4.1 The following are the prices (in dollars) of the six all-terrain truck tires rated most highly by a magazine in 2018.
159.00 196.00 159.00 127.35 122.99 125.00
Calculate the value of the mean. (Round your answers to the nearest cent.)
$
Calculate the value of the median. (Round your answers to the nearest cent.)
$

B. Why are these values so different?
There are only 6 values.
The distribution of this set is approximately normal.
The distribution of this set is negatively skewed.
The distribution of this set is positively skewed.

C. Which of the two—mean or median—appears to be better as a description of a typical value for this data set?

The median is always better as a description of a typical value.
The mean is always better as a description of a typical value.
The mean is better since it is not influenced by the extreme value.
The median is better since it is not influenced by the extreme value.
The mean is as good a description of a typical value as the median.

4.2 An article gave the following data on caffeine concentration (mg/ounce) for eight top-selling energy drinks.

Energy Drink Caffeine Concentration
(mg/oz)
Energy Drink #1 9.9
Energy Drink #2 10.0
Energy Drink #3 10.0
Energy Drink #4 10.0
Energy Drink #5 11.5
Energy Drink #6 8.7
Energy Drink #7 9.5
Energy Drink #8 9.4

(a) What is the value of the mean caffeine concentration (in mg/oz) for this set of top-selling energy drinks? mg/oz

4.3 An article reported the sodium content (mg) per 2 tablespoon serving for each of 11 different peanut butters as given below.
125 50 135 125 150 150 150 65 170 250 105

Calculate the mean sodium content (in mg) for the peanut butters in this sample. (Round your answer to four decimal places.)

Calculate the median sodium content (in mg) for the peanut butters in this sample.

4.4 The article “The Wedding Industry’s Pricey Little Secret”† stated that the widely reported average wedding cost is grossly misleading. The article reports that in 2012, the average wedding cost was $27,427 and the median cost was $18,086.

A) Do you agree with the statement that the average wedding cost is misleading? Explain why or why not.

4.5 Each student in a sample of 20 seniors at a particular university was asked if he or she was registered to vote. With R denoting registered and N denoting not registered, the sample data are as given below.
R N R N R R N N R R
N N N R N R N R R N

(a) If being registered to vote is considered a “success,” what is the value of the proportion of successes for this sample? (Enter your answer to two decimal places.)

(b) When would it be reasonable to generalize from this sample to the population of all seniors at this university?
This generalization could be made if the sample were made up of the first 20 registered voters who were asked if they were seniors at the university.
This generalization could be made if the sample were a simple random sample of registered voters at the university.
This generalization could be made if the sample were made up of the first 20 seniors at the university who were asked if they were registered to vote.
This generalization could be made if the sample were a simple random sample of seniors at the university.

4.6 Suppose that the number of speed-related crash fatalities for the 15 U.S. states that had the highest numbers of these fatalities in a certain year is given below.
State Speeding-Related
Traffic Fatalities
Texas 1,248
California 916
Pennsylvania 614
North Carolina 441
Illinois 387
Florida 361
New York 359
Ohio 356
Missouri 326
South Carolina 316
Arizona 297
Alabama 272
Virginia 271
Michigan 251
Oklahoma 217

a) the mean number of speeding-related fatalities for these 15 states. (Round your answer to four decimal places.)
fatalities

b) Calculate the median number of speeding-related fatalities for these 15 states.
fatalities

(c) Explain why it is not reasonable to generalize from this sample of 15 states to the other 35 states.

We would only be able to generalize from this sample to the other 35 states if the sample contained all 50 states.
The distribution of the number of fatalities appears positively skewed, and so it is not reasonable to generalize from this sample to the other 50 states.
The distribution of the number of fatalities appears negatively skewed, and so it is not reasonable to generalize from this sample to the other 50 states.
This sample represents the 15 states with the highest number of speeding-related fatalities, and so it is not reasonable to generalize from this sample to the other 35 states.
The distribution of the number of fatalities appears roughly symmetric, and so it is not reasonable to generalize from this sample to the other 50 states.

4.8 The following data are costs (in cents) per ounce for nine different brands of sliced Swiss cheese.
27 62 36 41 70 83 46 52 49
Calculate the variance for this data set. (Round your answer to four decimal places.)

Calculate the standard deviation for this data set. (Round your answer to four decimal places.)

