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MAT 155 – Lab
Normal Distribution on TI 83/84+
The normal distribution is a continuous probability distribution determined by the value of μ and σ. For continuous
distributions, the height of the bell-shaped curve does not represent the probability. The area under the curve
represents the probability that an individual outcome, X, is in the interval. If the mean of the distribution is 0 and
the standard deviation is 1, then the distribution is called the Standard Normal distribution. We can convert normal
x
distributions to the standard normal distribution by using the z-score formula z 
.

Complete Parts I and IV using the Normal Distribution Lab Answer Key posted below the document link on
Moodle. Work through parts I, II and III as an example of how to use the calculator to complete these
exercises.
I. Prerequisite Information on the Standard Normal Distribution
You need to know how to find the area under a standard normal curve before you can calculate normal
probabilities in applications correctly on the calculator. The following are examples of how to do this.
The main function on the calculator that calculate probabilities from a normal distribution is
normalcdf(LB,UB,  ,  )
a) Finding area under the Standard Normal Curve to the left of the z-score z = 1.5
This means that we are finding the probability that a randomly selected “individual” has a z-score
less than 1.5. The statement would be P(z < 1.5). This is shown by the shaded blue region on the
figure below.
We find this area by asking the calculator to find the area from a very, very small number (a very,
very large negative number) all the way up to 1.5 – the region shaded in blue. On the calculator, we
use
-1E99
as this “very, very large negative number”. It really means  1 10 99 , which is -1 with 99 zeros next
to it. (I’d say that’s a pretty large negative number!)
So this “-1E99” is our lower bound (the left bound of the region), and 1.5 is our upper bound (the
right bound of the region). For the standard normal curve, we always have
 = 0 and
 =1
nd
Press 2 VARS on your calculator to get to the DISTR menu. Now scroll down and press ENTER
on 2:normalcdf(. You should see “normalcdf(“ on your home screen. Type the following:
(-) 1 2nd , (Comma Key to get EE) 99 , 1.5 , 0 , 1 )
What we type into the calculator should look like this:
normalcdf(-1E99,1.5,0,1)
Now, hit ENTER. Your answer should be 0.9332 (rounded to 4 decimal places).
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b) Finding area under the Standard Normal Curve to the right of the z-score z = 1.5
This means that we are finding the probability that a randomly selected “individual” has a z-score
greater than 1.5. The statement would be P(z > 1.5). This is shown by the shaded blue region on the
figure below.
We find this area by asking the calculator to find the area from 1.5 all the way up to a very, very
large positive number – the region shaded in blue. On the calculator, we use
1E99
as this “very, very large positive number”. It really means 1 10 99 , which is 1 with 99 zeros next to
it.
So 1.5 is our lower bound, and this “1E99” is our upper bound. Again, for the standard normal
curve, we always have
 = 0 and
 =1
nd
Press 2 VARS on your calculator to get to the DISTR menu. Now scroll down and press ENTER
on 2:normalcdf(. You should see “normalcdf(“ on your home screen. Type the following:
1.5 , 1 2nd , (Comma Key to get EE) 99 , 0 , 1 )
What we type into the calculator should look like this:
normalcdf(1.5,1E99,0,1)
Now, hit ENTER. Your answer should be 0.0668 (rounded to 4 decimal places).
c) Finding area under the Standard Normal Curve in between the z-scores -1.5 and 1.5.
This means that we are finding the probability that a randomly selected “individual” has a z-score
in between -1.5 and 1.5. The statement would be P(-1.5 < z < 1.5). This is shown by the shaded blue
region on the figure below.
We find this area by asking the calculator to find the area from -1.5 all the way up to 1.5.
Therefore, -1.5 is our lower bound, and 1.5 is our upper bound. Once again, for the standard normal
curve, we always have
 = 0 and
 =1
nd
Press 2 VARS on your calculator to get to the DISTR menu. Now scroll down and press ENTER
on 2:normalcdf(. You should see “normalcdf(“ on your home screen. Type the following:
-1.5 , 1.5 , 0 , 1 )
What we type into the calculator should look like this:
normalcdf(-1.5,1.5,0,1)
Now, hit ENTER. Your answer should be 0.8664 (rounded to 4 decimal places).
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Compute the following areas using your calculator.
Determine the area under the standard normal curve that lies:
1. to the left of z = -2.45
2. to the right of z = -3.01
3. between z = -2.04 and z = 2.04
4. to the right of z = 1.78
5. between z = -1.67 and z = 0
6. to the left of z = 1.35
7. between z = -1.04 and z = 2.76
8. to the right of z = -0.55
9. to the left of z = 0.92
10. to the right of z = 3.45
STOP HERE!! Write down your answers for questions 1 – 10 from above because you will enter them on
the Normal Lab Answer Key on Moodle. You may want to go ahead and enter them to see if your
answers are correct.
Applications of the Normal Distribution
In the following examples, you will be applying what you know from the steps above to any normal curve.
These normal curves will not have a mean of 0 and standard deviation of 1. However, you are still finding area
under the curve, and the calculator steps are very similar.
