1. Applications of the Normal Distribution
Chapter 7 Section 3
Applications of the
Normal Distribution

2. 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

3. 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

4. Chapter 7 – Section 3
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

5. Chapter 7 – Section 3
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

6. Chapter 7 – Section 3 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

7. Chapter 7 – Section 3 For example, if
μ = 3
σ = 2
then a value of x = 4 for X corresponds to
a value of z = 0.5 for Z
For example, if
μ = 3
σ = 2
For example, if
μ = 3
σ = 2
then a value of x = 4 for X corresponds to

8. Chapter 7 – Section 3 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
Because of this relationship
Values of X  Values of Z
then
P(X < x) = P(Z < z)

9. Chapter 7 – Section 3
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

10. Chapter 7 – Section 3 A different way to illustrate this relationshipX
a μ b Z
a – μ σ
b – μ σ

11. Chapter 7 – Section 3
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
Solve the problem for the standard normal Z
The answer will be the same for the general normal 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
Solve the problem for the standard normal Z
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
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

12. Chapter 7 – Section 3 Examples For a general random variable X with
μ = 3
σ = 2
calculate P(X < 6)
This corresponds to
so P(X < 6) = P(Z < 1.5) =
Examples
For a general random variable X with
μ = 3
σ = 2
calculate P(X < 6)

13. Chapter 7 – Section 3 Examples For a general random variable X with
μ = –2
σ = 4
calculate P(X > –3)
This corresponds to
so P(X > –3) = P(Z > –0.25) =
Examples
For a general random variable X with
μ = –2
σ = 4
calculate P(X > –3)

14. Chapter 7 – Section 3 Examples For a general random variable X with
μ = 6
σ = 4
calculate P(4 < X < 11)
This corresponds to
so P(4 < X < 11) = P(– 0.5 < Z < 1.25) =
Examples
For a general random variable X with
μ = 6
σ = 4
calculate P(4 < X < 11)

15. Chapter 7 – Section 3
Technology often has direct calculations for the general normal probability distribution
Excel
The function NORMDIST (instead of NORMSDIST) calculates general normal probabilities
StatCrunch
Entering the values for the mean and standard deviation into the window turns the standard into a general normal calculator
Technology often has direct calculations for the general normal probability distribution
Technology often has direct calculations for the general normal probability distribution
Excel
The function NORMDIST (instead of NORMSDIST) calculates general normal probabilities

16. 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

17. Chapter 7 – Section 3 The inverse of the relationship
is the relationship
With this, we can compute value problems for the general normal probability distribution

18. Chapter 7 – Section 3
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
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
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
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

19. Chapter 7 – Section 3 Examples For a general random variable X with
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
Examples
For a general random variable X with
μ = 3
σ = 2
find the value x such that P(X < x) = 0.3

20. Chapter 7 – Section 3 Examples For a general random variable X with
μ = –2
σ = 4
find the value x such that P(X > x) = 0.2
Since P(Z > ) = 0.2, we calculate
so P(X > 1.37) = P(Z > ) = 0.2
Examples
For a general random variable X with
μ = –2
σ = 4
find the value x such that P(X > x) = 0.2

21. Chapter 7 – Section 3 Examples We know that z.05 = 1.28, so
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
Examples
We know that z.05 = 1.28, so
P(–1.28 < Z < 1.28) = 0.90
Examples
We know that z.05 = 1.28, so
P(–1.28 < Z < 1.28) = 0.90
Thus for a general random variable X with
μ = 6
σ = 4

22. Chapter 7 – Section 3
Technology often has direct calculations for the general normal probability distribution
Excel
The function NORMINV (instead of NORMSINV) calculates general normal probabilities
StatCrunch
Entering the values for the mean and standard deviation into the window turns the standard into a general normal calculator
Technology often has direct calculations for the general normal probability distribution
Technology often has direct calculations for the general normal probability distribution
Excel
The function NORMINV (instead of NORMSINV) calculates general normal probabilities

23. 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

24. 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: The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source:
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)

25. 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: The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source:
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.)

26. Example: Chapter 7 – Section 3
The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is centimeters (cm) and the standard deviation is cm. The product specifications require that the diameter of each ball bearing be between and cm.
a. What proportion of ball bearings can be expected to have a diameter under cm? (0.9987)
b. What proportion of ball bearings can be expected to have a diameter over cm? (0.0013)
c. What proportion of ball bearings can be expected to have a diameter between and cm? That is, what proportion of ball bearings can be expected to meet the specifications? (0.9986)

27. Example: Chapter 7 – Section 3
The diameters of ball bearings produced at a factory are approximately normally distributed. Suppose the mean diameter is centimeters (cm) and the standard deviation is cm. The product specifications require that the diameter of each ball bearing be between and cm.
d. What is the probability that the diameter of a randomly selected ball bearing will be over cm? (0.6306)
e. What is the probability that the diameter of a randomly selected ball bearing will be under cm? (0.1217)
f. What is the percentile rank of a ball bearing that has a diameter of cm? (3rd percentile)
g. What is the percentile rank of a ball bearing that has a diameter of 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 cm)