4.9 Cost per serving (in cents) for six high-fiber cereals rated very good and for nine high-fiber cereals rated good are shown below.
Cereals Rated Very Good
45 48 62 43 16 77
Cereals Rated Good
72 30 51 51 69 41 47 26 56

a) Compute the lower quartile, median, upper quartile, and interquartile range for the cereals rated very good.
lower quartile cents
median cents
upper quartile cents
interquartile range cents

b) Compute the lower quartile, median, upper quartile, and interquartile range for the cereals rated good.
lower quartile cents
median cents
upper quartile cents
interquartile range cents

4.10 Calculate the quartiles and the interquartile range for this combined data set.
The given data are as follows.
46 49 61 42 17 78 72 30
53 53 69 43 48 28 54
Begin by arranging the combined data in numerical order, smallest to largest.
17 28 30 42 46 48 49 53 54 61 69 72 78

The next step is to locate the median. The median, or middle number, divides the data into two equally sized halves. In other words, 50% of the data are less than the median and 50% of the data are greater. The number of observations, n, in the data set influences how to find the value of the median. When n is odd, the single middle value is the median. When n is even, the median is the average of the middle two values. For this data set, n is odd, so the median is the single middle value. The median for this data set is .

4.12 (c)Were there any mild or extreme outliers in the data set? How can you tell? (Hint: See Example 4.11.)
Recall that an observation more than one and a half times the interquartile range, 1.5(iqr), away from the nearest quartile is considered a mild outlier while an observation that is more than 3 times the interquartile range, 3(iqr), away from the nearest quartile is considered an extreme outlier. Earlier it was given that the lower quartile is 6 and the upper quartile is 36.
Find the iqr, 1.5(iqr), and 3(iqr).
iqr = upper quartile − lower quartile
iqr = − 6

iqr =

1.5(iqr) =

3(iqr) =

Mild outliers are at least minutes from the nearest quartile and extreme outliers are at least minutes from the nearest quartile.

4.13 Fiber content (in grams per serving) for 18 high fiber cereals are shown below.
Fiber Content
6 10 10 6 9 6 12 12 9
13 10 9 12 6 14 6 9 9

a) Find the median, quartiles, and interquartile range for the fiber content data set.
median=
lower quartile=
upper quartile=
interquartile range=

4.14 The average playing time of music albums in a large collection is 34 minutes, and the standard deviation is 3 minutes.

(a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean?

1 standard deviation above the mean
1 standard deviation below the mean
2 standard deviations above the mean
2 standard deviations below the mean

b) Without assuming anything about the distribution of times, at least what percentage of the times are between 28 and 40 minutes? (Round the answer to the nearest whole number.)
At least %
(b) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than 25 minutes or greater than 43 minutes? (Round the answer to the nearest whole number.)
No more than %
c) Assuming that the distribution of times is approximately normal, about what percentage of times are between 28 and 40 minutes? (Round the answers to two decimal places, if needed.)
%
Less than 25 min or greater than 43 min?
%
Less than 25 min?
%

4.15 (a)What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean?
To find the value that is some number of standard deviations above or below the mean, we will add or subtract some multiple of the standard deviation, s, to the sample mean, x. The average playing time was given to be 35 minutes with a standard deviation of 4 minutes, so we have

x = and s = .

4.16 In a study investigating the effect of car speed on accident severity, 5,000 reports of fatal automobile accidents were examined, and the vehicle speed at impact was recorded for each one. For these 5,000 accidents, the average speed was 44 mph and the standard deviation was 14 mph. A histogram revealed that the vehicle speed at impact distribution was approximately normal. (Use the Empirical Rule.)

(a)Approximately what percentage of these vehicle speeds were between 30 and 58 mph?
approximately %

b)Approximately what percentage of these vehicle speeds exceeded 58 mph? (Round your answer to the nearest whole number.)
approximately %

Assignment 5.0

5.6 (a)Calculate and interpret the value of the correlation coefficient for this data set.

