M116 – NOTES – CH 6
Normal Distributions (Chapter 6)

Finding probabilities (areas)
Step 1: Draw the graph shading the area desired. Label the mean and the specific xvalues 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 5 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, , )

Finding Values from Know 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
x    z

(score = mean + z-score * standard deviation)
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, , )
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
Example 3) - 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?
a) Show all your work, along with the diagram and the shading of the area you are
calculating.
Population: women
Variable: weight in pounds (quantitative, continuous, ratio level of measurement)
For women: X~N(μ = 143 lb, σ = 29 lb)
P(140 < x < 211) = P(
140  143
211  143
)=
z
29
29
P( - 0.10 < z < 2.34) = 0.9904 – 0.4602 = 0.5302
b) Use a feature of the calculator to find the answer.
Normalcdf(140, 211, 143, 29) = .5317
About 53.1 % of women have weights between 140 and 211 pounds.
c) OPTIONAL (ITP) Now we’ll simulate the problem by generating 50 numbers that
come from a normal distribution with a mean of 143 and a standard deviation of 29
pounds.
(We’ll clear List 1, generate the numbers and store them into List 1, we’ll sort the
list and then explore the editor)
STAT 4:ClrList L1 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L1 :
STAT 3:SortA(L1)
Go to the editor, explore the list and count how many women have weights in the given
interval. Then determine the probability.
How does the experimental probability compare with the theoretical probability from
part (a)? Comment on the law of large numbers.
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M116 – TI 83/84 CALCULATOR – CH 6
(Example 3 - Continued-ii) If a man is randomly selected, what is the probability that his
weight is between 140 and 211 pounds?
a) Show all your work, along with the diagram and the shading of the area you are
calculating.
Population: men
Variable: weight in pounds (quantitative, continuous, ratio level of measurement)
For men: X~N(μ = 170 lb, σ = 40 lb)
P(140 < x < 211) = P(
140  170
211  170
)=
z
40
40
P( - 0.75 < z < 1.03) = 0.8485 – 0.2266 = 0.6219
b) Use a feature of the calculator to find the answer.
Normalcdf(140, 211, 170, 40) = .6207
About 62.0 % of men have weights between 140 and 211 pounds.
c) OPTIONAL (ITP) Now we’ll simulate the problem by generating 50 numbers that
come from a normal distribution with a mean of 170 and a standard deviation of 40
pounds.
(We’ll clear List 2, generate the numbers and store them into List 2, we’ll sort the
list and then explore the editor)
STAT 4:ClrList L2 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L2 :
STAT 3:SortA(L2)
Go to the editor, explore the list and count how many men have weights in the given
interval. Then determine the probability.
How does the experimental probability compare with the theoretical probability of part
(a)? Comment on the law of large numbers.
d) Are women at greater risk of being injured? Do you think it is necessary to redesign the
ejection seats?
Yes, about 47% of women have weights outside of the interval [140, 211].
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M116 – TI 83/84 CALCULATOR – CH 6
Example 4) 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.
a) Show all steps to get the answer.
Population: men
Variable: head-breaths in inches (quantitative, continuous, ratio level of
measurement)
For men: X~N(μ = 6.0 in., σ = 1.0 in.)
Area to the left
.025
.975
z-score
-1.96
1.96
X = μ + z* σ
6 + (-1.96)(1) = 4.04
6 + (1.96)(1) = 7.96
b) Now use a feature of the calculator to answer.
invNorm(.025, 6, 1) = 4.04
invNorm(.975, 6, 1) = 7.96
c) OPTIONAL (ITP) We are going to use simulation to find the probability that a man
selected at random has a head breath smaller than the MINIMUM found in part (a).
Generate 50 numbers that come from a normal distribution with a mean of 6 and a
standard deviation of 1.
STAT 4:ClrList L3 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L3 :
STAT 3:SortA(L3)
Explore the list to count how many men in the sample have head breaths smaller than
the minimum found in part (a). What percent of the sample is this? How does it
compare to 2.5%?
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
How do we fit in?
ERGONOMICS is the study of people fitting into their environment
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.
If a man is randomly selected, find the probability that he has a sitting height less than 38.8 in.
What percentage will be “left out” (will not seat comfortably in the car?)
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 _________th percentile of the distribution which is the score that separates the ones
who will fit on the seat from the ones who will not fit.
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 _____ th percentile of this distribution which is the score that separates the ones who
will reach from the ones who will not reach.
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M116 – NOTES – CH 6
Normal Approximation to Binomial Distributions (Section 6.4)
When can we use the Normal distribution to approximate a Binomial distribution?

