chapter 3

Question 1

Using Chebyshev’s Theorem, find the interval that contains at least 93.75 percent of all measurements when the mean = 2.549 and s = 1.828

a) [−1.107, 6.205]

b) [2.435, 2.663]

c) [−2.935, 8.033]

d) [−4.763, 9.861]

e) [−26.699, 31.797]

Answer 1

d) [−4.763, 9.861]

Explanation
1 − (1/k2) = .9375; 1/k2 = 1 − .9375; 1/k = √.0625; k = 4

Question 2

Using Chebyshev’s theorem, approximate the minimum proportion of the data that will be within μ ± kσ for k = 1.6.

a) 92%

b) 61%

c) 68%

d) 58%

e) 39%

Answer 2

b) 61%

Explanation
100(1 − 1 /k2)% = 100(1 − 1 /1.62)% = 61%

Question 3

In a statistics class, 10 scores were randomly selected with the following results: 74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the IQR?

a) 10

b) 5.00

c) 5.25

d) 11.00

e) 12.00

Answer 3

a) 10

Explanation
IQR = Q3− Q1 = 77 − 67 = 10

Question 4

The average life of Canadian women is 73.75 years, and the standard deviation of the life expectancy of Canadian women is 6.5 years. Using Chebyshev’s theorem, determine the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years.

a) 33.33%

b) 55.56%

c) 88.89%

d) 68.26%

e) 93.17%

Answer 4

b) 55.56%

Explanation
Determine the value of k: (83.5, 64) = 73.75 ± k(6.5); k = 1.5

1 − (1/k2) = 1 − (1/2.25) = .5556, or 55.56%

Question 5

A quantity that measures the variation of a population or a sample relative to its mean is called the ________.

a) interquartile range

b) range

c) variance

d) coefficient of variation

e) standard deviation

Answer 5

d) coefficient of variation

Explanation
The formula is the standard deviation divided by the mean then multiplied by 100.

Question 6

If the median of a data set is 760, the third quartile is 950, and the first quartile is 650, what is the interquartile range?

a) 150

b) 190

c) 910

d) 110

e) 300

Answer 6

e) 300

Explanation
Interquartile range = 950 − 650 = 300

Question 7

The average lateness for one of the top airline companies is 10 minutes. The variance of the lateness measure is calculated as 9. An airplane arrived 13 minutes after the stated arrival time. Calculate the z-score for the lateness of this particular airplane.

a) 1.00

b) 1.33

c) .58

d) .44

e) .33

Answer 7

a) 1.00

Explanation
Z=13−109√=1

Question 8

A measurement located outside the upper limits of a box-and-whiskers display is ________.

a) always the largest value in the data set

b) always in the first quartile

c) within the lower limits

d) an outlier

Answer 8

d) an outlier

Explanation
The box gives you the range from the first to the third quartile.

Question 9

The average of the squared deviations of the individual population measurement from the population mean is the ________.

a) median

b) range

c) mean

d) variance

e) standard deviation

Answer 9

d) variance

Explanation
This number is represented by sigma and is calculated via the standard deviation.

Question 10

The ________ is the positive square root of the sample variance.

a) range

b) median

c) sample mean

d) population standard deviation

e) sample standard deviation

Answer 10

e) sample standard deviation

Explanation
It is represented by sigma and shows how far values are from the mean.

Question 11

In a statistics class, 10 scores were randomly selected, with the following results: 74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the 90th percentile?

a) 73

b) 74

c) 67

d) 77

e) 65.9

Answer 11

d) 77

Explanation
65, 66, 67, 68, 71, 73, 74, 77, 77, 77

(90/100)n = (90/100)(10) = 9th position = 77

Question 12

Compute the sample standard deviation of the data set 6,4,2,1,4,1.

a) 1.41

b) 2.00

c) 4.00

d) 3.33

e) 1.83

Answer 12

b) 2.00

Explanation
Std Dev = √Variance = √[Σ(x−mean)2/(n−1)]; mean = 3; √(20/5) = √4 = 2

Question 13

In a statistics class, 10 scores were randomly selected with the following results (mean = 71.5): 74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the range?

