³
´
1
0.198
.34
2.331
z
± ± ±
±
, percentile for
–
.34 is .3669.
61 days x .3669 = 22.38, or about 22
days.
13.
z for 19
th
percentile (from Table A) is
–
0.88.
So
–
0.88(0.821) + 0.078 =
–
0.644 or
about a 0.64% decrease.
14.
Mean = 36.5(1.8) + 32 = 97.7 ºF.
Standard deviation = 0.3(1.8) =
0.54 ºF.
15.
z for 60
th
percentile is 0.25, and
0.69
0.25(0.16)
0.73
²
ounces.
16
.
1
0.69
1.94
0.16
z
±
, which has a proportion of 1
–
0.9738 = .0262 letters above it requiring
additional postage.
17.
The width of this region is 20 seconds, which is onethird the 60second
range of values.
Hence the proportion is 0.33.
See figure below.
30
30
0

x
x
10
10
92
The Practice of Statistics, 4/e Chapter 2
© 2011 BFW Publishers
Test 2B
Part 1
1.
d
The distribution is clearly symmetric, but the shape is nonNormal.
2.
c
Whenever a distribution is symmetric, then mean and median are equal.
3.
c
By the 689599.7 rule, 68% of the scores are within 1 standard deviation from the mean.
The area from 16.0 to 16.1 is exactly half of this area (just the half below the mean).
4.
b
100
–
68 = 34% of the area is more than 1 standard deviation from the mean, and half of
that is below 1 standard deviation, or below 16 oz.
5.
d
The rightmost points on this plot would have to be farther to the left
—
lower
—
in order
for this Normal probability plot to be straight.
Hence these values are closer to the mean
than they would be in a Normal distribution, and the distribution is skewed right.
6.
a
Statement I is true for all density curves, and statement II is true for all symmetric
distributions.
68% of the scores (not 50%) are in the interval
P
V
±
to
P
V
²
.
7.
a
Percentile of
z
= 1.25 is .8944, so proportion above 1.25 is 1
–
.8944 = .1056.
8.
c
Using the graph, cumulative relative frequency for 2.15 is approximately 0.55, so
1
–
0.55 = 0.45 of the time were longer than 2.15.
9.
d
The distribution
z

scores is a linear transformation, so it won’t change the s
hape of the
distribution.
Whenever a distribution is standardized with
z
scores, the mean becomes 0
and the standard deviation becomes 1.
10. b
10
10
9
0.40
2.5
z
±
,
which is the 65.54
th
percentile of the Standard Normal curve, so
about 34.5% of the times are above that value.
Part 2
11.
The
z
score for a fouryearold car with 40 thousand miles is
40
56.68
0.94
17.82
±
±
.
I The
z

score for a threeyearold car with 30 thousand miles is
30
33.33
0.26
12.70
±
±
.
This means that the
three year old car had been driven more miles, relative to other cars the same age.
(The
percentiles are about 17.5% for the four year old car with 40 thousand miles and 39.7% for the
three year old car with 30 thousand miles.)
12.
42
56.68
0.82
17.82
z
±
±
, which is the 20.61
percentile of the Standard Normal curve.
About 80% of 30 cars
—
or 24 cars
—
have been driven
more than 42 thousand miles.
13.
z
for 60
th
percentile = 0.25.
0.25(17.82) + 56.68 = 61.1
thousand miles.
14.
Mean = 0.01(156
–
2) = 1.54 meters; standard deviation = 0.01(5) = 0.05
meters.
15.
z for 33
rd
percentile is
–
0.44, and 99.2 + (
–
0.44
)(10.5) = 94.58 ≈ 95 points per
game.
16
.
101.7
99.2
0.24
10.5
z
±
, which has a proportion of 1
–
0.5948 = .4052 of the scores
above it.
17.
The width of this region is 2 minutes, which is 2/5 of the range of values.
Hence
the proportion is 0.40.
See figure below.
1
4
0
1
© 2011 BFW Publishers
The Practice of Statistics, 4/e Chapter 2
93
Test 2C
Part 1
1.
a
Annual income in any company is likely to be skewed right, with upper level
administrators on the right tail.
The other four variables are more likely to be Normal.