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Transcript
Chapter 2 Review
Answers
39.
The distribution is skewed to the right since most of the values are 25 minutes or
less, but the values stretch up to about 90 minutes. The data are centered roughly
around 20 minutes and the range of the distribution is close to 90 minutes. The
two largest values appear to be outliers
40.
a)
b)
75.
See next slide
6%. Since 6% (3/50) of the sample was left-handed, our best guess for the population is
that 6% is left-handed.
For both kinds of cars, the highway MPG is higher than the city MPG. The twoseater cars have a wider spread of MPG than the minicompact cars do, both in the
city and on the highway. The minicompact cars get slightly better gas mileage,
and this difference is more pronounced on the highway. All four distributions
appear to be symmetric.
76.
a)
a)
Cold
Neutral
Hot
Hatched
16
38
75
Not hatched
11
18
29
Total
27
56
104
59.26% in a cold nest. 67.86% in a neutral nest, and 72.12% in a hot nest. Hatching
percentage increases with temperature; the cold nest did not prevent hatching, but did
make it less likely.
40A (from previous slide)
Handed-ness
50
45
40
35
30
25
20
15
10
5
0
R
L
A
Chapter 2 AP Statistics Practice Test
1.
E
2.
D
3.
B
4.
B
5.
A
6.
D
7.
C
8.
E
9.
E
10.
C
11.
a)
Jane’s performance was better. She did more curl-ups than 85% of girls
her age. Matt did more curl-ups than less than 50% of boys his age.
b)
We can confidently say that Jane would have the higher Z-score, since
her percentile was so much higher
12.
a)
His Z-score would be 1. So the proportion below this is about .84,
meaning that he is at the 84th percentile
b)
The Z-score for 20 is -2.55 ((20-22.8)/1.1), and the area below Z=-2.55
is .0054 (using Table A). The Z-score for 26 is 2.91 ((26-22.8)/1.1), and
the area below Z=2.91 is .9982 (using Table A). Because we want the
area in the tails, we must do 1-.9982, to give us .0018 in the right tail.
Then we add the two tails together, to get .0072 (.0054+.0018), or .72%.
So less than 1% of soldiers requires custom helmets
c)
Q1 is 22.062 (Z=-.67) and Q3 is 23.537 (Z=.67). 23.537-22.062 gives us
an IQR of 1.474 inches.
13.
These data do not appear to be normally distributed. The median and the
mean are not close to each other here, and they would be if the distribution
were symmetric. Because the mean is much larger than the median, we can
conclude that the distribution is probably skewed to the right. This is further
supported by the fact that the median is much closer to the minimum value
than the maximum value.
Chapter Review Exercises
1.
a)
The only way to obtain a z-score of 0 is if the x-value equals the mean. Thus, the mean
is 170 cm. 1=(177.5-170)/Stdev, so the standard deviation equals 177.5-170, or 7.5 cm.
b)
2.5 =
𝑥−170
7.5
X=188.75. A height of 188.75 cm has a z-score of 2.5
2.
a)
Z=1.20. Paul is somewhat taller than average for his age. His height is 1.2 standard
deviations above the mean for his age.
b)
85% of boys Paul’s age are shorter than Paul
3.
a)
Approximately the 70th percentile
b)
Median is about 5, Q1 is about 2.5, and Q3 is about 11. There are outliers according to
the 1.5(IQR) rule, because values exceeding 23.75 clearly exist.
4.
5.
a)
The shape would not change. The new mean would be 13.32 meters, the median would
be 12.8 meters, the standard deviation would be 3.81 meters, and the IQR would be
3.81 meters.
b)
The mean error would be 43.7-42.6=1.1 feet. The standard deviation of the errors would
be the same as the standard deviation of the guesses, 12.5 feet, because we have just
shifted the distribution
Both lines will be to the right of the peak, with the mean (B) being further to the
right than the median (A)
6.
a)
(327,345)
b)
339 is one standard deviation above the mean, so 16% of the horse pregnancies last
longer than 339 days. This is because 68% are within one standard deviation of the
mean, so 32% are more than one standard deviation from the mean. Because we only
want those that are one standard deviation ABOVE the mean, it is 16%.
7.
a)
.0122 (proportion) or 1.22%
(percent)
b)
.9878
c)
.0384
d)
.9494
8.
a)
Z=.84
b)
Z=.39
9.
a)
Z=-2.29. Approximately 1% of
babies will be identified as having
a low birthweight
b)
Z= -.67 and Z=.67. For Z=-.67,
X=3325.63. For Z= .67, X=4010.37.
Therefore, the quartiles of the
birthweight distribution are
3325.6 and 4010.4
10.
If the distribution is normal, it
must be symmetric about its
mean—and in particular, the 10th
and 90th percentiles must be equal
distances above and below the
mean—so the mean is 250 points. If
225 points below (above) the mean
is the 10th (90th) percentile, this is
1.28 standard deviations below
(above) the mean, so the
distribution’s standard deviation is
225/1.28=175.8 points.