a) 22.72

b) 516.20

c) 4.77

d) 12.00

e) 144.00

Answer 13

d) 12.00

Explanation
65, 66, 67, 68, 71, 73, 74, 77, 77, 77

77 − 65=12

Question 14

The average life of Canadian women is 73.75 years, and the standard deviation of the life expectancy of Canadian women is 6.5 years. Based on Chebyshev’s theorem, determine the upper and lower bounds on the average life expectancy of Canadian women such that at least 90 percent of the population is included.

a) [8.75 138.75]

b) [53.20 94.30]

c) [67.25 80.25]

d) [66.38 81.13]

e) [12.09 135.41]

Answer 14

b) [53.20 94.30]

Explanation
1−1k2=.901k2=.1k2=1.1=10 ; k=10−−√=3.162

lower bound = 73.75 − (3.162)(6.5) = 53.2

upper bound = 73.75 + (3.162)(6.5) = 94.3

Question 15

Another name for the 50th percentile is the ________.

a) median

b) mode

c) first quartile

d) mean

e) third quartile

Answer 15

a) median

Explanation
The median is the middle number of a sample or population.

Question 16

In a statistics class, 10 scores were randomly selected with the following results (mean = 71.5): 74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the variance?

a) 516.20

b) 4.77

c) 12.00

d) 22.72

e) 144.00

Answer 16

d) 22.72

Explanation
(74 − 71.5)= 2.5 squared = 6.25

(73 − 71.5)= 1.5 squared = 2.25

(77 − 71.5)= 5.5 squared = 30.25

(77 − 71.5)= 5.5 squared = 30.25

(71 − 71.5)= 0.5 squared = 0.25

(68 − 71.5)= −3.5 squared = 12.25

(65 − 71.5)= −6.5 squared = 42.25

(77 − 71.5)= 5.5 squared = 30.25

(67 − 71.5)= −4.5 squared = 20.25

(66 − 71.5)= −5.5 squared = 30.25

(6.25 + 2.25 + 30.25 + 30.25 + 0.25 + 12.25 + 42.25 + 30.25 + 20.25 + 30.25) = 204.5 / 9 = 22.71

Question 17

In a statistics class, 10 scores were randomly selected with the following results (mean = 71.5): 74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the standard deviation?

a) 144.00

b) 516.20

c) 4.77

d) 22.72

e) 12.00

Answer 17

c) 4.77

Explanation
You take the square root of 22.71 = 4.77

Question 18

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 18

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 18

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 19

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 19

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 20

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 20

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Question 21

A portfolio’s annual total returns (in percent) for a five-year period are:

–7.14 1.62 2.50 –2.50 9.27

The median and the standard deviation for this sample are the closest to ______________.

a) 1.62 and 5.46

b) 0.75 and 5.46

c) 1.62 and 6.11

d) 2.50 and 6.11

Answer 21

c) 1.62 and 6.11

Explanation
For the median arrange data in ascending order, then the median is middle value. For sample standard deviation, first calculate the sample mean, then calculate sample variance, and finally, the sample standard deviation.

Question 22

A disadvantage of using grouping (a frequency table) with sample data is that

a) the interpretation of the grouped data descriptive statistics is meaningless.

b) calculations involving central tendency and variation are more complicated than central tendency and variation calculations based on ungrouped data.

c) the descriptive statistics are less precise than the descriptive statistics obtained using ungrouped data.

d) it is much more difficult to summarize the information than it is with the ungrouped data.

Answer 22

c) the descriptive statistics are less precise than the descriptive statistics obtained using ungrouped data.

Explanation
This is because we do not have access to the individual values; only the grouped data values.

Question 23

According to Chebyshev’s theorem, at least what proportion of the data will be within μ ± kσ for k = 2?

a) 50%

b) 34%

c)25%

d) 68%

e) 75%

Answer 23

e) 75%

Explanation
For any values of k greater than 1, at least 100(1 − 1 /k2)% of the population measurements lie in the interval.

