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Normal Distribution Exercises
Turn in your answers on the Normal Distribution Exercises Answer Form. This is covered
in B&M Chapter 11. For use of the JMP calculator to compute the standard normal distribution
and standard normal quantiles, read Sall and Lehman (2001) Chapter 3. In JMP, see Help,
Contents, User's Guide, Using the Formula Editor. But use of the JMP calculator is not
required or necessary. All of this can be done using a standard normal CDF table.
0.0013
1. Standard Normal Distribution, Use a standard normal CDF table or the JMP calculator
PROBABILITY function NORMAL DISTRIBUTION to "compute" the probability that
a standard normal random variable is
a. less than −3,
b. less than −2,
c. between −3 and −2,
d. less than −1,
e. between −2 and −1,
f. less than 0,
g. between −1 and 0,
h. between 0 and 1,
i. between 1 and 2,
j. between 2 and 3,
k. greater than 3.
2. If you haven't already done so, draw a
standard normal curve and divide the
area under the curve into eight (8)
regions with vertical lines through the Z
axis at −3, −2, ..., 3. In each region write the probability of each interval: {Z < −3}(done
as an example), {−3 < Z < −2}, {−2 < Z < −1}, . . ., {2 < Z < 3}, {Z > 3}, which you've
already computed. This can be done in JMP calculator by using the PROBABILITY
function NORMAL DENSITY. But it's best to just draw the curve by hand.
3. Now use the picture you've just drawn to compute the probability that a standard normal
random variable is
a. between −1 and +1,
b. within two (2) standard deviations of 0,
c. within three (3) standard deviations of the mean,
d. not between −1 and +1,
e. not within two (2) standard deviations of 0,
f. not within
three (3)
standard
deviations
of the
mean.
4. Suppose that Z is
a standard normal
random variable.
a. Using a standard normal table, or the JMP PROBABILITY function NORMAL
QUANTILE to find the 1-st percentile, i.e., the number z0.01, that satisfies the
equation
P{Z < z 0.01} = 0.01
(Answer: z 0.01 = −2.326 or −2.33, done as follows, as shown in the graph: Look for
the probability 0.0100 in the standard normal CDF table. The closest to 0.0100 in
the standard normal CDF table is 0.0099. This is in the row for z = −2.3 and in the
column for 0.03, so the Z value corresponding to 0.0099 is z = −2.33, and that’s a
valid answer. We can also write z 0.01 = −2.33 to denote the 0.01 quantile, i.e., the 1.0
percentile. And we say, “−2.33 is the 0.01 quantile, i.e., the 1st percentile, of the
standard normal distribution.” The more precise value of −2.326 can be obtained
using the JMP calculator.)
b. Find the 5-th percentile, z0.05.
c. Find the 10-th percentile, z0.10.
d. Find the 90-th percentile, z0.90.
e. Find the 95-th percentile, z0.95.
f. Find the 99-th percentile, z0.99.
5. Assume that Z is standard normal.
a. Find a number z such that P{−z < Z < z} = 0.90.
Answer: z = 1.645 because P{−1.645 < Z < +1.645} = 0.90
b. For that number z compute: P{Z < −z} =
c. For that number z compute: P{Z < z} =
d. For that number z compute: P{Z > z} =
e. The number z is which percentile of the standard normal distribution?
6. Assume that Z is standard normal.
a. Find a number z such that P{−z < Z < z} = 0.95.
b. Compute: P{Z < −z} =
c. Compute: P{Z < z} =
d. Compute: P{Z > z} =
e. The
number z
is which
percentile
of the
standard
normal
distributio
n?
7. The Normal
Family. The
yearly growth of
dwarf-apple-tree
seedlings can be
measured by the increase in the length of the central leader. Suppose that the second-year
growth of such trees is normally distributed with a mean of 20 cm and a standard
deviation of 6 cm.
a. Compute the probability that the second-year growth of a randomly selected twoyear-old dwarf-apple-tree seedling is less than 15 cm.
b. Compute the percentage of such dwarf-apple-tree seedlings that would grow more
than 25 cm.
c. Compute the fraction of such dwarf-apple-tree seedlings that would be expected
to have a second-year growth of between 10 and 30 cm.
d. Find a number x such that the second-year growth of 90% of the seedlings is more
than x.
e. Find two numbers a and b such that the second-year growth of 90% of the
seedlings is between a and b and such that the second-year growth of 5% is less
than a and the second-year growth of 5% is more than b.
8. A veterinarian
found that the
average time it
takes residents to
perform a certain
procedure is 12
minutes. Assume
that the time it
takes residents to
perform the
procedure is
normally
distributed with a
mean of 12 minutes and a standard deviation of 2 minutes.
a. Compute the fraction of residents that you would expect to perform the procedure
within two minutes of the expected time of 12 minutes.
b. Compute the proportion of residents that you would expect to take less than 10
minutes or more than 14 minutes to perform the procedure.
c. Compute the percentage of residents that you would expect to take between 8 and
10 minutes to perform the procedure.
d. Compute the probability that a randomly selected resident would take between 9
and 11 minutes to perform the procedure.
e. Compute the proportion of residents that you would expect to take between 10
and 12 minutes to perform the procedure.
f. Compute the probability that a randomly selected resident would take more than
15 minutes to complete the procedure.
g. For purposes of planning and scheduling, find the time before which 99% of
residents would be expected to complete the procedure.
h. If 50 residents were randomly selected, how many would you expect to be able to
perform the procedure in 8 minutes or less? (Hint: The answer need not be an
integer)
Golde I. Holtzman, Department of Statistics, College of Arts and Sciences, Virginia Tech (VPI)
Last updated: September 6, 2011 © Golde I. Holtzman, all rights reserved.