TOPIC 12
Normal Distributions
In Class Activities
Activity 12-1: Body Temperatures and Jury Selection
11-4, 12-1, 12-19, 15-3, 15-12, 15-18, 15-19, 19-3, 19-7, 20-11, 22-10, 23-3
a.
Both are roughly symmetric and mound-shaped.
b.
[normal.pdf]
c.
A
C
B
20
30
40
Mean, SD
Within One SD of Mean
Within Two SD of Mean
Within Three SD of Mean
50
60
70
Exam Scores
80
90
100
110
[labeledexamscores.pdf]
Body Temperature
Dataset
Number of Senior
Citizens Dataset
98.249, 0.733
90/130 (69.2%)
123/130 (94.6%)
129/130 (99.2%)
14.921, 3.336
703/1000 (70.3%)
957/1000 (95.7%)
998/1000 (99.8%)
Empirical Rule
68%
95%
99.7%
d.
Yes – these predictions are quite close to each other and to what the empirical rule would predict.
e.
z = (97.5 - 98.249) / .733 = -1.02
f.
z = (11.5 – 14.921) / 3.336 = -1.025
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
1
g.
They are almost identical.
h.
Both indicate that the observations are just over one standard deviation below their respective
means.
i.
area = .1515
j.
yes – this value is reasonably close to .146 and .153.
Activity 12-2: Birth Weights
[insert computer screen icon]
12-2, 14-9, 15-15
1500
a.
2000
2500
3000
3500
4000
Birth Weight (in grams)
4500
5000
5500
[shadedbirthweight.pdf]
b.
Student guess.
c.
z = (2500-3300) / 570 = -1.40
d.
P(Z < -1.40) = .0808
e.
Applet gives probability of .0802.
f.
P(X > 4536) = P(Z > 2.17) = .0151 (from applet)
g.
1) Look up the z-score and subtract the given area from 1.000 or
2) Look up the opposite z-score (-2.17) since the curve is symmetric.
h.
P(3000 < X < 4000) = P(-.53 < Z < 1.23) = .5910 (applet)
= .8907 - .2981 = .5926 (Table II)
i.
low birth weight: (331772 / 4112052) = .081 which is very close to the predicted .08 from our
calculations in part d.
between 3000 and 4000 kg: (2697819 / 4112052) = .656. This is not as close to our prediction in
part h – but it is not unreasonably far from it.
j.
The lightest 2.5% corresponds to a z-score of -1.96 (Table II or technology).
z = -1.96 = (x -3300) / 570. weight = x = 2182.8 kg
k.
The heaviest 10% (or bottom 90%) corresponds to a z-score of 1.28 (Table II or technology).
z = 1.28 = (x -3300) / 570. weight = x = 4029.6 kg
Activity 12-3: Blood Pressure and Pulse Rate Measurements
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
2
a.
The diastolic blood pressures are the most symmetric and mound-shaped of these three, so it is
most likely to come from a normal population.
b.
The systolic blood pressures is least likely to have come from a normal distribution because its
distribution is not symmetric or mound-shaped.
c.
Probability Plots
Normal - 95% CI
Systolic Blood Pressure
Percent
99.9
99
Diastolic Blood Pressure
99.9
99
90
90
50
50
10
10
1
0.1
1
0.1
50
100
150
200
30
60
90
120
PulseRate
99.9
99
90
50
10
1
0.1
30
60
90
120
[bloodpressureprobplot.pdf]
These probability plots do confirm that the systolic pressures were unlikely to have come from a
normal distribution and the diastolic pressures could quite possibly have come from a normal
distribution.
Activity 12-4: Criminal Footprints
[insert self-check icon]
a.
[Pick up art WS3_CSE_3_12_17 and change PMS 199 to PMS 165.]
b.
Because you are dealing with a male footprint, the z-score is (22 – 25)/4 or –0.75. Using Table II
(or technology), the probability of the footprint being less than 22 centimeters is .2266. So,
roughly 23% of men have a footprint smaller than 22 centimeters and would be misclassified as
female.