Question 24

What is an advantage of the correlation coefficient over the covariance?

a) Both answers-that it falls between –1 and 1 and that it is a unit-free measure-are correct.

b) It falls between –1 and 1.

c) It falls between 0 and 1.

d) It is a unit-free measure, therefore making it easier to interpret.

Answer 24

a) Both answers-that it falls between –1 and 1 and that it is a unit-free measure-are correct.

Explanation
The correlation coefficient is preferred in evaluating the direction and strength of the linear relationship between two variables. It is a unit-free measure, assuming the values from the interval [–1, 1].

Question 25

You buy 50 stocks of Company A, 30 of Company B, and 20 of Company C. The annual returns of these companies are 8%, 12%, and 10% respectively. The average return for one year is the closest to _____.

a) 9.6%

b) 10.5%

c) 9.1%

d) 10.0%

Answer 25

a) 9.6%

Explanation
The mean for a frequency distribution for grouped data is defined as x¯=Σmifin.

Question 26

Is it possible for a data set to have more than one mode?

a) Yes, if two or more values in a data set occur the same number of times.

b) Yes, if two or more values in a data set occur with the most frequency and the frequency is greater than one.

c) Yes, if there are at least two different values in a data set, there is always more than one mode.

d) No, there must always be a single mode, or else there is no mode.

Answer 26

b) Yes, if two or more values in a data set occur with the most frequency and the frequency is greater than one.

Explanation
If multiple data values occur with the same frequency (greater than one), and that frequency is the greatest seen in the data, then the data has multiple modes. If every value occurred only once, then there would be no mode.

Question 27

Is it possible for a data set to have no mode?

a) No, unless there is an odd number of observations.

b) Yes, if there are no observations that occur more than once.

c) Yes, if two observations occur twice.

d) No, if the data set is nonempty, there is always a mode.

Answer 27

b) Yes, if there are no observations that occur more than once.

Explanation
If every value in a data set occurs only once, there is no mode.

Question 28

Find the coefficient of variation for IQ tests with a mean of 100 and a standard deviation of 15.

a) 67

b) 6.7

c)1.5

d) 15.0

Answer 28

d) 15.0

Explanation
Coefficient of variation = (Std dev/mean) × 100 = (15/100) × 100 = 15

Question 29

The following table shows the Price-to-Earnings ratio for a stereo equipment manufacturing company between 1998 and 2002.

Year P/E Ratio
1998 12.4
1999 14.6
2000 11.1
2001 8.2
2002 6.8

Determine the percentage change in the P/E ratios from 1998 to 1999.

a) −15.07%

b) 17.74%

c) 20.72%

d) −17.74%

e) 15.07%

Answer 29

b) 17.74%

Explanation
R1=14.6 -12.412.4 X 100 = 17.74%

Question 30

In the calculation of a mean for grouped data, ________ are used.

a) total sample size and sum of the midpoints of each class

b) sum of the frequency of each class and the sum of the midpoints of each class

c) sum of the frequency of each class and the sample midpoint

d) total sample size and sum of the weighted midpoints

Answer 30

d) total sample size and sum of the weighted midpoints

Explanation
Total sample size and sum of the weighted midpoint are used so therefore we cannot make many predictions about the individual data.

Question 31

Amounts spent by a sample of 50 customers at a retail store are summarized in the following relative frequency distribution.

Amount Spent (in $) Relative Frequency
0 up to 10 0.20
10 up to 20 0.40
20 up to 30 0.30
30 up to 40 0.10

a) The median amount will fall in the following class interval _____________.

b) 0 up to 10

c) 30 up to 40

d) 20 up to 30

e) 10 up to 20

Answer 31

e) 10 up to 20

Explanation
Given 50 observations, the median will be between the 25th and the 26th observations in the sorted data. The range 0-10 has 50 × 0.20 = 10 observations. The range 10-20 has 50 × 0.40 = 20 = observations.

Question 32

When using Chebyshev’s Theorem to obtain the bounds for 99.73 percent of the values in a population, the interval generally will be ________ the interval obtained for the same percentage if a normal distribution is assumed (Empirical Rule).

a) wider than

b) shorter than

c) the same as

Answer 32

a) wider than

Explanation
This is due to the fact that Chebyshev’s Theorem is used for non-mound shaped populations.

