MATH E306  Assignment 7 Solutions
Topic 12: Normal Distributions
Activity 126, 127, 128, 1212
Activities using Fathom: 1217, 1218, 1219
Activity 126: Pregnancy Durations
a. Pr(duration < 244) = Pr(Z < (244  270)/17) = Pr(Z < 1.53) = .0630 (Table II) or .0631 (applet)
b. Pr(duration > 275) = Pr(Z > .29) = .3859 (Table II) or .3843 (applet)
c. Pr(duration > 300) = Pr(Z > 1.76) = .0392 (Table II) or .0388 (applet)
d. Pr(260 < duration < 280) = Pr(.59 < Z < .59 ) = .7224  .2776 = .4448 (Table II) or .4436 (applet)
e. 508356/4112052 = .124; Pr(duration < 252) = Pr(Z < 1.06) = .1446.
The proportion predicted by the model (.1446) is very close to the actual proportion
of preterm deliveries (.124).
Activity 127: Professors’ Grades
a. The following sketch shows both teachers’ grade distributions:
b. Savage gives the higher proportion of As a more than 25% of his grades are As.
For Fisher, you calculate Pr(grade > 90) = Pr(Z > 2.29) = .0111.
For Savage, you calculate Pr(grade > 90) = Pr(Z > 0.67) = .2525.
c. Savage also gives a higher proportion of Fs as almost 16% of his grades are Fs.
For Fisher, you calculate Pr(grade < 60) = Pr(Z < 2.00) = .0228.
For Savage, you calculate Pr(grade < 60) = Pr(Z < 1.00) = .1587.
d. You have z = 1.28. So you want to solve (grade  69)/9 = 1.28; grade = 80.52.
Therefore, you would need to score above 80.52 in order to earn an A in Professor
DeGroot’s class.
Activity 128: Professors’ Grades
a. A score of 90 is more impressive with Professor Zeddes because it is rarer in
his class.
For Zeddes, you calculate Pr(grade > 90) = Pr(Z > 3.00) = .0013.
For Wells, you calculate Pr(grade > 90) = Pr(Z > 1.50) = .0668.
b. A score of 60 is also more discouraging with Professor Zeddes because it is rarer
in his class. For Zeddes, you calculate Pr(grade < 60) = Pr(Z < 3.00) = .0013. For Wells, you calculate Pr(grade < 60) = Pr(Z < 1.50) = .0668.
Activity 1212: Heights
a. Pr(height < 66) = Pr(Z < 1.33) = .0918 (using Table II) or .0912 (applet)
b. Pr(height > 72) = Pr(Z >.67) = .2514 (Table II) or .2525 (applet)
c. z = 1.28
1.28 = (height  70)/3 so height = 73.84 inches
d. Pr(height < 66) = Pr(Z <.33) = .6293 (Table II) or .6306 (applet)
Pr(height > 72) = Pr(Z > 2.33) = .0099 (Table II) or .0098 (applet)
z = 1.28
1.28 = (height  70)/3 so height = 68.84 inches
e. Yes, these are generally consistent with your calculations. The calculations are
much closer for men (9% vs. 11.7%) than for women (63% vs. 74%).
f. Yes, these are generally consistent with your calculations  even more so than the
previous calculations. For men, you have 25% vs. 29.9%, and for women, you have .1% vs. .5%.
Activities using Fathom:
1217, 1218, 1219
Activity 1217: Empirical Rule
a. Pr(–1 < Z < 1) = .8413  .1587 = .6826
b. Pr(–2 < Z < 2) = .9773  .0228 = .9545
c. Pr(–3 < Z < 3) = .9987  .0014 = .9973
d. To find the middle 50%, you need 25% of each side. Looking up .25 and .75
using Fathom, you find z = ± .675. You calculate IQR = .675  (.675) = 1.35.
e. The zscores for outliers are:
.675  1.5 X 1.35 = 2.7 and .675 + 1.5 X1.35 = 2.7.
Using Fathom, Pr(2.7 < Z < 2.7 ) = .9965  .0035 = .9930.
So, the probability that an observation from the normal distribution will be classified
as an outlier using the 1.5 X IQR rule is 1  .993 or .007.
Activity 1218: Critical Values
a. z* = 1.28
b. z* = 1.645
c. z* = 1.96
d. z* = 2.33
e. z* = 2.575
Activity 1219: Body Temperatures
a. The following histogram and normal probability plot display the data on body
temperatures:
These data seem fairly normally distributed.
b. The following histogram and normal probability plot display body temperature
data for men and women:
Based on these plots, the female body temperatures seem to more closely follow a
normal model.
