47
iii.
161.34
iv.
0.84 std. dev. below the mean
j.
Young
k.
The mean is most appropriate. From the boxplot the data appear to be relatively symmetric. When the
data are symmetric, it is appropriate to use the mean because it incorporates more information from
the data. (If the data were skewed, then it would be more appropriate to use the median; but these
data are not skewed.)
l.
IQR = 272.5 – 202.5 = 67; Q1 – 1.5*IQR = 205.5 – 1.5(67) = 105; Q3 + 1.5*IQR = 272.5 + 1.5(67) = 373. All
weights are between 105 and 373. There are no outliers.
m.
Mean – 2(standard deviation) = 240.08 – 2(44.38) = 151.32 ; Mean + 2(standard deviation) = 240.08 +
2(44.38) = 328.84 ; All players’ weights are between 2 standard deviations above and below the mean.
Solution to Exercise 2.9 (p. 33)
Kamala
Solution to Exercise 2.15 (p. 38)
a.
True
b.
True
c.
True
d.
False
Solution to Exercise 2.17 (p. 39)
b.
4,3,5
c.
4
d.
3
e.
f.
3,5
g.
3.94
h.
1.28
i.
3
j.
mode
Solution to Exercise 2.19 (p. 40)
c.
Maybe
Solution to Exercise 2.21 (p. 41)
a.
more children
b.
62.4%
Solution to Exercise 2.23 (p. 42)
b.
51,99
Solution to Exercise 2.24 (p. 42)
A
Solution to Exercise 2.25 (p. 43)
A
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CHAPTER 2. DESCRIPTIVE STATISTICS
Solution to Exercise 2.26 (p. 43)
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 Fall '08
 Ripol
 Statistics, Standard Deviation

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