chapter 7 section 3

Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 1 of 23

Chapter 7Section 3

Applications of theNormal Distribution


Chapter 7 – Section 3

● Learning objectives Find and interpret the area under a normal curve Find the value of a normal random variable

1

2




1

2



● We want to calculate probabilities and values for general normal probability distributions

● We can relate these problems to calculations for the standard normal in the previous section



● For a general normal random variable X with mean μ and standard deviation σ, the variable

has a standard normal probability distribution

● We can use this relationship to perform calculations for X

X

Z



● Values of X Values of Z● If x is a value for X, then

is a value for Z● This is a very useful relationship

x

z



● For example, if μ = 3 σ = 2

x

z


then a value of x = 4 for X corresponds to


then a value of x = 4 for X corresponds to

a value of z = 0.5 for Z

502

34.z



● Because of this relationship

Values of X Values of Z

then

P(X < x) = P(Z < z)

x

z

● Because of this relationship

Values of X Values of Z

then

P(X < x) = P(Z < z)● To find P(X < x) for a general normal random

variable, we could calculate P(Z < z) for the standard normal random variable



● This relationship lets us compute all the different types of probabilities

● Probabilities for X are directly related to probabilities for Z using the (X – μ) / σ relationship



● A different way to illustrate this relationship

a μ b

X

Z

a – μσ

b – μσ



● With this relationship, the following method can be used to compute areas for a general normal random variable X

● With this relationship, the following method can be used to compute areas for a general normal random variable X Shade the desired area to be computed for X

● With this relationship, the following method can be used to compute areas for a general normal random variable X Shade the desired area to be computed for X Convert all values of X to Z-scores using

x

z


Solve the problem for the standard normal Z


Solve the problem for the standard normal Z The answer will be the same for the general normal X



● Examples● For a general random variable X with

μ = 3 σ = 2

calculate P(X < 6)


μ = 3 σ = 2

calculate P(X < 6)● This corresponds to

so P(X < 6) = P(Z < 1.5) = 0.9332

512

36.z




μ = –2 σ = 4

calculate P(X > –3)


μ = –2 σ = 4

calculate P(X > –3)● This corresponds to

so P(X > –3) = P(Z > –0.25) = 0.5987

2504

23.

)( z




μ = 6 σ = 4

calculate P(4 < X < 11)


μ = 6 σ = 4

calculate P(4 < X < 11)● This corresponds to

so P(4 < X < 11) = P(– 0.5 < Z < 1.25) = 0.5858

2514

61150

464

.. zz



● Technology often has direct calculations for the general normal probability distribution


● Excel The function NORMDIST (instead of NORMSDIST)

calculates general normal probabilities


● Excel The function NORMDIST (instead of NORMSDIST)


● StatCrunch Entering the values for the mean and standard

deviation into the window turns the standard into a general normal calculator




1

2



● The inverse of the relationship

is the relationship

● With this, we can compute value problems for the general normal probability distribution

X

Z

ZX



● The following method can be used to compute values for a general normal random variable X

● The following method can be used to compute values for a general normal random variable X Shade the desired area to be computed for X

● The following method can be used to compute values for a general normal random variable X Shade the desired area to be computed for X Find the Z-scores for the same probability problem

● The following method can be used to compute values for a general normal random variable X Shade the desired area to be computed for X Find the Z-scores for the same probability problem Convert all the Z-scores to X using

ZX

● The following method can be used to compute values for a general normal random variable X Shade the desired area to be computed for X Find the Z-scores for the same probability problem Convert all the Z-scores to X using