II. Computing the Area to the Left
Example: The number of chocolate chips in an 18-ounce bag of Chips Ahoy! chocolate chip cookies is
approximately normally distributed with a mean of 1,262 chips and standard deviation 118 chips according to a
study by cadets of the U.S. Air Force Academy. What is the probability that a randomly selected 18-ounce bag
of Chips Ahoy! contains fewer than 1,000 chocolate chips? (#19 p. 402 of textbook)
1. First you must know  ,  .
 = 1,262
 = 118
2. Now we are ready to use the calculator. For this example, it is important to note that the question is
asking for P(X < 1000).
 Press 2nd VARS on your calculator to get to the DISTR menu.
 Now scroll down and press ENTER on 2:normalcdf(
 In the parentheses, you will type the lower bound (LB), the upper bound (UB), the mean (  ),
and the standard deviation (  ).
 So for this example, you will have normalcdf(-1E99,1000,1262,118) and press ENTER.
3. Your answer rounded to 4 decimal places should be 0.0132. This means there is a 1.32% chance that a
randomly selected 18-ounce bag of Chips Ahoy! has less than 1,000 chocolate chips. Since this
probability is less than 0.05 (or 5%), we would consider this event to be unusual.
III. Computing the Area to the Right
Example: The mean incubation time of fertilized chicken eggs kept at 100.5oF in a still-air incubator is 21
days. Suppose that the incubation times are approximately normally distributed with a standard deviation of
1 day. What is the probability that a randomly selected fertilized chicken egg takes over 22 days to hatch?
 = 21
1. First you must know  ,  .
=1
2. Now we are ready to use the calculator. For this example, it is important to note that the question is
asking for P(X > 22).
 Press 2nd VARS on your calculator to get to the DISTR menu.
 Now scroll down and press ENTER on 2:normalcdf(
 In the parentheses, you will type the lower bound (LB), the upper bound (UB), the mean (  ),
and the standard deviation (  ). (Note that your calculator will calculate the range above, unlike
the binomial probability distribution calculations.)
 So for this example, you will have normalcdf(22,1E99,21,1) and press ENTER.
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3. Your answer rounded to 4 decimal places should be 0.1587. This means there is a 15.87% chance that a
randomly selected fertilized chicken egg will take over 22 days to hatch. Since this probability is greater
than 0.05 (or 5%), we would not consider this event to be unusual.
IV. Computing the Area in Between
Example: The lengths of human pregnancies are approximately normally distributed, with a mean  = 266
days and standard deviation  = 16 days. What proportion of pregnancies lasts between 240 and 280 days?
1. First you must know  ,  .  = 266
 = 16
2. Now we are ready to use the calculator. For this example, it is important to note that the question is
asking for P(240 < X < 280).
 Press 2nd VARS on your calculator to get to the DISTR menu.
 Now scroll down and press ENTER on 2:normalcdf(
 In the parentheses, you will type the lower bound (LB), the upper bound (UB), the mean (  ),
and the standard deviation (  ). (Note that your calculator will calculate the range in between,
unlike the binomial probability distribution calculations.)
 So for this example, you will have normalcdf(240,280,266,16) and press ENTER.
3. Your answer rounded to 4 decimal places should be 0.7571. This means that 75.71% of pregnancies last
between 240 and 280 days. Another way to interpret this number is that there is a 75.71% probability
that a randomly selected pregnancy will last between 240 and 280 days.
V. Follow-Up Questions (Round all answers to 4 decimal places!!!)
1. The reading speed of sixth-grade students is approximately normal, with a mean speed of 125 words per
minute and a standard deviation of 24 words per minute.
You must know the following from reading the problem:
 = 125
 = 24
a. Find the probability that a randomly selected sixth-grade student reads less than 100 words per
minute.
b. Find the probability that a randomly selected sixth-grade student reads more than 140 words per
minute.
c. Find the probability that a randomly selected sixth-grade student reads between 110 and 130 words
per minute.
d. Would it be unusual for a sixth-grader to read more than 200 words per minute? (This will be a T/F
question on the answer key.)
2.
Fast-food restaurants spend quite a bit of time studying the amount of time cars spend in their drivethroughs. Certainly, the faster the cars get through the drive-through, the more the opportunity for
making money. In 2007, QSR Magazine studied drive-through times for fast-food restaurants and
Wendy’s had the best time, with a mean time spent in the drive-through of 138.5 seconds. Assuming
drive-through times are normally distributed with a standard deviation of 29 seconds, answer the
following.
a. Find the probability that a randomly selected car will get through Wendy’s drive-through in less
than 100 seconds.
b. Find the probability that a randomly selected car will spend more than 160 seconds in Wendy’s
drive-through.
c. Find the proportion of cars that spend between 2 and 3 minutes in Wendy’s drive-through.
d. Would it be unusual for a car to spend more than 3 minutes in Wendy’s drive-through? (This will
be a T/F question on the answer key.)
3.
Finish the Normal Distribution Lab Answer Key on Moodle to receive a grade for this Lab.
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