Question 1

At a tennis tournament a statistician keeps track of every serve. The statistician reported that the mean servespeed of a particular player was 104 miles per hour (mph) and the standard deviation of the serve speeds was 8mph. Assume that the statistician also gave us the information that the distribution of the serve speeds was bellshaped. What proportion of the playerʹs serves are expected to be between 112 mph and 120 mph?
A) 0.1350
B) 0.270
C) 0.95
D) 0.68

Accepted Answer

Given the distribution is bell-shaped (normal distribution), we can use the empirical rule or z-scores to find the proportion of serves between 112 mph and 120 mph. The mean is 104 mph, and the standard deviation is 8 mph. - 112 mph is 1 standard deviation above the mean (104 + 8 = 112).- 120 mph is 2 standard deviations above the mean (104 + 16 = 120).According to the empirical rule (68-95-99.7 rule), about 68% of data falls within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations. However, we are interested in the proportion between 1 and 2 standard deviations above the mean.The area between the mean and 1 standard deviation (either side) is approximately 34% of the data (half of 68%). The area between the mean and 2 standard deviations (either side) is approximately 47.5% of the data (half of 95%). Therefore, the proportion of serves between 112 mph and 120 mph, which is between 1 and 2 standard deviations above the mean, is the difference between these two percentages: 47.5% - 34% = 13.5% or 0.1350.

Question 2

A study was designed to investigate the effects of two variables - (1) a studentʹs level of mathematical anxietyand (2) teaching method - on a studentʹs achievement in a mathematics course. Students who had a low levelof mathematical anxiety were taught using the traditional expository method. These students obtained a meanscore of 460 with a standard deviation of 40 on a standardized test. Assuming a bell-shaped distribution,where would approximately 68% of the students score?
A) between 420 and 500
B) between 380 and 540
C) below 380 or above 540
D) below 420 or above 500

Accepted Answer

Approximately 68% of the scores in a bell-shaped (normal) distribution fall within one standard deviation of the mean. Given a mean of 460 and a standard deviation of 40, 68% of the scores would fall between 420 (460 - 40) and 500 (460 + 40).

Question 3

True or False: Variance is the square root of standard deviation.

Accepted Answer

Variance is the square of the standard deviation, not the square root.

Question 4

The scores from a state standardized test have a bell-shaped distribution with a mean of 100 and a standarddeviation of 15. Use the Empirical Rule to find the percentage of students with scores between 70 and 130.
A) 95%
B) 68%
C) 99.7%
D) 100%

Accepted Answer

The Empirical Rule states that for a bell-shaped distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations. Scores between 70 and 130 are within two standard deviations (15*2 = 30) from the mean (100), so 95% of the scores fall within this range.

Question 5

Last year batting averages in professional baseball averaged 0.266 with a high of 0.334 and a low of 0.228(minimum 250 at-bats). Based on this information, which measure of variation could be calculated?
A) range
B) variance
C) standard deviation
D) percentile

Accepted Answer

The range can be calculated as it is the difference between the highest and lowest values, which are provided (0.334 - 0.228).

Question 6

A small computing center has found that the number of jobs submitted per day to its computers has adistribution that is approximately bell shaped, with a mean of 84 jobs and a standard deviation of 10. Where dowe expect most (approximately 95%) of the distribution to fall?
A) between 64 and 104 jobs per day
B) between 74 and 94 jobs per day
C) between 54 and 114 jobs per day
D) between 64 and 114 jobs per day

Accepted Answer

The 95% confidence interval for a bell-shaped (normal) distribution is given by the mean plus or minus two standard deviations. Therefore, 84 ± 2(10) = 64 to 104 jobs per day.

Question 7

Health care issues are receiving much attention in both academic and political arenas. A sociologist recentlyconducted a survey of citizens over 60 years of age whose net worth is too high to qualify for governmenthealth care but who have no private health insurance. The ages of 25 uninsured senior citizens were as follows: $$\begin{array} { l l l l l l l l l l l l l } 
68 & 73 & 66 & 76 & 86 & 74 & 61 & 89 & 65 & 90 & 69 & 92 & 76 \
62 & 81 & 63 & 68 & 81 & 70 & 73 & 60 & 87 & 75 & 64 & 82 &
\end{array}$$ Suppose the mean and standard deviation are 74.0 and 9.7, respectively. If we assume that the distribution ofages is bell shaped, what percentage of the respondents will be between 64.3 and 93.4 years old?
A) approximately 81.5%
B) approximately 68%
C) approximately 95%
D) approximately 83.9%

Accepted Answer

The answer of Health care issues are receiving much attention...