Cost per Serving (x) Fiber per Serving (y) zx =
x − x
sx
zy =
y − y
sy
zxzy
47 7 zx = 47 − 48.556
15.209

 =  −0.1023
zy   =  7 − 9.333

2.351

 =  −0.9923
(−0.1023)(−0.9923) = 0.1015

47 10 zx = 47 − 48.556
15.209

 =   

zy   =  10 − 9.333

2.351

 =  0.2835

48 10
0.2835

53 8 0.2922 −0.5670 −0.1657
zxzy = 2.9546

Use the sum of zxzy products and n = 18 to calculate the value of the correlation coefficient. (Round your answer to four decimal places.)

r = zxzy

n − 1

 =  2.9546

− 1

 =   

5.7 The authors of the paper “Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement” compared two different instruments for measuring a subject’s ability to breathe out air.† (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen subjects participated in the study, and for each subject air flow was measured once using the Wright meter and once using the mini-Wright meter.

Subject Mini-
Wright
Meter Wright
Meter Subject Mini-
Wright
Meter Wright
Meter
1 512 494 10 445 433
2 430 395 11 432 417
3 520 516 12 626 656
4 428 434 13 260 267
5 500 476 14 477 478
6 600 557 15 259 178
7 364 413 16 350 423
8 380 442 17 451 427
9 658 650

a) Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce different readings but there is a strong relationship between the readings, it would be possible to use a reading from the mini-Wright meter to predict the reading that the larger Wright meter would have given. Use the given data to find an equation to predict y = Wright meter reading using a reading from the mini-Wright meter. (Round your values to four decimal places.)
ŷ = + x

b)What would you predict for the Wright meter reading of a subject whose mini-Wright meter reading was 532? (Round your answer to three decimal places.)

(c) What would you predict for the Wright meter reading of a subject whose mini-Wright meter reading was 394? (Round your answer to three decimal places.)

5.8 Does it pay to stay in school? A report looked at the median hourly wage gain per additional year of schooling in 2007. The report states that workers with a high school diploma had a median hourly wage that was 16% higher than those who had only completed 11 years of school. Workers who had completed 1 year of college (13 years of education) had a median hourly wage that was 17% higher than that of the workers who had completed only 12 years of school. The added gain in median hourly wage for each additional year of school is shown in the accompanying table. The entry for 15 years of schooling has been intentionally omitted from the table.
Years of Schooling 2007 Median Hourly Wage Gain
for the Additional Year (percent)
12 16
13 17
14 19
16 22
17 24
18 25

(a)Use the given data to predict the median hourly wage gain (in percent) for the 15th year of schooling. (Round your answer to one decimal place.)
%
(b)The actual wage gain for 15th year of schooling was 20%. How close (in percent) was the actual value to the predicted wage gain percent from part (a)? (Use as predicted − actual. Round your answer to one decimal place.)
%

5.9 Two hundred and eighty boys completed a test that measures the distance that the subject can walk on a flat, hard surface in 6 minutes. For each age group shown in the table, the median distance walked by the boys in that age group is also given.
Age Group Representative
Age (Midpoint
of Age Group) Median Six-Minute
Walk Distance
(meters)
3–5 4 545.3
6–8 7 583.0
9–11 10 668.3
12–15 13.5 700.1
16–18 17 728.6

(b)Find the equation of the least-squares line that describes the relationship between median distance walked in 6 minutes and representative age. (Round your values to four decimal places.)
ŷ =

c)culate the five residuals. (Round your answers to three decimal places.)
Representative
Age (x) Residual
4

7

10

13.5

17

5.10

Representative Age (x) Median 6-minute Walk Distance (y) xy x2
4.0 542.3 (4.0)(542.3) = 2,169.2 (4.0)(4.0) = 16
7.0 586.0 (7.0)(586.0) =
(7.0)(7.0) =

10.0 665.3

13.5 703.1 9,491.85 182.25
17.0 725.6 12,335.2 289
x = 51.5

     y = 3,222.3


     xy =  



     x2 =

5.11 Acrylamide is a chemical that is sometimes found in cooked starchy foods and which is thought to increase the risk of certain kinds of cancer. A paper describes a study to investigate the effect of x = frying time (in seconds) and y = acrylamide concentration (in micrograms per kg) in French fries. The data in the accompanying table are approximate values read from a graph that appeared in the paper.
Frying
Time Acrylamide
Concentration
150 160
240 120
240 195
270 185
300 145
300 270

a) When the potentially influential observation is deleted from the data set, the equation of the least-squares line using the remaining five observations is
ŷ = −42 + 0.83x. Use this equation to predict acrylamide concentration (in micrograms per kg) for a frying time of 270 seconds. (Round your answer to two decimal places.)

