Chapter 6 Normal
Distributions
What is Normal Distribution?
Characteristics of Normal
Distribution
 The curve is bell-shaped, with the highest point over the
mean 𝜇
 The curve is symmetrical about a vertical line through 𝜇
 The curve approaches the horizontal axis but never
touches or crosses it
 The inflection (transition) points between cupping
upward and downward occur above 𝜇 + 𝜎 𝑎𝑛𝑑 𝜇 − 𝜎
Let’s test your knowledge
A) Do these
distribution
s have the
same
mean?
B) Compare
the two
graphs,
what can
you tell me
about the
standard
deviation?
Answer
 A) Both graphs have the same mean
 B) Red curve has a smaller standard deviation because
the data is more compressed. Black curve has a bigger
standard deviation because the data is more spread out.
Empirical rule
 For a distribution that is symmetrical and bell-shaped
(normal distribution):
 Approximately 68% of the data values will lie within one
standard deviation on each side of the mean
 Approximately 95% of the data values will lie within two
standard deviations on each side of the mean
 Approximately 99.7% of the data values will lie within three
standard deviations on each side of the mean
Again look at the graph
Group Work
 Check Chebyshev’s theorem. What are the similarities
and differences between them?
Potential Answer
 Empirical rule gives a stronger statement than
Chebyshev’s theorem in it gives definite percentages,
not just lower limits.
Group Work: Creating the normal
distribution curve
 The yearly corn per acre on a particular farm is normally
distributed with mean = 60 and standard deviation of 7.
Create the normal distribution curve.
Answer
 49
56
63
60
67
74
81
Group Work: Empirical Rule
 The playing life of a Sunshine radio is normally
distributed with mean 600 hours and standard deviation
of 100 hours. What is the probability that a radio
selected at random will last from 600 to 700 hours?
 34.1%
Technology
 If you have TI-83/TI-84
 Press the Y= key. Then, under DISTR, select
1:normalpdf (𝑥, 𝜇, 𝜎) and fill in desired 𝜇 𝑎𝑛𝑑 𝜎 values.
Press the WINDOW key. Set Xmin to 𝜇 −
3𝜎 𝑎𝑛𝑑 𝑋𝑚𝑎𝑥 𝑡𝑜 𝜇 + 3𝜎. Finally, press the ZOOM key and
select option 0: ZoomFit
What are some applications of
normal distribution curve?
 It gives us how reliable are the data
 Used for control charts
Control Charts
 It is useful when we are examining data over a period of
equally spaced time interval or in some sequential order.
Example of Control Chart
How to Make a control chart of the
random variable x
 1) Find the mean and standard deviation of the x
distribution by
 A) Using past data from a period during which the process
was “in control” or
 B) using specified “target” values for 𝜇 𝑎𝑛𝑑 𝜎
 2) Create a graph in which the vertical axis represents x
values and the horizontal axis represents time.
 3) Draw a horizontal line at height
𝜇 𝑎𝑛𝑑 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙, 𝑑𝑎𝑠ℎ𝑒𝑑 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 − 𝑙𝑖𝑚𝑖𝑡 𝑙𝑖𝑛𝑒𝑠 𝑎𝑡 𝜇 ± 2𝜎 𝑎𝑛𝑑 𝜇 ±
3𝜎
 4) Plot the variable x on the graph in time sequence
order. Use line segments to connect the points in time
sequence order
Example:
Day 1
x
2
3
11 20 25
4
5
6
7
8
9
10 11 12 13 14 15
23
16
19
8
25
17
20
23
29
A) Find the mean and the standard deviation (population)
B) Find control limit 𝜇 ± 2𝜎
C) Find control limit 𝜇 ± 3𝜎
18
14
10
Answer
 Mean = 18.53
 Standard deviation = 5.82
𝜇 ± 2𝜎 = 6.89 𝑎𝑛𝑑 30.17
𝜇 ± 3𝜎 = 1.07 𝑎𝑛𝑑 35.99
Interpreting control chart
 Ms. Tamara of the Antlers Lodge examines the control
chart for housekeeping. During the staff meeting, she
makes recommendations about improving service or, if
all is going well, she gives her staff a well-deserved “pat
in the back”. Determine if the housekeeping process is
out of control
Answer
 Graph to the left: there are 9 consecutive days on one
side of the mean. It is a warning signal. It would mean
that the mean is slowly drifting upward from 19.3.
