College Prep. Stats.
Chapter 3 Review ANSWER KEY
Name: ____________________________________
1.
a) x = 918.8 mm
b) x = 928 mm
c) mode: 936 mm
d) midrange: 907 mm
e) range: 58 mm
f) s = 23 mm
g) s2 = 529 mm2
h) Q1 = 901.5 mm
i) Q3 = 935 mm
j) CV = 2.5%
2.
Enter data in L1 and calculate 1 – Var Stats.
Enter data in L1 and calculate 1 – Var Stats.
Look for the data that value that occurs the most often.
878  936
2
936 – 878
Enter data in L1 and calculate 1 – Var Stats.
232
Enter data in L1 and calculate 1 – Var Stats.
Enter data in L1 and calculate 1 – Var Stats.
23
CV 
 100%
918.8
z = –1.77. This is a usual value
because it is between –2 and 2.
3.
936
878
928 935
901.5
z
878  918.8
23
Minimum: 878
Q1: 901.5
Median: 928
Q3: 935
Maximum: 936
The distribution is not normal.
It is skewed to the left.
Enter data in L1 and calculate 1 – Var Stats.
Then set STAT PLOT on boxplot.
Skewed left because of left tail
4.
The ACT score of 14.0 is a better
score.
1030  1518
14  21.1
ACT: z 
325
4.8
(The ACT z score is a higher number) It is more
standard deviations above the mean.
SAT: z 
5.
About 3.5 years
s
15  1
4
6.
Minimum usual height: 83.7cm
Maximum usual height: 111.3 cm
No, since 83.7 < 83.8 < 111.3, a
height of 87.8 cm is not considered
unusual. On this basis, the
physician should not be concerned.
7. a)
Word Counts
from Men
0 – 9,999
10,000 – 19,999
20,000 – 29,999
30,000 – 39,999
40,000 – 49,999
Frequency (f)
46
90
40
7
3
Total = 186
Minimum usual value = x  2s  97.5  2  6.9  83.7 cm
Maximum usual value = x  2s  97.5  2  6.9  111.3 cm
Class
midpoints (x)
4999.5
14999.5
24999.5
34999.5
44999.5
f*x
229977
1349955
999980
244996.5
134998.5
Total = 2959907
x
2959907
 15913.5
186
b)
s
Word Counts
from Men
0 – 9,999
10,000 – 19,999
20,000 – 29,999
30,000 – 39,999
40,000 – 49,999
Frequency (f)
46
90
40
7
3
Total = 186
Class
midpoints (x)
4999.5
14999.5
24999.5
34999.5
44999.5
f*x
x2
f* x2
229977
1349955
999980
244996.5
134998.5
Total = 2959907
24995000.25
224985000.2
624975000.3
1224965000
2024955000
1149770012
20248650020
24999000010
8574755002
6074865001
Total = 61047040050
186  61047040050  29599072
 8682
186  185
8.
P22 = 56
L
9.
30
97
58
72 81
125
22
40  8.8 , so take the 9th data value
100
Q1 = 58; Q3 = 81; IQR = 23; 1.5IQR = 34.5
Q3 + 1.5IQR = 115.5 (anything greater is an outlier)
Q1 – 1.5IQR = 23.5 (anything lower is an outlier)
125 is an outlier
Enter data in L1. Set
STAT PLOT on modified
boxplot.
10.
CV = 26.1%
CV 
58th percentile (P58)
241
100
422
There are 241 data values smaller than74.7. Therefore, 241 is your numerator)
65.4% of adults in town are
within 1.7 standard deviations
away from the mean
1 

1  2  100
 1.7 
x  3.45
38
11
18.43629016
70.525
11.
12.
13.
w
4
5
x
1 (D)
3 (B)
wx
4
15
1
4 (A)
4
5
3 (B)
15
11
38
14.
68%
Draw a bell shaped curve. Label the center point with
the mean (138). Label the bars to the left by subtracting
the standard deviation (12) . Label the bars to the right
by adding the standard deviation. The amounts 126 and
150 are one standard deviation away from the mean,
which is 68% using the Empirical Rule.
102
114
126
138
150
162
174
15.
z = –0.88
x = 72; x  79 ; s = 8 ; z = ?
Use the z-score formula
x  x 72  79
z

s
8
16.
x = 19.1
x = ?; x  14 ; s = 3 ; z = 1.7
Use the z-score formula to solve for x.
xx
z
s
x  14
1.7 
3