[Pick up art WS3_CSE_3_12_18 and change PMS 199 to PMS 165.]
c.
Now you are dealing with a female footprint, so the z-score is (22 – 19)/3 or 1.00. Using Table II
(or technology), the probability of the footprint being longer than 22 centimeters is 1 – .8413 or
.1587. This indicates that about 16% of females would be mistakenly identified as male.
[Pick up art WS3_CSE_3_12_19 and change PMS 199 to PMS 165.]
d.
Using Table II or technology, the z-score to produce a probability of .08 is –1.41. To find the
corresponding male foot length, you need to solve –1.41 = (x – 25)/4, which gives you x = 25 –
1.41(4) = 25 – 5.64 = 19.36 centimeters. You can also think of this as subtracting 1.41 standard
deviations of 4 from the mean of 25. Notice that this new cut-off value (19.36 centimeters) is
much smaller than before (22 centimeters) in order to reduce the probability of classifying a male
footprint as having come from a woman.
[Pick up art WS3_CSE_3_12_20 and change PMS 199 to PMS 165.]
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
3
e.
Using this new cut-off value of 19.36 centimeters, the z-score for a female footprint is (19.36 –
19)/3 or 0.12. The probability of a female footprint being longer than 19.36 centimeters is 1 –
.5478 or .4522 (using Table II or technology).
[Pick up art WS3_CSE_3_12_21 and change PMS 199 to PMS 165.]
f.
The probability of misclassifying a female footprint as a male footprint is much greater than
before (.4522 as opposed to .1587). In order to reduce the probability of one type of error
(misclassifying a man’s footprint as having come from a woman) from .2266 to .08, the
probability of making the other kind of error increases substantially. This exercise reveals that
there is a trade-off between the probabilities of making the two kinds of errors that can occur: you
can reduce one error probability but only by increasing the other error probability.
Homework Activities
Activity 12-5: Normal Curves
a.
mean ≈ 50, standard deviation ≈ 5
b.
mean ≈ 1100, standard deviation ≈ 300
c.
mean ≈ -20, standard deviation ≈ 40
d.
mean ≈ 225, standard deviation ≈ 75
Activity 12-6: Pregnancy Durations
9-14, 12-6
a.
P(X < 244) = P(Z < (244-270)/17) = P(Z < -1.53) = .0630 (Table II)
.0631 (applet)
b.
P(X > 275) = P(Z > .29) = .3859 (Table II)
.3843 (applet)
c.
P(X > 300) = P(Z > 1.76) = .0392 (Table II)
.0388 (applet)
d.
P(260 < X <280) = P(-.59 < Z < .59 ) = .7224 - .2776 = .4448 (Table II)
e.
508356/4112052 = .124; P(X < 252) = P(Z< -1.06) = .1446. The proportion predicted by the
model (.1446) is very close to the actual proportion of preterm deliveries (.124).
.4436 (applet)
Activity 12-7: Professors’ Grades
12-7, 12-8
0.06
Fisher
0.05
Density
0.04
0.03
0.02
Savage
0.01
0.00
a.
b.
20
40
60
80
Grades
100
120
140
[fisher.pdf]
Savage gives the higher proportion of A’s as more than 25% of his grades are A’s:
Fisher: P(X >90) = P(Z >2.29) = .0111
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
Savage: P(X >90) = P(Z >0.67) = .2525
4
Savage also gives a higher proportion of F’s as almost 16% of his grades are F’s:
c.
Fisher: P(X <60) = P(Z<-2.00) = .0228
d.
Savage: P(X <60) = P(Z <-1.00) = .1587
z = 1.28. (grade– 69) / 9 = 1.28; grade = 80.52, so you would need to score above 80.52 1in
order to earn an A.
Activity 12-8: Professors’ Grades
12-7, 12-8
Zeddes
Wells
40
a.
50
60
70
80
Final Exam Score
90
100
110
[zeddes.pdf]
A score of 90 is more impressive with Professor Zeddes because it is rarer in his class.
