Download Chapter 2

Chapter 2 Online Quiz
Chapter 2: Modeling Distributions of Data
Relative cum. freq.
1. Suppose that a particular set of observations has the cumulative relative frequency graph
(ogive) shown below. Recall that for each point (x, y) in the ogive, x = the left endpoint of a
class, and y = the percentage of observations in all classes below that class.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
18, 1
16, 0.8
12, 0.25
14, 0.325
10, 0.15
8, 0.1
6, 0
6
8
10
12
14
16
18
Value
The median of the distribution is located
A. in the class from 6 up to but not including 8.
AR. Incorrect. This class holds only the smallest 10% of the observations. Because the median
is the 50th percentile, it cannot be in this class.
*B. in the class from 14 up to but not including 16.
BR. Correct. In this class, the relative cumulative frequency increases from 0.325 to 0.8.
Therefore, the median must have been passed (50th percentile) in this class.
C. in the class from 16 up to but not including 18.
CR. Incorrect. This class holds only the largest 20% of the observations. Because the median is
the 50th percentile, it cannot be in this class.
2. For the density curve shown, which of the following statements is true?
A. The mean is larger than the median.
AR. Incorrect. Because the density curve is symmetric, the mean and median must coincide.
B. The proportion of outcomes between 0.2 and 0.5 is equal to 0.3.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
1
Chapter 2 Online Quiz
BR. Incorrect. The proportion that we seek is the area under the density curve between 0.2 and
0.5. The length of this interval is 0.3, but the height of the density curve is only 0.5, so the
desired proportion is (0.3)(0.5) = 0.15.
*C. The proportion of outcomes greater than 1.5 is equal to 0.25.
CR. Correct. The proportion that we seek is the area under the density curve between 1.5 and 2.
The length of this interval is 0.5, and the height of the density curve is 0.5, so the desired
proportion is (0.5)(0.5) = 0.25.
3. Suppose that the distribution of scores on a certain standardized exam is Normal, with a mean
of 500 and a standard deviation of 100. The third quartile of the score distribution
A. could be anywhere between 400 and 600.
AR. Incorrect. For a symmetric density curve like the Normal curve, the third quartile must be
at a specific point relative to the median and cannot vary as stated.
*B. is greater than 500.
BR. Correct. Because the third quartile is the 75th percentile and the median (50th percentile) is
located at the peak of the Normal curve, the third quartile must be to the right of the peak, that is,
greater than 500.
C. is less than 500.
CR. Incorrect. Because the third quartile is the 75th percentile and the median (50th percentile)
is located at the peak of the Normal curve, the third quartile must be to the right of the peak, that
is, greater than 500.
4. What effect will decreasing the value of μ and increasing the value of σ have on the
appearance of a Normal density curve?
*A. The peak will move to the left, and the curve will become shorter and broader.
AR. Correct. Decreasing μ shifts the peak to the left, whereas increasing σ broadens the “bell”
of the curve. To keep the total area under the curve the same (= 1), the curve must shrink in
height.
B. The peak will move to the right, and the curve will become shorter and broader.
BR. Incorrect. Decreasing μ shifts the peak to the left, not to the right.
C. The peak will move to the left, and the curve will become taller and narrower.
CR. Incorrect. Increasing σ causes the “bell” of the curve to broaden, rather than narrow.
5. The weights of cockroaches living in a university dormitory follow a Normal distribution with
mean 80 grams and standard deviation 5 grams. The percentage of cockroaches having weights
between 72 grams and 88 grams must be
A. less than 68%.
AR. Incorrect. The interval 72 to 88 contains all values within 1.6 standard deviations (= 8/5) of
the mean. Since 68% of all values lie within 1 standard deviation of the mean, the percentage of
all values lying within 1.6 standard deviations of the mean must be larger than 68%.
*B. between 68% and 95%.
