Download Unit 2 Review Key File

AP Statistics
Unit 2 Review
Refer to the following setting for problems 1 and 2.
According to the National Center for Health Statistics, the distribution of heights for 15-year-old males is
symmetric, single-peaked, and bell-shaped. For this distribution, a z-score of 0 corresponds to a height of 170
centimeters (cm) and a z-score of 1 corresponds to a height of 177.5 cm.
1. Consider the height distribution for 15-year-old males.
A. Find its mean and standard deviation. Show your method clearly.
The mean is 170 and the standard deviation is 7.5
B. What height would correspond to a z-score of 2.5? Show your work.
x  170
2.5 
Therefore, x = 188.75 cm
7.5
2. Paul is 15 years old and 179 cm tall.
A. Find the z-score corresponding to Paul’s height. Explain what this value means.
179  170
z
 1.20 Paul is taller than average for his age. His height is 1.2 standard deviations above
7.5
the average male height for his age
B. Paul’s height puts him at the 85th percentile among 15-year-old males. Explain what this means to
someone who knows no statistics.
85% of boys Paul’s age are shorter than Paul’s height.
3. Mrs. Causey asked her students how much time they had spent using a
computer during the previous week. The figure shows a cumulative
relative frequency graph of her students’ responses.
A. At what percentile does a student who used her computer for 10
hours last week fall?
Reading up from 10 hours on the x-axis to the graphed line and
then across to the y-axis, we see that 10 hours corresponds to
about the 70th percentile.
B. Estimate the five number summary from the graph. Are there any outliers?
min. = 0, Q1 = 2.5, median = 5, Q3 = 11, max = 30
1.5(IQR) = 1.5(11 – 2.5) = 12.75
Q1 – 1.5(IQR) = -10.25
Q3 + 1.5(IQR) = 23.75, Since the max is greater than 23.75, there are outliers
4. A group of Australian students were asked to estimate the width of their classroom in feet. Use the
dotplot and summary statistics below to answer the following questions.
A. Suppose we converted each student’s guess from feet to meters (3.28 ft = 1 m). How would the shape
of the distribution be affected? Find the mean, median, standard deviation, and IQR for the
transformed data.
If we converted the guesses from feet to meters the shape of the distribution would not change. The
43.7
42
 13.32 meters, the median would be
 12.80 meters, the standard
new mean would be
3.28
3.28
12.5
12.5
 3.81 meters, and the IQR would be
 3.81 meters.
deviation would be
3.28
3.28
B. The actual width of the room was 42.6 feet. Suppose we calculated the error in each student’s guess
as follows: guess – 42.6. Find the mean and standard deviation of the errors. How good were the
students’ guesses? Justify your answer.
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,
but not changed its width by subtracting 42.6 from each guess.
5. The figure is a density curve. Sketch the curve onto your paper.
A. Mark the approximate location of the median. Justify your choice of
location.
B. Mark the approximate location of the mean. Justify your choice of
location.
6. Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from
conception to birth varies according to a roughly Normal distribution with mean 336 days and standard
deviation 3 days. Use the 68-95-99.7 rule to answer the following questions.
A. Almost all (99.7%) horse pregnancies fall in what range of lengths?
336 ± 3(3) = (327, 345)
B. What percent of horse pregnancies are longer than 339 days? Show your work.
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 and half of those are greater than 339.
7. Use the Standard Normal Table to find the proportion of observations from a standard Normal distribution
that falls in each of the following regions. In each case, sketch a standard Normal curve and shade the
area representing the region.
A. z < -2.25
0.0122
B. z > -2.25
1 – 0.0122 = 0.9878
C. z > 1.77
1 – 0.9616 = 0.0384
D. -2.25 < z < 1.77
0.9616 – 0.0122 = 0.9494
8. A. Find the number z at the 80th percentile of a standard Normal distribution.
0.84
B. Find the number z such that 35% of all observations from a standard Normal distribution are greater
than z.
0.39
9. Researchers in Norway analyzed data on the birth weights of 400,000 newborns over a 6-year period. The
distribution of birth weights is Normal with a mean of 3668 grams and a standard deviation of 511 grams.
Babies that weigh less than 2500 grams at birth are classified as “low birth weight.”