Zeddes: P(X >90) = P(Z >3.00) = .0013
b.
Wells: P(X >90) = P(Z >1.50) = .0668
A score of 60 is also more discouraging with Professor Zeddes because it is rarer in his class.
Zeddes: P(X <60) = P(Z <-3.00) = .0013
Wells: P(X <60) = P(Z <-1.50) = .0668
Activity 12-9: IQ Scores
12-9, 14-13, 15-20
a.
0.035
0.030
Density
0.025
0.020
0.015
0.010
0.005
0.000
70
80
90
100
110
120
IQ Scores
130
140
150
160
[iqscores.pdf]
b.
Student guess.
c.
P(X < 100) = P(Z < -1.25) = .1057
d.
P(X > 130) = P(Z > 1.25) = .1057
e.
P(110 < X < 130) = P(-.42 < Z < 1.25) = .8944 - .3372 = .5572
f.
P(X < 75) = P(Z < -3.33) = .0004
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
5
g.
The top 1% (or bottom 99%) corresponds to z = 2.33, so setting (IQ-115)/12 = 2.33 and solving
for IQ< we find IQ = 142.96, so need an IQ above 142.96 to be in the top 1%.
Activity 12-10: Candy Bar Weights
12-10, 14-10, 15-10
a.
P(X < 2.13) = P(Z < -1.75) = .0401
b.
P(X > 2.25) = P(Z > 1.25) = .1057
c.
P(2.2 < X < 2.3) = P(0 < Z < 2.5) = .9938 - .5 = .4938
d.
We want P(X < 2.13) = .001, so Z = (2.13-mean) / .04 = -3.085. Thus mean  = 2.2534 oz.
e.
We want P(X < 2.13) = .001, so Z = (2.13-2.2) / σ = -3.085. Thus standard deviation σ = .02269
oz.
f.
We want P(X < 2.13) = .001, so Z = (2.13-2.15) / σ = -3.085. Thus standard deviation σ =
.006483 oz.
Activity 12-11: SATs and ACTs
9-5, 12-11
a.
P(X > 1740) = P(Z > 1) = .1587
b.
P(X > 30) = P(Z > 1.5) = .0668
c.
Kathy seems to be stronger because a smaller proportion of her peers outperformed her.
Activity 12-12: Heights
a.
P(X < 66) = P(Z < -1.33) = .0918 (Table II)
.0912 (applet)
b.
P(X > 72) = P(Z > .67) = .2514 (Table II)
.2525 (applet)
c.
z = 1.28
d.
P(X < 66) = P(Z < .33) = .6293 (Table II)
.6306 (applet)
P(X > 72) = P(Z > 2.33) = .0099 (Table II)
.0098 (applet)
z = 1.28
1.28 = (height – 70)/3
1.28 = (height – 70)/3
height = 73.84 inches
height = 68.84 inches
e.
Yes – these are generally consistent with our calculations. The calculations are much closer for
men (9% vs 11.7%) than for the women (63% vs 74%).
f.
Yes – these are generally consistent with our calculations – even more so than the previous
calculations. For men 25% vs 29.9%, and for women .1% vs .5%.
Activity 12-13: Weights
a.
b.
P(X < 150) = P(Z < -0.71) = 23.89% (Table II)
23.75%(applet)
P(X < 200) = P(Z < 0.71) = 76.11% (Table II)
76.25%(applet)
P(X < 250) = P(Z < 2.14) = 98.38% (Table II)
98.39% (applet)
P(X < 150) = P(Z < 0.33) = 62.93% (Table II)
63.06%(applet)
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
6
c.
P(X < 200) = P(Z < 2.00) = 97.73% (Table II)
97.72%(applet)
P(X < 250) = P(Z < 3.67) = 99.99% (Table II)
99.99% (applet)
The normal model does a reasonable job of predicting these percentages. It tends to under
estimate the percentage of both men and women who weigh less than 150 and 200 lbs somewhat,
but it is very close with the percentages who weigh less than 250 lbs.