BR. Correct. The interval 72 to 88 contains all values within 1.6 standard deviations (= 8/5) of
the mean. Since 68% of all values lie within 1 standard deviation of the mean and 95% of all
values lie within 2 standard deviations of the mean, the percentage of observations within 1.6
standard deviations of the mean must be between 68% and 95%.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
2
Chapter 2 Online Quiz
C. between 95% and 99.7%.
CR. Incorrect. The interval 72 to 88 contains all values within 1.6 standard deviations (= 8/5) of
the mean. Since 95% of all values lie within 2 standard deviation of the mean, the percentage of
observations within 1.6 standard deviations of the mean must be less than 95%.
6. Scores on the American College Testing (ACT) college entrance exam follow a Normal
distribution with mean 18 and standard deviation 6. Lisa’s standardized score on the ACT was z
= −0.7. What was her actual ACT score?
A. 4.2.
AR. Incorrect. Check the formula you used for the z-score.
*B. 13.8.
BR. Correct. Solving (x – μ)/σ = −0.7 for x using the given information yields x = (−0.7)(6) +
18 = 13.8.
C. 22.2.
CR. Incorrect. Check the value you used for z.
7. An office uses two brands of fluorescent light bulbs in its overhead light fixtures. From past
experience, it is known that Brand A bulbs have a mean life length of 3000 hours and a standard
deviation of 200 hours, while Brand B bulbs have a mean life length of 2700 hours and a
standard deviation of 250 hours. Which bulb has a longer life relative to its brand, a Brand A
bulb that lasts 3150 hours or a Brand B bulb that lasts 2850 hours?
*A. The Brand A bulb has a longer life relative to its brand.
AR. Correct. We compare the two measurements using z-scores. The z-score for the Brand A
bulb is z = (3150 – 3000)/200 = 0.75, while the z-score for the Brand B bulb is z = (2850 –
2700)/250 = 0.60. This means that the Brand A bulb lies 0.75 standard deviations above its
group’s mean, while the Brand B bulb lies 0.60 standard deviations above its group’s mean.
Since the z-score is larger for the Brand A bulb, the Brand A bulb has a longer relative life.
B. The Brand B bulb has a longer life relative to its brand.
BR. Incorrect. Check how you calculated the z-scores when comparing the two measurements.
C. The two bulbs have equally long lives.
CR. Incorrect. Check how you calculated the z-scores when comparing the two measurements.
8. The lifetime of a 2-volt nonrechargeable battery in constant use has a Normal distribution
with a mean of 516 hours and a standard deviation of 20 hours. The proportion of batteries with
lifetimes that exceed 520 hours is approximately
A. 0.2000.
AR. Incorrect. The z-score corresponding to 520 is (520 – 516)/20 = 0.20. The z-score is not
the same as the proportion we seek.
B. 0.5793.
BR. Incorrect. The z-score corresponding to 520 is (520 – 516)/20 = 0.20. The area to the left
of z = 0.20 is 0.5793, which corresponds to the proportion with lifetimes below 520 hours.
*C. 0.4207.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
3
Chapter 2 Online Quiz
CR. Correct. The z-score corresponding to 520 is (520 – 516)/20 = 0.20. The area to the right
of z = 0.20 is 1 – 0.5793 = 0.4207, which corresponds to the proportion with lifetimes greater
than 520.
9. The lifetime of a 2-volt nonrechargeable battery in constant use has a Normal distribution
with a mean of 516 hours and a standard deviation of 20 hours. Of all batteries, 90% have a
lifetime shorter than
*A. 541.6 hours.
AR. Correct. The z-score corresponding to the 90th percentile of the lifetime distribution is
1.28. If we add 1.28 standard deviations or (1.28)(20) = 25.6 hours to the mean of 516 hours, we
get 541.6 hours.
B. 517.28 hours.
BR. Incorrect. The z-score corresponding to the 90th percentile is 1.28. We have to multiply
this z-score by the standard deviation and add the product to the mean. Just adding the z-score
itself to the mean is not sufficient.
C. 490.4 hours.
CR. Incorrect. The z-score corresponding to the 90th percentile is 1.28. In this case, you have
found the z-score corresponding to the 10th percentile, that is, −1.28.
10. A company that manufactures and bottles apple juice has a machine that automatically fills
16-ounce bottles. There is some variation, however, in the exact amount of juice dispensed into
each bottle. From a large number of observations taken over a long period of time, it was found
that the actual amount of juice dispensed into each bottle was Normally distributed with a mean
of 16 ounces and a standard deviation of 1 ounce. Find the percentage of all bottles that are
either underfilled or overfilled by at least 0.25 ounce.