Question 8

Each year advertisers spend billions of dollars purchasing commercial time on network sports television. In thefirst 6 months of 1988, advertisers spent $1.1 billion. Who were the largest spenders? In a recent article, listedthe top 10 leading spenders (in million of dollars): $$\begin{array} { l r l r } 
\text { Company A } & \$ 70.7 & \text { Company F } & \$ 27.3 \
\text { Company B } & 60.7 & \text { Company G } & 26.1 \
\text { Company C } & 55.4 & \text { Company H } & 22.8 \
\text { Company D } & 54.8 & \text { Company I } & 23 \
\text { Company E } & 28.7 & \text { Company J } & 19.4
\end{array}$$ Calculate the sample variance.
A) 367.36
B) 19.17
C) 38.89
D) 51.30

Accepted Answer

The sample variance is calculated by taking the sum of the squared differences between each value and the mean, then dividing by the number of observations minus one. Given the data, the mean expenditure is $38.89 million. The squared differences from the mean are then summed and divided by the number of observations minus one (10-1=9), resulting in a sample variance of 367.36.

Question 9

A study was designed to investigate the effects of two variables - (1) a studentʹs level of mathematical anxietyand (2) teaching method - on a studentʹs achievement in a mathematics course. Students who had a low levelof mathematical anxiety were taught using the traditional expository method. These students obtained a meanscore of 460 with a standard deviation of 50 on a standardized test. Assuming a bell-shaped distribution, whatpercentage of scores exceeded 360?
A) approximately 97.5%
B) approximately 95%
C) approximately 2.5%
D) approximately 84%

Accepted Answer

The mean score is 460 and the standard deviation is 50. The z-score for a score of 360 is (360 - 460) / 50 = -2.0. The percentage of scores that exceed 360 is the area under the normal curve to the right of z = -2.0, which is approximately 97.5%.

Question 10

For the data set given below the sample variance will be computed. How many degrees of freedom are there?6 9 6 1 5 11 5 9 8
A) 8
B) 9
C) 1
D) 10

Accepted Answer

The degrees of freedom for sample variance is calculated as n - 1, where n is the number of observations. Here, n = 9, so the degrees of freedom are 9 - 1 = 8.

Question 11

The amount of television viewed by todayʹs youth is of primary concern to Parents Against WatchingTelevision (PAWT). 300 parents of elementary school-aged children were asked to estimate the number ofhours per week that their child watched television. The mean and the standard deviation for their responseswere 16 and 4, respectively. PAWT constructed a stem-and-leaf display for the data that showed that thedistribution of times was a bell-shaped distribution. Give an interval around the mean where you believe most(approximately 95%) of the television viewing times fell in the distribution.
A) between 8 and 24 hours per week
B) less than 12 and more than 20 hours per week
C) between 4 and 28 hours per week
D) between 12 and 20 hours per week

Accepted Answer

For a bell-shaped distribution, approximately 95% of the data falls within two standard deviations of the mean. Given a mean of 16 hours and a standard deviation of 4 hours, two standard deviations from the mean would be from 16 - 2(4) = 8 to 16 + 2(4) = 24.

Question 12

Which is not a measure of dispersion?
A) Mean
B) Standard deviation
C) Variance
D) Range

Accepted Answer

Mean is not a measure of dispersion; it is a measure of central tendency, which describes the center of a data set. Measures of dispersion, such as standard deviation, variance, and range, describe the spread of the data.

Question 13

Many firms use on-the-job training to teach their employees new software. Suppose you work in the personneldepartment of a firm that just finished training a group of its employees in new software, and you have beenrequested to review the performance of one of the trainees on the final test that was given to all trainees. Themean and standard deviation of the test scores are 80 and 3, respectively, and the distribution of scores is bellshaped. What percentage of test-takers scored better than a trainee who scored 71?
A) approximately 99.85%
B) approximately 84%
C) approximately 95%
D) approximately 97.5%

Accepted Answer

Given the distribution is bell-shaped (normal distribution) and the standard deviation is 3, a score of 71 is 3 standard deviations below the mean (80 - 71 = 9; 9/3 = 3). In a normal distribution, about 99.85% of the data lies within three standard deviations above the mean, meaning approximately 99.85% scored better than a score of 71.

Question 14

The _______________ is the easiest measure of dispersion to calculate.
A) Range
B) Standard deviation
C) Variance
D) Mean absolute deviation

Accepted Answer

The range, calculated as the difference between the maximum and minimum values in a dataset, is the simplest measure of dispersion to compute.