A. What percent of babies will be identified as low birth weight? Show your work.
2500  3668
z
 2.29 The z-value that corresponds to a baby weight less than 2500 grams at birth
511
is -2.29. The percent of babies weighing less than this is the area to the left. According to the table or
calculator, this is 0.0110. So approximately 1% of babies will be identified as having low birth weight.
B. Find the quartiles of the birth weight distribution. Show your work.
The z-scores corresponding to the quartiles are -0.67 and 0.67. To find the x-values corresponding to
the quartiles, we solve the following equations for x.
x  3668
0.67 
 x  0.67(511)  3668  3325.63 Therefore Q1 = 3325.63
511
x  3668
0.67 
 x  0.67(511)  3668  4010.37 Therefore Q3 = 4010.37
511
10. A fast-food restaurant has just installed a new automatic ketchup dispenser for use in preparing its
burgers. The amount of ketchup dispensed by the machine follows a Normal distribution with mean 1.05
ounces and standard deviation 0.08 ounce.
A. If the restaurant’s goal is to put between 1 and 1.2 ounces of ketchup on each burger, what percent of
the time will this happen? Show your work.
1  1.05
 0.625
z-score for 1 z 
0.08
1.2  1.05
 1.875
z-score for 1.2 z 
0.08
the area to the left of 1.875 is approximately 0.9699 and the area to the left of -0.625 is approximately
0.2643. The difference of these areas is the area (or percent) between 1 and 1.2.
0.9699 – 0.2643 = .7056
70.56% of the time there is between 1 and 1.2 ounces of ketchup on a burger.
B. Suppose that the manager adjusts the machine’s settings so that the mean amount of ketchup
dispensed is 1.1 ounces. How much does the machine’s standard deviation have to be reduced to
ensure that at least 99% of the restaurant’s burgers have between 1 and 1.2 ounces of ketchup on
them?
The middle 99% of the data occurs between z-score -2.575 and 2.575.
1  1.1
2.575 
s
2.575s  0.1
0.1
s
 0.388
2.575
The machine’s standard deviation should be approximately 0.388 to ensure that at least 99% of the
restaurant’s burgers have between 1 and 1.2 ounces of ketchup on them.
11. Many companies “grade on a bell curve” to compare the performance of their managers and professional
workers. This forces the use of some low performance ratings, so that not all workers are listed as “above
average.” Ford Motor Company’s “performance management process” for a time assigned 10% A grades,
80% B grades and 10% C grades to the company’s 18,000 managers. Suppose that Ford’s performance
scores really are Normally distributed. This year, managers with scores less than 25 received C’s, and
those with scores above 475 received A’s. What are the mean and standard deviation of the scores?
Show your work.
If the distribution is Normal, it must be symmetric about its mean and in this case 10% and 90% should be
475  25
 250 . The 10th
about the same distance from the mean. The mean is approximately 25 
2
percentile is about 250 – 25 = 225 points below the mean. The 10th percentile also has a z score of -1.28,
225
 175.8
so 1.28 standard deviations is about 225 points. The distribution’s standard deviation is
1.28
points.
12. Here are the lengths in millimeters of the thorax for 49 male fruit flies.
0.64, 0.64, 0.64, 0.68, 0.68, 0.68, 0.72, 0.72, 0.72, 0.72, 0.74, 0.76, 0.76, 0.76, 0.76, 0.76, 0.76, 0.76, 0.76,
0.78, 0.80, 0.80, 0.80, 0.80, 0.80, 0.82, 0.82, 0.84, 0.84, 0.84, 0.84, 0.84, 0.84, 0.84, 0.84, 0.84, 0.84, 0.88,
0.88, 0.88, 0.88, 0.88, 0.88, 0.88, 0.88, 0.92, 0.92, 0.92, 0.94
Are these data approximately Normally distributed? Why or why not?
My first step is to graph the data
Since a graph of the data is not symmetric, the
distribution is not Normally distributed.
13. A Normal probability plot of a set of data is shown here. Would you say that
these measurements are approximately Normally distributed? Why or why
not?
Since the normal probability plot does not show a straight (or nearly straight)
line, the data are not Normally distributed.