 Graph to the right: We have data value beyond 𝜇 + 3𝜎.
We have two of three data values beyond 𝜇 − 2𝜎. Both
are warning signals.
Homework Practice
 Pg 244 #1-7, 11,13
Standard Units and Areas Under
the Standard Normal Distribution
Note:
 Normal distributions vary from one another in two ways:
The mean 𝜇 may be located anywhere on the x axis, and
the bell shape may be more or less spread according to
the size of the standard deviation 𝜎
Scenario
 Suppose you have two students comparing their test
result in two different classes.
 Jack got a 76 while the class average was a 66
 Jill got an 82 while the class scored 72
 Both scored 10 points above the mean.
 How do we determine who did better in respect to their
peers in the class?
The Answer
 You have to use the z test or z score because you want
to know the position in term of the standard deviation
away from the mean.
What is z test or z score?
 Z value or z score gives the number of standard
deviations between the original measurement x and the
mean 𝜇 of the x distribution
𝑧 =
𝑥−𝜇
𝜎
Note:
 Unless otherwise stated, in the remainder of the book
we will take the word average to be either the sample
arithmetic mean or population mean.
Calculating z scores
 Suppose the company states that their average large
pizza is 10 inches in diameter and the standard
deviation is 0.7 inches. A customer ordered a pizza and
found out it is of 7 inches. Assume the pizza follows a
normal distribution. If the size of the pizza is below
three standard deviation, the company would be in
danger of getting customer complaints and have a bad
name in its company.
 How many standard deviation s from the mean is 7? Is
the company going to be in trouble?
Answer
𝑧 =
𝑥−𝜇
𝜎
=
7−10
.7
= −4.28
It is 4.28 below the mean, and
therefore the company is in
trouble.
Group Work
 (Fake Data) National weigh average for men at 5’8 is
139 lbs with a standard deviation of 15.7. If you are 5’8
and weigh 171 lbs. Find the z score. Should you be
concerned?
Answer
 𝑧=
𝑥−𝜇
𝜎
=
171−139
15.7
= 2.03
 Answer may vary depending how you think of it. If you
consider outlier is when z=2.5 or more then no, but if
you say that he is not with the 95% (within 2 standard
deviation) then he should. (based on empirical rule)
Question:
 How would you find the raw score x if you are given the
z score, mean and standard deviation?
Answer
𝑥 = 𝑧𝜎 + 𝜇
Group Work
 Based on your experience, it takes you on average of 15
minutes to walk to school with standard deviation of 2
minutes.
 A)On one particular day, it took you 12 minutes to get to
school. What is the z score? Is the z value positive or
negative? Why?
 B) What would be the commuting time corresponding to
a standard score of z=-2.5?
Answer
 A) z=-1.5. It is negative because it is shorter than the
expected time.
 B) 10 minutes
Standard normal distribution
 Standard normal distribution is a normal distribution
with mean 𝜇 = 0 and standard deviation 𝜎 = 1
Area under the curve. The left
tail rule
 Look at Appendix II Table 5 on A22
Group Activity
 What is the value for and sketch the graph:
 A) z=1.18
 B) z=-2.35
 C) z=0.56
 D) z=-3.34
Group Activity: challenge
 What’s the value for:
 A) z>1.25
 B) values in between z=1.21 and z=2.36
 C) values to the right of 0.95
 D) values in between z=-2.75 and z=1.38
Homework practice
 Pg 256 #1-48 eoe
Areas under any normal curve
How do you find the z score/z value?
Answer
 𝑧=
𝑥−𝜇
𝜎
Example:
 Let x have a normal distribution with 𝜇 = 7 𝑎𝑛𝑑 𝜎 = 3. Find
the probability that an x value selected at random is
between (5 and 8). 𝑃(5 ≤ 𝑥 ≤ 8)
Answer
 You have to convert the x into z score:
 𝑍=
5−7
3
= −3
2
 𝑍=
8−7
3
=3
1
2
 𝑃 𝑧 = − 3 = .2546
1
 𝑃 𝑧 = 3 = .6293
 .6293-.2546=.3747
Group Work
 A typical iphone have a battery life that is normally
distributed with a mean of 18 hours with a standard
deviation of 2 hours. What is the probability that an
iphone will have a battery life that is greater than 24
hours?