Normal Distributions as Approximation to Binomial Distributions
If
np  5 and nq  5 , then the binomial random variable has a probability
distribution that can be approximated by a normal distribution with the mean and
standard deviation given as
  n* p
  n* p*q
10) For each of the following problems indicate whether the normal distribution is appropriate
to approximate the binomial distribution. If so, calculate the probability of exactly 5 successes
using the binomial distribution and then estimate the probability using the normal distribution.
a) Case I: n = 12, p = 0.5
(i) n*p = 12 * .5 = 6,
n*q = 12 * .5 = 6
Both np, and nq are ≥ 5, so the normal approximation is appropriate with
  n * p  12 *.5  6
  n * p * q  12 *.5*.5  3 ~ 1.7
(ii) Calculate probability using the binomial distribution
P(x = 5) = binompdf(12, .5, 5) = .1934
This is the same as the area of the rectangle centered at 5.
(iii) Estimate the probability using the normal distribution
This is the same as the area under the normal curve between 4.5 and 5.5
P(x = 5) = normalcdf(4.5, 5.5, 6, 3 ) = .1932 (Notice that this value is very close to the
probability found in part (ii))
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b) Case II: n = 12, p = 0.2
(i) In this case the normal approximation is not appropriate because n*p = 12 * 0.2 is
lower than 5.
(ii) Calculate probability using the binomial distribution
P(x = 5) = binompdf(12, .2, 5) = .0532
c) Case III: n = 12, p = 0.9
(i) In this case the normal approximation is not appropriate because n*q = 12 * 0.1 is
lower than 5.
(ii) Calculate probability using the binomial distribution
P(x = 5) = binompdf(12, .9, 5) = .00005
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M116 – NOTES – CH 6
Summary

Identifying Unusual Results with the Range Rule of Thumb
According to the range rule of thumb, most values should lie within 2 standard
deviations from the mean. We can therefore identify “unusual” values by determining if
they lie outside these limits:
Minimum usual value = μ - 2σ
Maximum usual value = μ + 2σ

Identifying Unusual Results with Probabilities
Unusually high: x successes among n trials is an unusually high number of successes
if
P(x or more) is very small (such as 0.05 or less).
Unusually low: x successes among n trials is an unusually low number of successes if
P(x or fewer) is very small (such as 0.05 or less).

Rare Event Rule
If, under a given assumption the probability of a particular observed event is extremely
small, we conclude that the assumption is probably not correct.

