Math 117 - Normal Distributions (Chapter 6.2)
Problem-type 1 - Finding probabilities, areas, proportions, %
Step 1: Draw the graph shading the area desired. Label the mean and the specific
x-values being considered.
Step 2: Find the z-Score for each x-value involved.
Z-score = (score – mean) / standard deviation
Z-SCORE: z 
x

Step 3: Use table to find the cumulative left area bounded by z.
Step 4: Answer the problem.

Using the TI-83/84 to obtain probabilities, percentages, areas
Press
2nd, VARS
Select
2:normalcdf(
Type
left endpoint, right endpoint, , )
Problem-Type 2 - Finding Values from Known Areas (Probabilities)
Working BACKWARDS
Step 1: Draw the graph, shade and label the given area, and identify the
location of the x-value being sought.
Step 2: Find the cumulative left area bounded by x.
Step 3: Use table 5 to find the z-score.
(Go backwards! From the main body of the table to the z-score)
Step 4 Find the score, x, by using the formula
(score = mean + z-score * standard deviation)
x    z

Using the TI-83/84 to obtain normal scores, percentiles
Press
2nd, VARS
Select
3:invNorm(
Type
total area to the left of the desired value, , )
1
ERGONOMICS: How do we fit in the environment?
(1) - The United States Air Force ACES-II ejection seat used in fighter jets have been originally
designed for men whose weight is between 140 and 211 pounds. Nowadays many women are
joining the air force and we wonder if it is necessary to re-design the ejection sits.
We’ll take into consideration that weights of women are normally distributed with a mean of
143 and a standard deviation of 29 pounds, and that weights of men are normally distributed
with a mean of 170 and a standard deviation of 40 pounds.
(i) If a woman is randomly selected, what is the probability that her weight is between 140 and
211 pounds?
(ii) If a man is randomly selected, what is the probability that his weight is between 140 and
211 pounds?
(iii) Are women at greater risk of being injured? Do you think it is necessary to redesign the
ejection seats?
2) Designing Helmets
Engineers must consider the breadths of male heads when designing motorcycle helmets. Men
have head breaths that are normally distributed with a mean of 6.0 in. and a standard
deviation of 1.0 in. Due to financial constraints, the helmets will be designed to fit all men
except those with head breaths that are in the smallest 2.5% or largest 2.5 %. Find the minimum
and maximum head breaths that will fit men.
Answers: 4.04 in. and 7.96 in.
3) Designing cars
The sitting height of drivers must be considered in the design of a new car model.
Sitting heights of men are normally distributed with a mu of 36.0 in. and a sigma of 1.4 in.
A car that accommodates men with sitting heights up to 38.8 in. is being designed.
a) If a man is randomly selected, find the probability that he has a sitting height less than 38.8
in.
b) What percentage will be “left out” (will not seat comfortably in the car?)
Answer: (a) 0.9772; (b) 2.28%
4) Hip breadths and airplane seats
In designing seats to be installed in commercial aircraft, engineers want to make the seat wide
enough to fit 98% of all males.
Men hip breadths are normally distributed with a mean of 14.4 inches and a standard
deviation of 1.0 inches.
Find the score that separates the ones who will fit on the seat from the ones who will not fit.
Which percentile is this; the 98th o the 2nd?
Answer: 16.45 inches
5) Designing car dashboards
When designing the placement of a CD player in a new model car, engineers must consider the
forward grip reach of the driver. Design engineers decide that the CD should be placed so that
it is within the forward grip reach of 95% of women.
Women have forward grip reaches that are normally distributed with a mean of 27.0 in.
and a standard deviation of 1.3 in.
Find the score that separates the ones who will reach from the ones who will not reach. Which
percentile is this; the 95th or the 5th?
Answer: 24.86 inches
2
6) Why do doorways have a height of 6 ft 8 in.? (80 inches)
Heights of men are normally distributed with a mean of 69.0 in. and a standard deviation
of 2.8 in.
a) Find the percentage of men who can fit through the standard doorway without bending or
bumping their head.
b) When designing an environment, one common criterion is to use a design that
accommodates 95% of the population. How high should doorways be if 95% of men will fit
through without bending or bumping their head? What percentile is this score; the 95th or the
5th?
Answer: a) 99.99%; b) 73.6 in.
Other examples of normally distributed variables
7) Birth weights in the U.S. are normally distributed with a mean of 3420 grams and a
standard deviation of 495 grams.
a) The Newport General Hospital requires special treatment for babies that are less than 2450 g
(unusually light) or more than 4390 g. (unusually heavy). What is the percentage of babies who
do not require special treatment because they have birth weights between 2450 g. and 4390 g?
b) The Newport General Hospital wants to redefine the minimum and maximum birth weights
that require special treatment because they are unusually low or unusually high. After
considering relevant factors, a committee recommends special treatment for birth weights in the
lower 3% and the highest 1%. Find these boundary weights.
Answer: a) 95%; b) 2489.4 g. and 4573.35 g.
8) The Bean Stalk Club – Tall Club International
Heights of women are normally distributed with a mean of 63.6 in. and a standard
deviation of 2.5 in.
Heights of men are normally distributed with a mean of 69.0 in. and a standard deviation
of 2.8 in.
Tall Club International is a social organization for tall people. It has a requirement that men
must be at least 74 in. tall, and women must be at least 70 in. tall.
What percentage of men meets that requirement? What percentage of women? Are the height
requirements for men and women fair? Why or why not?
Answers: 3.67% of men and 0.52% of women. Not fair because a higher percentage of men are
allowed into the club.
3
Applications of the Empirical Rule (68% - 95% - 99.7% Rule)
This rule describes what percent of observations lie within one, two, and three standard
deviations of the mean.
Heights of women are normally distributed with a mean of 63.6 inches and a standard
deviation of 2.5 inches.
(1) Sketch the number line labeling the mean and the points that are one, two or three standard
deviations on either side of the mean.
(2) Approximately 68% of women have heights between _________ and ________
Approximately 95% of women have heights between _________ and ________
Approximately 99.7% of women have heights between _________ and ________
(3) What percent of women have heights between _______ and _______
What percent of women have heights lower than __________?
What percent of women are taller than ________ inches?
(4) What percent of women have heights that are more than ______ standard deviations away
from the mean?
(5) What is the area under the normal curve between ______ and ________
What is the area under the normal curve to the right of ________
........................To the left of __________
(6) If we select one woman at random, what is the probability that the woman selected is taller
than _________________
4