micrograms per kg

Assignment 6.0
6.6 A large department store offers online ordering. When a purchase is made online, the customer can select one of four different delivery options: expedited overnight delivery, expedited second-business-day delivery, standard delivery, or delivery to the nearest store for customer pick-up. Consider the chance experiment that consists of observing the selected delivery option for a randomly selected online purchase.
Suppose that the probability of an overnight delivery selection is 0.2, the probability of a second-day delivery selection is 0.3, and the probability of a standard-delivery selection is 0.2. Find the following probabilities.
(a)Find the probability that a randomly selected online purchase selects delivery to the nearest store for customer pick-up.

(b)Find the probability that the customer selects a form of expedited delivery.

(c)Find the probability that either standard delivery or delivery to the nearest store is selected.

6.7 The manager of an online music store has kept records of the number of songs downloaded in a single transaction by customers who make a purchase at the store. Consider the chance experiment of observing the number of songs downloaded by a randomly selected customer. The accompanying table gives six possible outcomes and the estimated probability of each of these outcomes.
Number of songs downloaded 1 2 3 4 5 6 or more
Estimated probability 0.35 0.20 0.15 0.15 0.09 0.06
(a) What is the estimated probability that a randomly selected customer downloads three or fewer songs?

(b) What is the estimated probability that a randomly selected customer downloads at most three songs?

6.9 Medical insurance status—covered (C) or not covered (N)—is determined for each individual arriving for treatment at a hospital’s emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients.
The simple events are O1 = (C, C), meaning that the first patient selected was covered and the second patient selected was also covered, O2 = (C, N), O3 = (N, C), and O4 = (N, N). Suppose that probabilities are P(O1) = 0.81, P(O2) = 0.09, P(O3) = 0.09, and P(O4) = 0.01.
(a) What simple events are contained in A, the event that at most one patient is not covered?

A = {(C, C), (C, N), (N, N)}
A = {(C, C), (N, C), (N, N)}
A = {(C, N), (N, C)}
A = {(C, N), (N, C), (N, N)}
A = {(C, C), (C, N), (N, C)}

Calculate P(A).
P(A) =

(b) What simple events are contained in B, the event that the two patients have different statuses with respect to coverage?
B = {(C, C), (C, N)}
B = {(C, C), (N, N)}
B = {(C, N), (N, C)}
B = {(C, N), (N, N)}
the empty set
Calculate P(B).
P(B) =

6.10 Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, as shown in the accompanying picture.

A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments.
(a) When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? (Round your answer to three decimal places.)

6.11 Phoenix is a hub for a large airline. Suppose that on a particular day, 8,000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1,400 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 410 connecting passengers missed their connecting flight and were delayed in Phoenix. Of the 410 who were delayed, 65 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8,000 passengers. Calculate the following probabilities. (Round your answers to three decimal places.)

(a)Calculate the probability that the selected passenger had Phoenix as a final destination.

(b)Calculate the probability that the selected passenger did not have Phoenix as a final destination.

(c) Calculate the probability that the selected passenger was connecting and missed the connecting flight.

(d)Calculate the probability that the selected passenger was a connecting passenger and did not miss the connecting flight.

(e)Calculate the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight in Phoenix.

6.12 Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Suppose that E denotes the event that the first airline’s flight is fully booked on a particular day, and F denotes the event that the second airline’s flight is fully booked on that same day. Suppose that P(E) = 0.7, P(F) = 0.6, and P(E ∩ F) = 0.5.
(a)Calculate P(E | F) the probability that the first airline’s flight is fully booked given that the second airline’s flight is fully booked. (Round your answer to three decimal places.)

(b)Calculate P(F | E). (Round your answer to three decimal places.)

6.13 Is ultrasound a reliable method for determining the gender of an unborn baby? There is accompanying data on 1,000 births in the table below.
Ultrasound
Predicted
Female Ultrasound
Predicted
Male
Actual Gender Is
Female 439 47
Actual Gender Is
Male 132 382

(a)Use the given information to estimate the probability that a newborn baby is female, given that the ultrasound predicted the baby would be female. (Round your answer to three decimal places.)

(b)Use the given information to estimate the probability that a newborn baby is male, given that the ultrasound predicted the baby would be male. (Round your answer to three decimal places.)

Sample Solution

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