?


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Normal approximation to the binomial

When you have a binomial distribution where n is large and p

isn’t too small (rule of thumb: mean>5), then the binomial starts

to look like a normal distribution

Recall: smoking example…

1 4 52 3 6 7 80

.27 Starting to have a normal shape

even with fairly small n. You can

imagine that if n got larger, the

bars would get thinner and thinner

and this would look more and

more like a continuous function,with a bell curve shape. Here

np=4.8.


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Normal approximation to binomial

1 4 52 3 6 7 80

.27

What is the probability of fewer than 2 smokers?

Normal approximation probability:=4.8

=1.39

239.1

8.2

39.1

)8.4(2

Z

Exact binomial probability (from before) = .00065 + .008 = 0.00865

P(Z


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A little off, but in the right ballpark… we could also use the value

to the left of 1.5 (as we really wanted to know less than but not

including 2; called the “continuity correction”)…

37.239.1

3.3

39.1

)8.4(5.1

Z

P(Z≤-2.37) =.0089A fairly good approximation of

the exact probability, .00865.


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Practice problem

1. You are performing a cohort study. If the probability of

developing disease in the exposed group is .25 for the study

duration, then if you sample (randomly) 500 exposed

people, What’s the probability that at most 120 people

develop the disease?


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Solution:

P(Z


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More Sample Problems

a. P(z ≤ 1.5) =

b. P(z ≤ 1.0) =

c. P(1 ≤ z ≤ 1.5) =

d. P(0 < z < 2.5) =


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a. P(z ≤ 1.5) = 0.9332

b. P(z ≤ 1.0) = 0.8413

c. P(1 ≤ z ≤ 1.5) = 0.9332 – 0.8413 = 0.09

d. P(0 < z < 2.5) = 0.9938 – 0.5000 = 0.4938


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a. P(z ≤ - 1.0) =

b. P(z ≥ - 1) =

c. P(z ≥ - 1.5) =

d. P(- 3 < z ≤ 0) =


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a. P(z ≤ - 1.0) = 0.1587

b. P(z ≥ - 1) = 1 – P(z ≤ - 1) = 1 – 0.1587 = 0.8413

c. P(z ≥ - 1.5) = 1 – P(z ≤ - 1.5) = 1 – 0.0668 = 0.9332

d. P(- 3 < z ≤ 0) = 0.5 – 0.0014 = 0. 4986


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Given: µ = 77 σ = 20

a. P(x < 50) = ?

Convert to z: z = (x - µ) / σ = (50 – 77) / 20 = - 1.35

P(x < 50) = P(z < - 1.35) = 0.0885

b. P(x > 100) = ?

z = (100 – 77) / 20 = 1.15P(x > 100) = P(z > 1.15) = 1 – P(z ≤ 1.15) = 1 – 0.8749 =

0.1251 or 12.51 %


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Continuation of Sample Problem

c. x = ? to be considered a heavy user

Upper 20% of the area is in the right tail of the normal curve.

80% of the area is to the left. Go to Table 1 and locate 0.8 (or

80%) as the table entry. The closest entry is 0.7995. That

point represents a z-value of 0.84. Use this value of z in the

following equation:

z = (x - µ) / σ0.84 = (x – 77)/ 20 x = 93.8 hours


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A statistics instructor grades on a curve. He does not want to

give more than 15 percent A in his class. If test scores of

students in statistics are normally distributed with a mean of

75 and a standard deviation of 10, what should be the cut-off

point for an A?

z = (x - µ) / σ

1.04 = (x – 75) / 10

x = 85.4 or 85


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Sample Problems

The service life of a certain brand of automobile battery is

normally distributed with a mean of 1000 days and a standarddeviation of 100 days. The manufacturer of the battery wants

to offer a guarantee, but does not know the length of the

warranty. It does not want to replace more than 10 percent of

the batteries sold. What should be the length of the warranty?

z = ( x - µ ) / σ

- 1.28 = (x – 1000) / 100x = 872 days


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Reference:

Anderson Sweeney Williams

continuous probability distribution.pdf

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