Activity 12-14: Dog Heights
a.
P(22.5<X<27.5) = P(-1<Z<1) = .8413 -.1587 = . .6826
b.
By the empirical rule, about 95% of all male Sheltie’s heights fall between 12 and 18 inches
(within 2 standard deviations).
c.
The top 10% (or bottom 90%) corresponds to z = 1.28 (table or technology), so then solving for
height, we find:
1.28 = (height – 15)/1.5
height = 16.92 inches
d.
Sheltie: P(X >18) = P(Z > 2.00) = .0228
Shepherd: P(X >28) = P(Z > 1.2) = .1151
So the Sheltie having a height of 18 inches is more unusual than the German Shepherd having a height
of 28 inches.
Activity 12-15: Baby Weights
a.
z = (13.9-12.5)/1.5 = .93. At 3 months, Benjamin Chance’s weight was about 9/10 of a standard
deviation above the average.
b.
P(X > 13.9) = P(Z > .93) = .1762 (Table II)
.1753 (applet)
We must assume that 3-month old American baby weights are normally distributed.
c.
By the empirical rule, to be in the middle 68% of weights, his weight would need to be within one
standard deviations of the mean, so within 17.25 ± 2 = 15.25 lbs and 19.25 lbs.
Activity 12-16: Coin Ages
12-16, 14-1, 14-2, 19-15
This distribution is not likely to be normally distributed – it is likely to be strongly skewed to the right
because no coin can have an age less than zero years, but there will be a few very old coins still in
circulation. If the mean is 12.3 years and the average deviation from the mean is 9 years – the distribution
cannot be symmetrically distributed about the mean and follow the empirical rule.
Activity 12-17: Empirical Rule
a.
P(-1 < Z < 1) = .8413 - .1587 = .6826
b.
P(-2 < Z < 2) = .9773 - .0228 = .9545
c.
P(-3 < Z < 3) = .9987 - .0014 = .9973
d.
To find the middle 50%, need 25% of each side. Looking up .25 and .75 in the table (or
technology), we find z = ±-.675.
IQR = .675 – (-.675) = 1.35
e.
z-score for outliers are -.675- 1.5×1.35 = -2.7 and
.675+1.5×1.35 = 2.7
Using Table II, P(-2.7 < Z < 2.7 ) = .9965 - .0035 = .9930. So the probability an observation from
the normal distribution will be classified an outlier using the 1.5 × IQR rule is 1- .993 or .007.
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
7
Activity 12-18: Critical Values
12-18, 16-2, 16-20, 19-13
a.
z* = 1.28 (probability below = .95)
b.
z* = 1.645
c.
z* = 1.96
d.
z* = 2.33
e.
z* = 2.575
Activity 12-19: Body Temperatures
12-1, 12-19, 15-3, 15-18, 15-19, 19-3, 19-7, 20-11, 22-10, 23-3
[insert computer screen icon]
a.
Normal Fit
25
Mean
StDev
N
98.25
0.7332
130
Frequency
20
15
10
5
0
96.75
97.50
98.25
99.00
Body Temperatures
99.75
100.50
[bodytempshistogram.pdf]
[bodytempsprobplot.pdf]
These data seem fairly normally distributed.
Probability Plots
Normal - 95% CI
Normal
97
Female Body Temperatures
12
Frequency
10
20
98
99
100
Male Body Temperatures
99.9
99
95
90
95
90
10
80
70
60
50
40
30
20
80
70
60
50
40
30
20
10
5
10
5
5
2
96.5 97.0 97.5 98.0 98.5 99.0 99.5
Female Body Temperatures
99
4
0
99.9
15
8
6
96.0
101
0
[bodytempshistogram.pdf]
Percent
b.
1
1
0.1
0.1
96
97
98
99
100
97.2
98.4
99.6
100.8
Male Body Temperatures
[bodytempsprobplot.pdf]
Based on these plots, the female body temperatures seem to more closely follow a normal model.
