Question 1

A store manager kept track of the number of newspapers sold each week over a seven-week period. The results are shown below. $$\begin{array} { l l l l l l l } 
81 & 71 & 202 & 113 & 269 & 248 & 242
\end{array}$$ Find the median number of newspapers sold.

Accepted Answer

To find the median, first, the numbers need to be arranged in ascending order: 71, 81, 113, 202, 242, 248, 269. The median is the middle number, which is 202 newspapers.

Question 2

The customer service department of a phone company is experimenting with two different systems. On Monday they try the first system which is based on an automated menu system.
On Tuesday they try the second system in which each caller is immediately connected with a
Live agent. A quality control manager selects a sample of seven calls each day. He records the
Time for each customer to have his or her question answered. The times (in minutes)are listed below. Find the coefficient of variation for each of the two sets of data, then compare the
Variation. Round results to one decimal place. 
Automated Menu: $11.7 \quad 7.4 \quad 3.9 \quad 2.9 \quad 9.2 \quad 6.3 \quad 5.5$
$\begin{array} { l l l l l l l l } \text { Live agent: } & 6.2 & 2.9 & 4.4 & 4.1 & 3.4 & 5.2 & 3.7 \end{array}$

Accepted Answer

The coefficient of variation is a measure of the relative variability of a data set. It is calculated by dividing the standard deviation by the mean. The higher the coefficient of variation, the more variable the data set.For the automated menu system, the mean is 6.8 minutes and the standard deviation is 3.1 minutes. The coefficient of variation is therefore 3.1 / 6.8 = 0.454, or 45.4%.For the live agent system, the mean is 4.2 minutes and the standard deviation is 1.2 minutes. The coefficient of variation is therefore 1.2 / 4.2 = 0.283, or 28.3%.Therefore, the automated menu system has a higher coefficient of variation than the live agent system, which means that there is more variability in the times for the automated menu system.

Question 3

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data. $$\begin{array} { l l l l l l } 
22.6 & 16.1 & 36.1 & 36.0 & 23.8 & 20.3
\end{array}$$

Accepted Answer

The standard deviation of a sample is calculated using the formula: $\sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$, where $x_i$ represents each value in the sample, $\bar{x}$ is the mean of the sample, and $n$ is the number of observations in the sample. First, calculate the mean: $\bar{x} = \frac{22.6 + 16.1 + 36.1 + 36.0 + 23.8 + 20.3}{6} = 25.82$. Then, calculate the sum of the squared deviations from the mean: $\sum (x_i - \bar{x})^2 = (22.6 - 25.82)^2 + (16.1 - 25.82)^2 + (36.1 - 25.82)^2 + (36.0 - 25.82)^2 + (23.8 - 25.82)^2 + (20.3 - 25.82)^2 = 419.12$. Finally, divide by $n-1 = 5$ and take the square root: $\sqrt{\frac{419.12}{5}} = \sqrt{83.824} \approx 9.2$. The correct answer should be approximately 9.2, but since this option is not available and there seems to be a typo in the options, the closest correct methodology points towards an answer that is not listed correctly. The process described is correct for calculating standard deviation, but the final answer provided in the options does not match the expected result based on the calculation.

Question 4

A student earned grades of B, B, A, C, and D. Those courses had these corresponding numbers of credit hours:  4,5,1,5,4 . The grading system assigns quality points to letter grades as follows:  A=4, B=3, C=2, D=1 , and  F=0 . Compute the grade point average (GPA) and round the result to two decimal places.

Accepted Answer

The answer of A student earned grades of B, B,...

Question 5

If the standard deviation for a set of data is 0, which of the following must be true?_

Accepted Answer

If the standard deviation for a set of data is 0, it means that all the data values are identical. The standard deviation measures how spread out the data is from the mean, so if there is no variation, then all the data values must be the same. Option A is incorrect because it is possible to have a standard deviation of 0 without all the data values being 0. Option C is also incorrect because the data values could be positive or zero. Option D is incorrect because it is possible for the standard deviation to be 0 in certain situations.

Question 6

Construct a boxplot for the given data. Include values of the 5-number summary in all_ boxplots. The test scores of 40 students are listed below. $\begin{array} { l l l l l l l l l l } 25 & 35 & 43 & 44 & 47 & 48 & 54 & 55 & 56 & 57 \end{array}$
$\begin{array} { l l l l l l l l l l } 59 & 62 & 63 & 65 & 66 & 68 & 69 & 69 & 71 & 72 \end{array}$
$\begin{array} { l l l l l l l l l l l } 72 & 73 & 74 & 76 & 77 & 77 & 78 & 79 & 80 & 81 \end{array}$
$\begin{array} { l l l l l l l l l l } 81 & 82 & 83 & 85 & 89 & 92 & 93 & 94 & 97 & 98 \end{array}$

Accepted Answer

The answer of Construct a boxplot for the given data....