Normal Distributions as Approximation to Binomial Distributions
If
np  5 and nq  5 , then the binomial random variable has a probability
distribution that can be approximated by a normal distribution with the mean and
standard deviation given as
8
CHAPTER 6
8) Singular is a medication whose purpose is to control asthma attacks. In clinical trials of
Singular, 18.4% of the patients in the study experienced headaches as a side effect.
PART 1:
a) Compute the mean and the standard deviation of a random variable X, the number of
patients experiencing headaches in 400 trials of the probability experiment.
Population: asthma sufferers who use Singular
Success attribute: Experience headache as a side effect
n = 400, p = .184
  n * p  400 *.184  73.6
  n * p * q  400 *.184 *.816  60.0576 ~ 7.7
b) Use the range rule of thumb to determine the usual range.
[73.6 – 2*7.7, 73.6 + 2*7.7] = [58.2, 89]
c) Would it be unusual to observe 86 patients who experience headaches in a random
sample of 400 patients who use this medication?
It’s usual to observe anywhere from 59 to 89 people experiencing headaches in groups of
400, so 86 is usual
d) Would it be unusual to observe 93 patients who experience headaches in a random
sample of 400 patients who use this medication?
It’s usual to observe anywhere from 59 to 89 people experiencing headaches in groups of
400, so 93 is an unusually high result.
e) If in a random sample of 400 patients who use this medication you actually observe
93 patients who experience headaches; what might you conclude about the actual
percentage of patients who experience headaches?
Either something very unusual happened or it may be an indication that, the percentage
of Singular users who experience headaches as a side effect is actually higher than the
posted 18.4%
9
PART 2
We would like to answer some probability questions related to the problem from the previous
page in order to use the probability rule to decide if certain outcomes are usual or unusual.
a) Is it appropriate to use a normal distribution to approximate the binomial distribution?
Both, n*p and n*q must be greater than or equal to 5
Since n*p = 400*0.184 = 73.6
and
n*q = 400*.816 = 326.4
Then the normal distribution is appropriate to estimate this probability
b) If we select a random sample of 400 patients who use this medication,
(i) Use the normal approximation to estimate the probability that 86 or more experience
headaches. Would it be unusual to observe 86 patients who experience headaches in a
random sample of 400 patients who use this medication?
Normalcdf(85.5,10^9, 73.6,
60.0576 ) = 0.0623 (approximation)
This probability is more than 0.05 which implies that 86 is usual.
(ii) Use the normal approximation to estimate the probability that 93 or more experience
headaches. Would it be unusual to observe 93 patients who experience headaches in a
random sample of 400 patients who use this medication?
Normalcdf(92.5,10^9, 73.6,
60.0576 ) = 0.007 (approximation) < 0.5, then 93 is unusual
d) The probability that in groups of 400 patients, 93 or more experience headache as a
side effect is ___.007____ This means, in 1000 trials of this experiment we expect
about ___7_____ trials to result in 93 or more experiencing headaches. Because this
event only happens _7______ out of __1000___ times, we consider it to be
usual/unusual
c) If in a random sample of 400 patients who use this medication you actually observe 93
patients who experience headaches; what might you conclude about the actual
percentage of patients who experience headaches?
Either something very unusual happened or it may be an indication that, the percentage
of Singular users who experience headaches as a side effect is actually higher than the
posted 18.4%.
RARE EVENT RULE: If under a given assumption (18.4% of the patients who
take Singular experience headaches as a side effect), the probability of a particular
observed event (such as 93 patients experiencing headaches) is extremely small, we
conclude that the assumption is probably not correct.
10
11
CHAPTER 6
9) Depakote is a medication whose purpose is to reduce the pain associated with migraine
headaches. In clinical trials and extended studies of Depakote, 2% of the patients in the study
experienced weight gain as a side effect.
a) Compute the mean and standard deviation of the random variable X, the number of
patients experiencing weight gain in 600 trials of the probability experiment.
Population: migraine headaches sufferers who use Depakote
Success attribute: Experience weight gain as a side effect
n = 600, p = .02
  n * p  600 *.02  12
  n * p * q  600 *.02 *.98  11.76 ~ 3.4
b) Use the range rule of thumb to determine the usual range.
Notice that np and nq are both >5, so the normal approximation is appropriate, which means we
use the 2sigma rule
[12 – 2*3.4, 12 + 2*3.4] = [5.2, 18.8]
c) Would it be unusual to observe 16 or more patients who experience weight gain in a
random sample of 600 patients who take the medication?
It’s usual to observe anywhere from 6 to 18 people experiencing weight gain in groups of
600, so 16 is usual
d) Would it be unusual to observe 21 patients who experience weight gain in a random
sample of 600 patients who take the medication?
It’s usual to observe anywhere from 6 to 18 people experiencing weight gain in groups of
600, so 21 is an unusually high result
e) If in a random sample of 600 patients who use this medication you actually observe
21 patients who experience weight gain; what might you conclude about the actual
percentage of patients who experience weigh gain? (Read rare event rule on page 7)
Either something very unusual happened or it may be an indication that, the percentage
of Depakote users who experience weight gain as a side effect is actually higher than the
posted 2%.
12
PART 2:
We would like to answer some probability questions related to the problem from the previous
page in order to use the probability rule to decide if certain outcomes are usual or unusual.
a) Is it appropriate to use a normal distribution to approximate the binomial distribution?
Both, n*p and n*q must be greater than or equal to 5
Since n*p = 600*0.02 = 12
and
n*q = 600*.98 = 588
Then the normal distribution is appropriate to estimate this probability
b) If we select a random sample of 600 patients who use this medication,
(i) Use the normal approximation to estimate the probability that 16 or more experience
weight gain. Would it be unusual to observe 16 patients who experience weight gain in
a random sample of 600 patients who take the medication?
P(x≥16) = Normalcdf(15.5,10^9, 12, 11.76 ) = 0.1537 (approximation)
Since this probability is higher than 0.05, according to the probability rule, it’s common
to observe 16 people experiencing weight gain in groups of 600
(ii) Use the normal approximation to estimate the probability that 21 or more experience
weight gain. Would it be unusual to observe 21 patients who experience weight gain in
a random sample of 600 patients who take the medication?
P(x ≥21) = Normalcdf(20.5,10^9, 12, 11.76 ) = 0.007 (approximation)
Since this probability is lower than 0.05, according to the probability rule, it’s unusual to
observe 21 people experiencing weight gain in groups of 600.
21 would be a more “common” outcome in a population in which more than 2%
experience weight gain as a side effect. That is why we conclude...*****(See ***)
d) The probability that in groups of 600 patients, 21 or more experience weight gain as
a side effect is ___.007____ This means, in 1000 trials of this experiment we expect
about __7___ trials to result in 21 or more experiencing weight gain. Because this event
only happens___7____ out of __1000___ times, we consider it to be usual/unusual
e) If in a random sample of 600 patients who use this medication you actually observe
21 patients who experience weight gain; what might you conclude about the actual
percentage of patients who experience weigh gain? (Read rare event rule on page 7)
Either something very unusual happened or it may be an indication that, the
percentage of Depakote users who experience weight gain as a side effect is
actually higher than the posted 2%.
RARE EVENT RULE: If under a given assumption (2% of the patients who take
this medication experience weight gain as a side effect), the probability of a
particular observed event (such as 21 patients experiencing weight gain) is
extremely small, we conclude that the assumption is probably not correct.
13