Answer
 𝑃 𝑥 > 24 = 𝑃 𝑧 > 3 =.0013
Tech notes finding area under the
curve in between z values
 For TI 83/TI 84
 Press DISTR key, select 2:normalcdf
(𝑙𝑜𝑤𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑧 𝑣𝑎𝑙𝑢𝑒, 𝑢𝑝𝑝𝑒𝑟 𝑏𝑜𝑢𝑛𝑑 𝑧 𝑣𝑎𝑙𝑢𝑒, 𝜇, 𝜎)
Find x given probability
 A company sells a hybrid battery that has a mean life of
8 years with standard deviation of 2 years. How long
can the company guarantee its warranty such that they
do not want to refund on more than 5% of the hybrid
battery?
Answer
 Since refund only occur when it falls below the mean. It
is on the left side of the normal distribution curve.
When is z such that it is .05?
 Z=-1.645
𝑥−𝜇
 Using the formula 𝑧 = 𝜎 = −1.645 =
4.71, 4.71 𝑦𝑒𝑎𝑟𝑠. 𝑜𝑟 55 𝑚𝑜𝑛𝑡ℎ𝑠
𝑥−8
2
,𝑥 =
Group Work
 A company sells a product that has a mean shell life of
60 month with the standard deviation of 5 months. How
long can the company guarantee its warranty such that
they do not want to refund on more than 7% of the
product?
Answer
 you have to use z=-1.48
 -1.48*5+60=52.6 months
Example: Find z
 Find the z value such that 90% of the area under the
standard normal curve lies between –z and z
Answer

1−.90
2
= .05, 𝑠𝑜 𝑧 = 1.645
 Z values between -1.645 and 1.645
Group work
 Find the z value such that 80% of the area under the
standard normal curve lies between –z and z
Group work
 Find the z value such that 75% of the area under the
standard normal curve lies between –z and z
Tech note for finding x value
 Example find an x value from a normal distribution with
mean 40 and standard deviation 5 such that 97% of the
area lies to left of x
 TI 83/TI 84: press DISTR key and select
3:invNorm(𝑎𝑟𝑒𝑎, 𝜇, 𝜎)
Check for normality
 How do we tell if data follow a normal distribution?
 1) Make a histogram, it should be roughly bell-shaped
 2) Outliers. Make a box-and-whiskers plot (5 number summary)
 Outliers are either greater than Q3+1.5(IQR) or below Q1-1.5(IQR)
 3) Skewness: Pearson’s index
 𝑃𝑒𝑎𝑟𝑠𝑜𝑛′ 𝑠 𝑖𝑛𝑑𝑒𝑥 =
3 𝑥−𝑚𝑒𝑑𝑖𝑎𝑛
𝑠
,
 𝑛𝑜 𝑠𝑘𝑒𝑤𝑛𝑒𝑠𝑠 𝑖𝑓 𝑖𝑛𝑑𝑒𝑥 𝑖𝑠 𝑖𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 − 1 𝑎𝑛𝑑 1
 Skewed distributions are not normal
4) Normal quantile plot (or normal probability plot)

for TI 83/TI 84, put all the data in the calculator under edit, then
turn on STATPLOT then highlight the sixth plot option
 It should look like a straight line
Check for normality
50
23
17
60
75
73
25
27
30
46
53
13
61
47
87
78
37
81
76
29
36
48
62
55
Homework Practice
 Pg 267 #1-30 eoe, 31
Normal Approximation to the
Binomial Distribution
Normal approximation to the
binomial distribution
 Consider a binomial distribution where
 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
 𝑟 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠
 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠 𝑜𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑟𝑖𝑎𝑙
 𝑞 = 1 − 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 𝑜𝑛 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑡𝑟𝑖𝑎𝑙
 If np>5 and nq>5, then r has a binomial distribution
that is approximated by a normal distribution with
 𝜇 = 𝑛𝑝 𝑎𝑛𝑑 𝜎 = 𝑛𝑝𝑞
 As n increases, the approximation becomes better
Example:
 Find the 𝑛, 𝑝, 𝑞, 𝜇, 𝜎 of the binomial distributions for which
p=0.25, q=0.75, and the number of trials is first n=3,
n=10, then n=25 and then n=50
Answer
 Find the 𝑛, 𝑝, 𝑞, 𝜇, 𝜎
Important note about binomial
approximation
 Remember how to graph the binomial distribution?
Remember how you have to incorporate the 0.5?