Question 7

The prices (in dollars)of 12 electric smoothtop ranges are listed below. Find the range for the given sample data. 
$\begin{array} { c c c c c c } 
865 & 1010 & 655 & 565 & 1465 & 1110 \
710 & 765 & 820 & 1310 & 555 & 1065
\end{array}$

Accepted Answer

The range is the difference between the highest and lowest values. The highest value is 1465 and the lowest value is 555. Therefore, the range is 1465 - 555 = 910.

Question 8

The number of vehicles passing through a bank drive -up line during each 15-minute period was recorded. The results are shown below. Find the median number of vehicles going
Through the line in a 15-minute period. $\begin{array} { l l l l } 25 & 27 & 25 & 28 \end{array}$
$\begin{array} { l l l l } 28 & 25 & 30 & 27 \end{array}$
$ \begin{array}{llll}35 & 31 & 31 & 29 \ 24 & 31 & 25 & 20\end{array} $
$ \begin{array}{llll}15 & 27 & 27 & 27\end{array} $

Accepted Answer

To find the median, first, list all the numbers in ascending order: 15, 20, 24, 25, 25, 25, 25, 27, 27, 27, 27, 28, 28, 29, 30, 31, 31, 31, 35. With 19 data points, the median is the 10th number, which is 27.

Question 9

The test scores of 40 students are listed below. Find P56._ $\begin{array} { l l l l l l l l l l } 30 & 35 & 43 & 44 & 47 & 48 & 54 & 55 & 56 & 57 \end{array}$
$\begin{array} { l l l l l l l l l l l } 59 & 62 & 63 & 65 & 66 & 68 & 69 & 69 & 71 & 72 \end{array}$
$\begin{array} { l l l l l l l l l l } 30 & 35 & 43 & 44 & 47 & 48 & 54 & 55 & 56 & 57 \ 59 & 62 & 63 & 65 & 66 & 68 & 69 & 69 & 71 & 72 \ 72 & 73 & 74 & 76 & 77 & 77 & 78 & 79 & 80 & 81 \ 81 & 82 & 83 & 85 & 89 & 92 & 93 & 94 & 97 & 98 \end{array}$
$\begin{array} { l l l l l l l l l l } 72 & 73 & 74 & 76 & 77 & 77 & 78 & 79 & 80 & 81 \end{array}$
$\begin{array} { l l l l l l l l l l } 81 & 82 & 83 & 85 & 89 & 92 & 93 & 94 & 97 & 98 \end{array}$

Accepted Answer

P56 is the 56th percentile, which means that 56% of the data is below this value. Since there are 40 data points, the 56th percentile is the 22nd data point. The 22nd data point is 74, so the answer is C.

Question 10

If your score on your next statistics test is converted to a z score, which of these z scores would you prefer?

Accepted Answer

A z-score of 2.00 indicates that your score is two standard deviations above the mean, which is generally considered good performance.

Question 11

Find the standard deviation for the given sample data. Round your answer to one more_ decimal place than the original data. $$\begin{array} { l l l l l l l l l } 
18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15
\end{array}$$

Accepted Answer

The sample standard deviation is calculated by first finding the mean, then the squared differences from the mean, averaging those, and taking the square root. The mean is 12.56, the squared differences average to 29.06, and the square root is approximately 5.4.

Question 12

Find the variance for the given data. Round your answer to one more decimal place than the_ original data. To get the best deal on a microwave oven, Jeremy called six appliance stores
And asked the cost of a specific model. The prices he was quoted are listed below: $\begin{array} { l l l l l l } \$ 663 & \$ 273 & \$ 410 & \$ 622 & \$ 174 & \$ 374 \end{array}$

Accepted Answer

The answer of Find the variance for the given data....

Question 13

Scores on the SAT test have a mean of 1518 and a standard deviation of 325. Scores on the ACT test have a mean of 21.1 and a standard deviation of 4.8. Which of the following
Choices is not true?

Accepted Answer

The z-score for an SAT score of 1490 is calculated as $ \frac{1490 - 1518}{325} = -0.09 $, indicating it is below average. For an ACT score of 17.0, the z-score is $ \frac{17.0 - 21.1}{4.8} = -0.85 $, also below average. Comparing the z-scores, -0.09 is closer to the mean than -0.85, making the SAT score of 1490 relatively better than the ACT score of 17.0. Therefore, statement D is not true as it incorrectly asserts the opposite.