A. 19.74%.
AR. Incorrect. Recall that you are looking for the percentage of bottles with fill amounts that
differ by at least 0.25 ounce from the observed mean of 16 ounces. The percentage you found
corresponds to the bottles with fill amounts that differ by at most 0.25 ounce from the observed
mean (that is, the percentage of bottles with fill amounts between 15.75 ounces and 16.25
ounces).
*B. 80.26%.
BR. Correct. You are seeking the percentage of bottles that contain either less than 15.75
ounces or more than 16.25 ounces. The z-scores corresponding to the values 15.75 and 16.25 are
z = −0.25 and z = 0.25, respectively. The area under the standard Normal curve to the left of z =
−0.25 is 0.4013, while the area under the standard Normal curve to the right of z = 0.25 is 1 –
0.5987 = 0.4013. Adding these two areas together gives 0.4013 + 0.4013 = 0.8026.
C. 40.13%.
CR. Incorrect. You are seeking the percentage of bottles that contain either less than 15.75
ounces or more than 16.25 ounces. You have included only one of these possibilities in
computing your answer.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
4
Chapter 2 Online Quiz
11. Suppose that a distribution has a mean of 10 and a standard deviation of 2. According to
Chebyshev’s inequality, the percentage of observations from the distribution that lie between 5
and 15 is
A. at most 84%.
AR. Incorrect. Recall that Chebyshev’s inequality says that at least (100)(1 – 1/k2)% of all
observations must lie within k standard deviations of the mean. Therefore, you are looking for a
lower bound on the percentage (“at least”), rather than an upper bound (“at most”).
*B. at least 84%.
BR. Correct. A value that lies within 5 units of the mean, 10, lies within 5/2 = 2.5 standard
deviations of the mean. By Chebyshev’s inequality, at least (100)(1 – 1/k2)% of all observations
must lie within k standard deviations of the mean. Here, k = 2.5, so the required percentage is
(100)[1 – 1/(2.5)2]% = 100(0.84)% = 84%.
C. at least 75%.
CR. Incorrect. A value that lies within 5 units of the mean, 10, lies within 5/2 = 2.5 standard
deviations of the mean. You used k = 2 in Chebyshev’s inequality rather than k = 2.5.
12. You have a set of data that you suspect may have come from a Normal distribution. To
assess Normality, you construct a Normal probability plot. Which of the following would
constitute evidence that the data did actually arise from a Normal distribution?
*A. a strongly linear relationship between the data values and their standardized values
AR. Correct. If the data are Normal, then since the standardized values are obtained from the
original Normal values by using a linear function of the original values (look at the formula for
z), the plot should display a linear pattern.
B. a bell-shaped (Normal) curve relationship between the data values and their standardized
values
BR. Incorrect. A Normal curve does not indicate Normality when a Normal probability plot is
used.
C. a random scattering of points when the standardized values are plotted against the original
data values
CR. Incorrect. We are looking for a specific pattern in the plot that will signify Normality,
namely, a linear pattern. A random scattering may result from non-Normal data.
13. Suppose that a set of data has the Normal probability plot pictured.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
5
Chapter 2 Online Quiz
From the plot, the distribution of the original data set is most likely to have which of the
following shapes?
A. approximately Normal (bell-shaped)
AR. Incorrect. A bell-shaped distribution of data would have an approximately linear Normal
probability plot. In this case, the plot is clearly nonlinear. The largest observations fall
systematically above the line drawn through the main body of points.
B. skewed to the left
BR. Incorrect. Recall that in the Normal probability plot of a left-skewed distribution, the
smallest observations fall below a line drawn through the main body of points. In this case, the
smallest observations (in the lower left-hand corner of the plot) are actually above the line.
*C. skewed to the right
CR. Correct. Recall that in the Normal probability plot of a right-skewed distribution, the
largest observations fall above a line drawn through the main body of points—clearly the case
here.
© W.H. Freeman/BFW Publishers 2011
The Practice of Statistics for AP*, 4e
6