Statistics & Probability
Normal Distribution:
Finding Probabilities and
Application of Z-score
PROBABILITY AND NORMAL
DISTRIBUTIONS
If a random variable x is normally distributed,
you can find the probability that x will fall in a
given interval by calculating the area under the
normal curve for that interval.
μ = 500
σ = 100
P(x < 600) = Area
μ = 500 600
x
PROBABILITY AND NORMAL
DISTRIBUTIONS
Normal Distribution
μ = 500 σ = 100
P(x < 600)
Standard Normal Distribution
μ=0 σ=1
x   600  500
z

1

100
P(z < 1)
z
x
μ =500 600
μ=0 1
Same Area
P(x < 600) = P(z < 1)
PROBABILITY AND NORMAL
DISTRIBUTIONS
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally
distributed, find the probability that a student receives
a test score less than 90.
μ = 78
σ=8
z  x - μ = 90 -78
σ
8
= 1.5
P(x < 90)
μ =78
90
μ =0
?
1.5
P(x < 90) = P(z < 1.5) = 0.9332
x
z
The probability that a
student receives a test
score less than 90 is 0.9332.
PROBABILITY AND NORMAL
DISTRIBUTIONS
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally
distributed, find the probability that a student receives
a test score greater than 85.
μ = 78
σ=8
P(x > 85)
μ =78 85
μ =0 0.88
?
z = x - μ = 85 -78
σ
8
= 0.875  0.88
x
z
The probability that a
student receives a test
score greater than 85 is
0.1894.
P(x > 85) = P(z > 0.88) = 1  P(z < 0.88) = 1  0.8106 = 0.1894
PROBABILITY AND NORMAL
DISTRIBUTIONS
The average on a statistics test was 78 with a standard
deviation of 8. If the test scores are normally distributed,
find the probability that a student receives a test score
between 60 and 80.
z = x - μ = 60 - 78 = -2.25
P(60 < x < 80)
μ = 78
σ=8
60
σ
8
z 2  x - μ = 80 - 78
σ
8
1
μ =78 80
2.25
μ =0 0.25
?
?
x
z
= 0.25
The probability that a
student receives a test
score between 60 and 80 is
0.5865.
P(60 < x < 80) = P(2.25 < z < 0.25) = P(z < 0.25)  P(z < 2.25)
= 0.5987  0.0122 = 0.5865
FINDING PROBABILITIES FOR
NORMAL DISTRIBUTIONS
A survey indicates that people use their cellular
phones an average of 1.5 years before buying a
new one. The standard deviation is 0.25 year. A
cellular phone user is selected at random. Find the
probability that the user will use their current
phone for less than 1 year before buying a new
one. Assume that the variable x is normally
distributed
FINDING PROBABILITIES FOR
NORMAL DISTRIBUTIONS
Normal Distribution
μ = 1.5 σ = 0.25
z
P(x < 1)
Standard Normal Distribution
μ=0 σ=1
x

1  1.5

 2
0.25
P(z < –2)
0.0228
z
x
1
1.5
P(x < 1) = 0.0228
–2
0
FINDING PROBABILITIES FOR
NORMAL DISTRIBUTIONS
EXERCISE
1. In a survey of U.S. men, the heights in the 20 –29 age
group were normally distributed, with a mean of 69.4
inches and a standard deviation of 2.9 inches. Find the
probability that a randomly selected study participant
has a height that is
(a) less than 66 inches,
(b) between 66 and 72 inches
(c) more than 72 inches, and
(d) identify any unusual events. Explain
your reasoning.
FINDING PROBABILITIES FOR
NORMAL DISTRIBUTIONS
EXERCISE
2. A survey indicates that for each trip to the
supermarket, a shopper spends an average of 45
minutes with a standard deviation of 12 minutes in
the store. The length of time spent in the store is
normally distributed and is represented by the
variable x. A shopper enters the store. Find the
probability that the shopper will be in the store for
between 24 and 54 minutes.
FINDING PROBABILITIES FOR
NORMAL DISTRIBUTIONS
EXERCISE
3. Triglycerides are a type of fat in the bloodstream. The
mean triglyceride level in the United States is 134
milligrams per deciliter. Assume the triglyceride levels
of the population of the United States are normally
distributed with a standard deviation of 35 milligrams
per deciliter. You randomly select a person from the
United States. What is the probability that the
person’s triglyceride level is less than 80?
FINDING A Z-SCORE GIVEN AN AREA
Find the z-score that corresponds to a
cumulative area of 0.3632.
Solution:
0.3632
z
z 0
FINDING A Z-SCORE GIVEN AN
AREA
•Locate 0.3632 in the body of the Standard
Normal Table.
The z-score
is –0.35.
• The values at the beginning of the corresponding row
and at the top of the column give the z-score.
FINDING A Z-SCORE GIVEN AN
AREA
EXERCISE
Answer the following.
1. Find a z-score that corresponds to an area of
0.05.
2. Find the z-value if the area to the left of the Z
score is 0.60.
3. Find the z-score that has 10.75% of the
distribution’s area to its right.
FINDING A Z-SCORE GIVEN A
PERCENTILE
Find the z-score that corresponds to P75.
Area = 0.75
μ =0
?
0.67
z
The z-score that corresponds to P75 is the same z-score
that corresponds to an area of 0.75.
The z-score is 0.67.
TRANSFORMING A Z-SCORE TO AN
X-SCORE
To transform a standard z-score to a data value x
in a given population, use the formula
x = μ + zσ
TRANSFORMING A Z-SCORE TO AN
X-SCORE
A veterinarian records the weights of cats treated at a
clinic. The weights are normally distributed, with a
mean of 9 pounds and a standard deviation of 2
pounds. Find the weights x corresponding to z-scores of
1.96, –0.44 and 0.
Solution: Use the formula x = μ + zσ
•z = 1.96:
x = 9 + 1.96(2) = 12.92 pounds
•z = –0.44: x = 9 + (–0.44)(2) = 8.12 pounds
•z = 0:
x = 9 + (0)(2) = 9 pounds
Notice 12.92 pounds is above the mean, 8.12 pounds is
below the mean, and 9 pounds is equal to the mean.
FINDING VALUES
EXERCISE
A. Find the z-score that corresponds to each
percentile.
1. P10
2. P20
3. P99
B. Solve the problems completely.
1. A vending machine dispenses coffee into an eightounce cup. The amounts of coffee dispensed into
the cup are normally distributed, with a standard
deviation of 0.03 ounce. You can allow the cup to
overflow 1% of the time. What amount should you
set as the mean amount of coffee to be dispensed?
FINDING VALUES
EXERCISE
1. In a large section of a statistics class, the points for
the final exam are normally distributed, with a
mean of 72 and a standard deviation of 9. Grades
are assigned according to the following rule.
• The top 10% receive A’s.
• The next 20% receive B’s.
• The middle 40% receive C’s.
• The next 20% receive D’s.
• The bottom 10% receive F’s.
Find the lowest score on the final exam that would
qualify a student for an A, a B, a C, and a D.