This is the answer for the original problem




μ = 3 σ = 2

find the value x such that P(X < x) = 0.3


μ = 3 σ = 2

find the value x such that P(X < x) = 0.3● Since P(Z < –0.5244) = 0.3, we calculate

so P(X < 1.95) = P(Z < –0.5244) = 0.3

9512524403 ..x




μ = –2 σ = 4

find the value x such that P(X > x) = 0.2


μ = –2 σ = 4

find the value x such that P(X > x) = 0.2● Since P(Z > 0.8416) = 0.2, we calculate

so P(X > 1.37) = P(Z > 0.8416) = 0.2

3714841602 ..x



● Examples

● We know that z.05 = 1.28, so

P(–1.28 < Z < 1.28) = 0.90

58124281658042816 21 .... xx

● Examples


P(–1.28 < Z < 1.28) = 0.90● Thus for a general random variable X with

μ = 6 σ = 4

● Examples


P(–1.28 < Z < 1.28) = 0.90● Thus for a general random variable X with

μ = 6 σ = 4

so P(–0.58 < X < 12.58) = 0.90





● Excel The function NORMINV (instead of NORMSINV)



● Excel The function NORMINV (instead of NORMSINV)


● StatCrunch Entering the values for the mean and standard

deviation into the window turns the standard into a general normal calculator


Summary: Chapter 7 – Section 3

● We can perform calculations for general normal probability distributions based on calculations for the standard normal probability distribution

● For tables, and for interpretation, converting values to Z-scores can be used

● For technology, often the parameters of the general normal probability distribution can be entered directly into a routine


Example: Chapter 7 – Section 3

● The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1066 and standard deviation 191. (Source: www.ets.org/Media/Tests/GRE/pdf/01210.pdf.) The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source: www.columbia.edu/cu/gsas/departments/psychology/department.html.)

a. What proportion of combined GRE scores can be expected to be under 1100? (0.5706)

b. What proportion of combined GRE scores can be expected to be over 1100? (0.4294)

c. What proportion of combined GRE scores can be expected to be between 950 and 1000? (0.0930)

d. What is the probability that a randomly selected student will score over 1200 points? (0.2415)

e. What is the probability that a randomly selected student will score less than 1066 points? (0.5000)



● The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1066 and standard deviation 191. (Source: www.ets.org/Media/Tests/GRE/pdf/01210.pdf.) The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source: www.columbia.edu/cu/gsas/departments/psychology/department.html.)

f. What is the percentile rank of a student who earns a combined GRE score of 1300? (89th percentile)

g. What is the percentile rank of a student who earns a combined GRE score of 1000? (36th percentile)

h. Determine the 70th percentile of combined GRE scores. (1166)

i. Determine the combined GRE scores that make up the middle 95% of all scores. (692 to 1440)

j. Compare the results in part (i) to the values given by the Empirical Rule. (684 to 1448; they are very close, since the Empirical Rule is based on the normal distribution.)



● The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is 1.002 centimeters (cm) and the standard deviation is 0.006 cm. The product specifications require that the diameter of each ball bearing be between 0.980 and 1.020 cm.

a. What proportion of ball bearings can be expected to have a diameter under 1.020 cm? (0.9987)

b. What proportion of ball bearings can be expected to have a diameter over 1.020 cm? (0.0013)

c. What proportion of ball bearings can be expected to have a diameter between 0.980 and 1.020 cm? That is, what proportion of ball bearings can be expected to meet the specifications? (0.9986)



● The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is 1.002 centimeters (cm) and the standard deviation is 0.006 cm. The product specifications require that the diameter of each ball bearing be between 0.980 and 1.020 cm.

● d. What is the probability that the diameter of a randomly selected ball bearing will be over 1.000 cm? (0.6306)

● e. What is the probability that the diameter of a randomly selected ball bearing will be under 0.995 cm? (0.1217)

● f. What is the percentile rank of a ball bearing that has a diameter of 0.991 cm? (3rd percentile)

● g. What is the percentile rank of a ball bearing that has a diameter of 1.011 cm? (93rd percentile)

● h. Determine the 10th percentile of the diameters of ball bearings. (0.994 cm)

● i. Determine the diameters of ball bearings that make up the middle 99% of all diameters. (0.987 to 1.017 cm)

chapter 7 section 3

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