 For example when r=5, you draw the binomial
distribution from 4.5 to 5.5.
How to make the Continuity
Correction
 Convert the discrete random variable r (number of
successes) to the continuous normal random variable x
by doing the following:
 1) If r is a left point of an interval, subtract 0.5 to obtain
the corresponding normal variable x; that is, x=r-0.5
 2) If r is a right point of an interval, add 0.5 to obtain
the correspoinding normal variable x: that is, x=r+0.5
 For instance, 𝑃 6 ≤ 𝑟 ≤ 10 , where r is a binomial random
variable is approximated by 𝑃(5.5 ≤ 𝑥 ≤ 10.5), where x is
the corresponding normal random variable.
Normal Approximation
 The owner of a new apartment building must install 25
water heaters. From past experience in other apartment
buildings, she knows that Quick Hot is a good brand. A
Quick hot heater is guaranteed for 5 years only, but from
the owner’s past experience, she knows that the
probability it will last 10 years is 0.25.
 A) What is the probability that 8 or more of the 25 water
heaters will last at least 10 years?
 B) How does this result compare with the result we can
obtain by using the formula for the binomial probability
distribution with n=25 and p=0.25?
 C) How do the results of part a and b compare?
Answer
 A) since n=25 and p=0.25, r is the binomial random
variable corresponding to the number of successes out
of n=25. We want to find 𝑃(𝑟 ≥ 8).
 𝜇 = 𝑛𝑝 = 25 ∗ 0.25 = 6.25
 𝜎 = 𝑛𝑝𝑞 = 25 ∗ 0.25 ∗ 0.75 = 2.17
 𝑠𝑖𝑛𝑐𝑒 𝑟 ≥ 8, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑥 ≥ 7.5
 𝑧=
𝑥−𝜇
𝜎
=
7.5−6.25
2.17
≈ 0.58 so, 1-0.7190=0.2810
 B) Using TI83/TI84 you get 0.2735
 C) The error is only .0075, which is small.
Group Work:
 The Denver Post stated that 80% of all new products
introduced in grocery stores fail (are taken off the
market) within 2 years. If a grocery store chain
introduces 66 new products, what is the probability that
within 2 years
 A) 47 or more fail?
 B) 58 of fewer fail?
 C) 15 or more succeed?
 D) fewer than 10 succeed?
Answer
 𝑛𝑞 = 52.8>5, 𝑛𝑝 = 13.2>5, yes I can use normal approx to
binomial distribution
 1) 𝑛𝑞 = 52.8 𝜎 = 3.25, 𝑧 =
.0262=.9738
 2) 𝑛𝑞 = 52.8 𝜎 = 3.25, 𝑧 =
46.5−52.8
=-1.94
3.25
= .0262, so 1-
58.5−52.8
=1.75=
3.25
 3) 𝑛𝑝 = 13.2, 𝜎 = 3.25, 𝑧 =
.3446
14.5−13.2
3.25
 4) 𝑛𝑝 = 13.2, 𝜎 = 3.25, 𝑧 =
9.5−13.2
3.25
.9599
= 0.4 = .6554, 1 − .6554 =
= −1.14 = .1271
Group Work
 It is estimated that 3.5% of the general population will
live past their 90th birthday. In a graduating class of 753
high school seniors, what is the probability that
 A) 15 or more will live beyond their 90th birthday?
 B) 30 or more will live beyond their 90th birthday?
 C) Between 25 and 35 will live beyond their 90th
birthday?
 D) more than 40 will live beyond their 90th birthday?
Answer
 𝑛𝑞 = 726.645>5, 𝑛𝑝 = 26.355>5, yes I can use normal
approx to binomial distribution
 1) 𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 =
.0094=.9906
 2) 𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 =
.7324=.2676
 3)𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 =
35.5−26.355
5.04
14.5−26.355
=-2.35=.0094
5.04
29.5−26.355
=.62=.7324
5.04
so, 1-
so, 1-
24.5−26.355
=-.37=.3557,
5.04
,𝑧 =
= 1.81 = .9649, 𝑠𝑜 .9649 − .3557 = .6092
 4)𝑛𝑝 = 26.355 𝜎 = 5.04, 𝑧 =
.9975=.0025
40.5−26.355
=2.81=.9975=15.04
Use the examples and my
answers to find out the pattern of
the wording!
Homework Practice
 Pg 278 #1-14 eoo