Question 33

If a population distribution is skewed to the right, then, given a random sample from that population, one would expect that the ________.

a) median would be less than the mean

b) mode would be equal to the mean

c) median would be equal to the mean

d) median would be greater than the mean

Answer 33

a) median would be less than the mean

Explanation
The median in this case would be a better representation of the population – showing where most of the numbers congregate.

Question 34

A college professor collected data on the number of hours spent by his 100 students over the weekend to prepare for Monday’s Business Statistics exam. He processed the data by Excel and the following incomplete output is available.

Mean 7
Sample Variance 7.84
Skewness 1.17

The coefficient of variation in the data is ______.

a) 90%

b) 40%

c) 111%

d) 243%

Answer 34

b) 40%

Explanation
Calculated by dividing a data set’s standard deviation by its mean, the coefficient of variation is a unit-free measure that allows for direct comparisons of mean-adjusted dispersion across different data sets.

Question 35

Which of the following is the most influenced by outliers?

a) Arithmetic mean

b) Median

c) 75th percentile

d) Mode

Answer 35

a) Arithmetic mean

Explanation
As opposed to the median, mode, and all percentiles, the arithmetic mean is the most affected by outliers because it is a calculated average.

Question 36

Suppose the average price for new cars has a mean of $30,100, a standard deviation of $5,600 and is normally distributed. Based on this information, what interval of prices would we expect at least 95% of new car prices to fall within?

a) $18,900 to $41,300

b) $13,300 to $46,900

c) $24,500 to $35,700

d) $7.700 to $52,500

Answer 36

a) $18,900 to $41,300

Explanation
The empirical rule dictates that 95% of all observations fall in the interval x− +− 2 s.

Question 37

Which of the following is a measure of the strength of the linear relationship between x and y that is dependent on the units in which x and y are measured?

a) correlation coefficient

b) slope

c) covariance

d) least squares line

Answer 37

c) covariance

Explanation
This tells you how close the relationship is to a straight line between x and y.

Question 38

Professors at a local university earn an average salary of $80,000 with a standard deviation of $6,000. The salary distribution is approximately bell-shaped. Because of budget limitations, it has been decided that only those whose salaries are approximately in the bottom 2.5% would get a raise. What is the maximum current salary that qualifies for the raise?

a) It is about $58,000.

b) It is about $74,000.

c) It is about $68,000.

d) It is about $62,000.

Answer 38

c) It is about $68,000.

Explanation
Because the data are bell-shaped, we apply the empirical rule. According to the empirical rule, approximately 95% of the observations fall within two standard deviations of the mean μ+−2σ and 5% fall above/below this range. So, for the lower 2.5%, the value is $80,000 − 2 × $6,000 = $68,000.

Question 39

The average class size this semester in the business school of a particular university is 38.1 students with a standard deviation of 12.9 students. The z-score for a class with 21 students is _____.

a) –1.33

b) 0

c) 1.51

d) 0.8

Answer 39

a) -1.33

Explanation
The z-score is computed as Z= x− x− s .

Question 40

In a statistics class, 10 scores were randomly selected, with the following results:
74, 73, 77, 77, 71, 68, 65, 77, 67, 66.
What is the 10th percentile?

a) 77.0

b) 65.5

c) 73.85

d) 67.3

e) 66.75

Answer 40

b) 65.5

Explanation
65, 66, 67, 68, 71, 73, 74, 77, 77, 77

(10/100)n = (10/100)(10) = 1th position = 65 + 66 / 2 = 65.5

Question 41

In its standard form, Chebyshev’s theorem provides a lower bound on __________________________________

a) the number of observations lying outside a certain interval

b) the number of observations lying within a certain interval

c) the proportion (or percentage) of observations lying outside a certain interval

d) the proportion (or percentage) of observations lying within a certain interval

Answer 41

d) the proportion (or percentage) of observations lying within a certain interval

Explanation
For every k > 1, Chebyshev’s theorem provides a lower bound on the proportion (or percentage) of the data that lie within k standard deviations from the mean