Chapter 7
Section 3
Applications of the
Normal Distribution
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 1 of 23
Chapter 7 – Section 3
● Learning objectives
1

Find and interpret the area under a normal curve
2 Find the value of a normal random variable
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 2 of 23
Chapter 7 – Section 3
● Learning objectives
1

Find and interpret the area under a normal curve
2 Find the value of a normal random variable
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 3 of 23
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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 4 of 23
Chapter 7 – Section 3
● For a general normal random variable X with
mean μ and standard deviation σ, the variable
Z
X 

has a standard normal probability distribution
● We can use this relationship to perform
calculations for X
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 5 of 23
Chapter 7 – Section 3
● Values of X  Values of Z
● If x is a value for X, then
z
x

is a value for Z
● This is a very useful relationship
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 6 of 23
Chapter 7 – Section 3
z
x

● For example, if
 μ=3
 σ=2
then a value of x = 4 for X corresponds to
43
z
 0.5
2
a value of z = 0.5 for Z
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 7 of 23
Chapter 7 – Section 3
● Because of this relationship
Values of X  Values of Z
z
x

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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 8 of 23
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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 9 of 23
Chapter 7 – Section 3
● A different way to illustrate this relationship
X
a
μ
b
Z
a–μ
σ
b–μ
σ
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 10 of 23
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
x
z

 Solve the problem for the standard normal Z
 The answer will be the same for the general normal X
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 11 of 23
Chapter 7 – Section 3
● Examples
● For a general random variable X with
 μ=3
 σ=2
calculate P(X < 6)
● This corresponds to
z
63
 1.5
2
so P(X < 6) = P(Z < 1.5) = 0.9332
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 12 of 23
Chapter 7 – Section 3
● Examples
● For a general random variable X with
 μ = –2
 σ=4
calculate P(X > –3)
● This corresponds to
 3  ( 2 )
z
 0.25
4
so P(X > –3) = P(Z > –0.25) = 0.5987
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 13 of 23
Chapter 7 – Section 3
● Examples
● For a general random variable X with
 μ=6
 σ=4
calculate P(4 < X < 11)
● This corresponds to
46
z
 0.5
4
11  6
z
 1.25
4
so P(4 < X < 11) = P(– 0.5 < Z < 1.25) = 0.5858
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 14 of 23
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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 15 of 23
Chapter 7 – Section 3
● Learning objectives
1

Find and interpret the area under a normal curve
2 Find the value of a normal random variable
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 16 of 23
Chapter 7 – Section 3
● The inverse of the relationship
Z
X 

is the relationship
X    Z
● With this, we can compute value problems for
the general normal probability distribution
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 17 of 23
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
X    Z
 This is the answer for the original problem
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 18 of 23
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
x  3  0.5244  2  1.95
so P(X < 1.95) = P(Z < –0.5244) = 0.3
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 19 of 23
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 > 0.8416) = 0.2, we calculate
x  2  0.8416  4  1.37
so P(X > 1.37) = P(Z > 0.8416) = 0.2
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 20 of 23
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
x1  6  1.28  4  0.58
x2  6  1.28  4  12.58
so P(–0.58 < X < 12.58) = 0.90
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 21 of 23
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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 22 of 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
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 23 of 23
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)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 24 of 23
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.)

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.)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 25 of 23
Example: Chapter 7 – Section 3
● 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)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 26 of 23
Example: Chapter 7 – Section 3
● 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)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 3 – Slide 27 of 23