Math e-306 Assignment 7 Solutions Topic 12: Normal Distributions




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MATH E-306 - Assignment 7 Solutions

Topic 12: Normal Distributions

Activity 12-6, 12-7, 12-8, 12-12


Activities using Fathom: 12-17, 12-18, 12-19
Activity 12-6: 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 12-7: 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 12-8: 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 12-12: 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:

12-17, 12-18, 12-19


Activity 12-17: 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 z-scores 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 12-18: Critical Values
a. z* = 1.28

b. z* = 1.645

c. z* = 1.96

d. z* = 2.33

e. z* = 2.575

Activity 12-19: 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.



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