Question 15

Solar energy is considered by many to be the energy of the future. A recent survey was taken to compare thecost of solar energy to the cost of gas or electric energy. Results of the survey revealed that the distribution ofthe amount of the monthly utility bill of a 3-bedroom house using gas or electric energy had a mean of $120and a standard deviation of $11. If the distribution can be considered bell shaped, what percentage of homeswill have a monthly utility bill of more than $109?
A) approximately 84%
B) approximately 95%
C) approximately 16%
D) approximately 32%

Accepted Answer

Using the empirical rule for a bell-shaped distribution, a score of $109 is one standard deviation below the mean ($120 - $11 = $109). According to the empirical rule, approximately 68% of the data falls within one standard deviation of the mean, leaving 32% outside. Since the question asks for the percentage of homes with bills more than $109, we consider the upper half of the remaining percentage, which is half of 32%, or 16%, added to the 68% within one standard deviation. This gives us 84%.

Question 16

In the computation of the sample standard variance, how many degrees of freedom are there in a data set thathas 15 data values?
A) 14
B) 15
C) 16
D) none of these

Accepted Answer

The degrees of freedom for the sample standard variance is calculated as the number of data values minus one, so for 15 data values, it is 15 - 1 = 14.

Question 17

A statistic is said to be biased if it
A) Systematically underestimates or overestimates a parameter
B) Has exactly the same value as the parameter
C) Is determined from a trimmed sample
D) Leads to an erroneous conclusion about the sample

Accepted Answer

A biased statistic is one that does not accurately represent the parameter it is estimating because it systematically overestimates or underestimates the true value.

Question 18

A severe drought affected several western states for 3 years. A Christmas tree farmer is worried about thedroughtʹs effect on the size of his trees. To decide whether the growth of the trees has been retarded, the farmerdecides to take a sample of the heights of 25 trees. Typically trees of this age have a mean height of 65 incheswith a standard deviation of 9 inches. Assuming the distribution is bell shaped, where do you expect middle95% of the tree heights to fall?
A) between 47 and 83 inches tall
B) between 56 and 74 inches tall
C) between 38 and 92 inches tall
D) over 56 inches tall

Accepted Answer

Using the empirical rule for a bell-shaped distribution, the middle 95% of the data falls within two standard deviations of the mean. The mean is 65 inches, and the standard deviation is 9 inches. Therefore, two standard deviations from the mean is 65 ± 2(9) = 65 ± 18, which gives us a range of 47 to 83 inches.

Question 19

True or False: The variance of a population is the arithmetic average of the squared deviations about thepopulation mean.

Accepted Answer

The variance of a population is indeed calculated as the arithmetic average of the squared deviations from the population mean, which measures the spread of the population data.

Question 20

The scores from a state standardized test have a mean of 80 and a standard deviation of 10. The distribution ofthe scores is roughly bell shaped. Use the Empirical Rule to find the percentage of scores that lie between 60and 80.
A) 47.5%
B) 68%
C) 34%
D) 95%

Accepted Answer

The Empirical Rule states that for a bell-shaped distribution, approximately 68% of the data falls within one standard deviation of the mean. Since the mean is 80 and the standard deviation is 10, the range from 70 to 90 (one standard deviation below and above the mean) contains about 68% of the scores. The question asks for the percentage between 60 and 80, which is two standard deviations below the mean to the mean itself. This range covers half of the 68% (from 70 to 80) plus half of the next 27% (from 60 to 70, since 95% of the data falls within two standard deviations, leaving 95%-68%=27% for the two tails beyond one standard deviation). Half of 68% is 34%, and half of 27% is 13.5%, so adding these gives 34% + 13.5% = 47.5%.

Quiz 3: Numerically Summarizing Data

True or False: Variance is the square root of standard deviation.

The scores from a state standardized test have a bell-shaped distribution with a mean of 100 and a standarddeviation of 15. Use the Empirical Rule to find the percentage of students with scores between 70 and 130.

Last year batting averages in professional baseball averaged 0.266 with a high of 0.334 and a low of 0.228(minimum 250 at-bats) . Based on this information, which measure of variation could be calculated?

A small computing center has found that the number of jobs submitted per day to its computers has adistribution that is approximately bell shaped, with a mean of 84 jobs and a standard deviation of 10. Where dowe expect most (approximately 95%) of the distribution to fall?

For the data set given below the sample variance will be computed. How many degrees of freedom are there?6 9 6 1 5 11 5 9 8

Which is not a measure of dispersion?

The _______________ is the easiest measure of dispersion to calculate.

In the computation of the sample standard variance, how many degrees of freedom are there in a data set thathas 15 data values?

A statistic is said to be biased if it

True or False: The variance of a population is the arithmetic average of the squared deviations about thepopulation mean.

The scores from a state standardized test have a mean of 80 and a standard deviation of 10. The distribution ofthe scores is roughly bell shaped. Use the Empirical Rule to find the percentage of scores that lie between 60and 80.