Activity 12-20: Natural Selection
10-1, 10-6, 10-7, 12-20, 22-21, 23-3
[insert computer screen icon]
a.
The normal model seems appropriate for these data – both the sparrows that survived and those
that died.
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
8
150
155
160
died
99
165
170
survived
95
90
Percent
80
70
60
50
40
30
20
10
5
1
150
b.
155
160
165
170
Total Length (mm)
The variables keel of sternum, tibiotarsus, humerus and femur seem to be well modeled by a
normal distribution for both sparrows that died and survived.
0.75
0.80
died
99
0.85
0.90
1.0
0.95
survived
95
95
90
90
80
80
70
60
50
40
70
60
50
40
30
30
20
20
10
10
5
5
1
0.75
0.80
0.85
0.90
1
0.95
Keel of Sternum (in)
0.65
died
0.70
0.75
0.65
95
90
90
0.70
0.75
0.80
survived
80
Percent
70
60
50
40
30
70
60
50
40
30
20
20
10
10
5
0.65
1.2
Tibiotarsus (in)
died
99
95
1
1.1
0.80
survived
80
Percent
1.0
1.2
survived
[bumpustibiotarsus.pdf]
[bumpushumerus.pdf]
99
1.1
died
99
Percent
Percent
[bumpusprobplot1.pdf]
5
0.70
0.75
0.80
Humerus (in)
[bumpuskeel.pdf]
1
0.65
0.70
0.75
0.80
Femur (in)
[bumpusfemur.pdf]
The variables length of beak and head, and alar extent are well modeled by a normal distribution for the
sparrows that died, but not well modeled for the sparrows that survived, while the variables weight and
skull width are in the reverse situation – they are well modeled by a normal distribution for the sparrows
that survived but not for the sparrows that died.
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
9
30
31
died
99
32
33
34
240
survived
95
260
survived
95
90
90
80
80
70
60
50
40
Percent
Percent
250
died
99
30
70
60
50
40
30
20
20
10
10
5
5
1
30
31
32
33
34
Length of Beak and Head (mm)
1
[bumpuslength.pdf]
250
260
Alar Extent (mm)
[bumpusalar.pdf]
0.550
0.575
died
99
240
0.600
0.625
22
0.650
survived
died
99
24
26
28
30
survived
95
95
90
90
80
Percent
Percent
80
70
60
50
40
70
60
50
40
30
20
30
10
20
5
10
5
1
1
0.550
0.575
0.600
22
24
26
28
30
Weight (g)
0.625
0.650
Skull Width (in)
[bumpusskull.pdf]
[bumpusweight.pdf]
Activity 12-21: Honda Prices
7-10, 10-19, 12-21, 28-14, 28-15, 29-10, 29-11
[insert computer screen icon]
The prices are approximately normally distributed, but the mileage and year variables are not
(skewed to the right and to the left respectively).
Probability Plot of Year, Price, Mileage
Year
Percent
99.9
99
90
90
50
50
10
10
1
0.1
1
0.1
1990
1995
2000
Price
99.9
99
2005
2010
0
10000
20000
30000
Mileage
99.9
99
90
50
10
1
0.1
-100000
0
100000
200000
300000
[hondaprobplots.pdf]
Activity 12-22: Your Choice
6-30, 12-22, 19-22, 21-28, 22-27, 26-23, 27-22
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
10
a.
Answers will vary, but some possibilities include: Dickinson placement exam scores, body
temperatures, birth weights, pregnancy durations, IQ scores, Professor’s grades, SAT scores, ACT
scores, candy bar weights, weights (of men and women), dog heights.
b.
Answers will vary but some possibilities include: lengths of reigns of British rulers, Olympic
rowers’ weights, college football scores (Activity 7-11), running times of Hitchcock films,
distances between exits on the Pennsylvania Turnpike, coin ages.
Rossman/Chance, Workshop Statistics, 3/e
Solutions, Unit 3, Topic 12
11