Question 14

Find the variance for the given sample data. Round your answer to one more decimal place_ than the original data.
 $\begin{array} { l l l l l } 
7 & 7 & 2 & 5 & 1
\end{array}$

Accepted Answer

To find the variance, first calculate the mean: (7+7+2+5+1)/5 = 4.4. Then, find the squared differences from the mean: (7-4.4)², (7-4.4)², (2-4.4)², (5-4.4)², (1-4.4)². These are 6.76, 6.76, 5.76, 0.36, and 11.56. The sum of these squared differences is 31.2. Divide by the number of data points minus one (5-1=4) to get the sample variance: 31.2/4 = 7.8.

Question 15

Listed below are the amounts of time (in months)that the employees of a restaurant have been working at the restaurant. Find the mean for the given sample data. Round your
Answer to one more decimal place than the original data values. $$\begin{array} { l l l l l l l l l l l l l l } 
1 & 5 & 6 & 8 & 11 & 14 & 17 & 46 & 61 & 90 & 99 & 126 & 143 & 167
\end{array}$$

Accepted Answer

The answer of Listed below are the amounts of time...

Question 16

A student earned grades of 84, 78, 84, and 72 on her four regular tests. She earned a grade of_ 78 on the final exam and 86 on her class projects. Her combined homework grade was 87. The
Four regular tests count for 40% of the course grade, the final exam counts for 30%, the project
Counts for 10%, and homework counts for 20%. What is her weighted mean grade? Round to
One decimal place.

Accepted Answer

The weighted mean grade is calculated by multiplying each component by its weight and summing the results: (84+78+84+72)/4 * 0.4 + 78 * 0.3 + 86 * 0.1 + 87 * 0.2 = 81.2.

Question 17

Use the given sample data to find $Q _ { 2 } .$ ._ $\begin{array} { l l l l l l l l } 
49 & 52 & 52 & 52 & 74 & 67 & 55 & 55
\end{array}$

Accepted Answer

The answer of Use the given sample data to find...

Question 18

When finding percentiles, if the locator L is not a whole number, one procedure is to interpolate so that a locator of 23.75, for example, leads to a value that is 3/4 of the way
Between the 23rd and 24th scores. Use this method of interpolation to find P75 for the set of
Test scores below. $$\begin{array} { l l l l l l } 
51 & 54 & 64 & 68 & 72 & 74
\end{array}$$
$\begin{array} { l l l l l } 76 & 83 & 94 & 94 & 99 \end{array}$

Accepted Answer

To find P75 (the 75th percentile) for the given set of 11 scores, we first calculate the locator $L$. Since there are 11 scores, $L = \frac{75}{100} \times (11 + 1) = 9$. This means we need to find a value 3/4 of the way between the 9th and 10th scores. The 9th and 10th scores are 94 and 94, respectively. Interpolating 3/4 of the way between two identical scores still results in the same score, so P75 is 94. However, given the options and the context of interpolation, it seems there might have been a misunderstanding in the calculation or interpretation of the question. Normally, if L is not a whole number, you would interpolate between the two closest ranks. But in this case, with the scores provided and the typical method of finding percentiles, the correct answer based on the usual interpretation of percentile calculation would directly be the value at the locator's position if the scores were different. Given the misunderstanding and the options provided, the correct interpretation based on the expected process of interpolation (and assuming a need to interpolate between different values) would not directly apply to the provided scores as they lead directly to a score of 94 without the need for interpolation. This indicates a discrepancy in the understanding or the provided options.

Question 19

Human body temperatures have a mean of 98.20° F and a standard deviation of 0.62° F. Sally's temperature can be described by z = -1.5. What is her temperature? Round your
Answer to the nearest hundredth.

Accepted Answer

Sally's temperature is found using the formula for a z-score: $Z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. Rearranging to solve for $X$ gives $X = Z\sigma + \mu$. Plugging in Sally's z-score (-1.5), the mean (98.20° F), and the standard deviation (0.62° F), we get $X = -1.5(0.62) + 98.20 = 97.27° F$.

Question 20

Approximately what percentage of apple diameters is greater than the 66th percentile?

Accepted Answer

The 66th percentile means that 66% of the data falls below it, so 34% is greater than it.

Quiz 3: Describing, Exploring, and Comparing Data

A store manager kept track of the number of newspapers sold each week over a seven-week period. The results are shown below. $\begin{array} { l l l l l l l } 81 & 71 & 202 & 113 & 269 & 248 & 242\end{array}$ Find the median number of newspapers sold.

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data. $\begin{array} { l l l l l l } 22.6 & 16.1 & 36.1 & 36.0 & 23.8 & 20.3\end{array}$

If the standard deviation for a set of data is 0, which of the following must be true?_

The prices (in dollars) of 12 electric smoothtop ranges are listed below. Find the range for the given sample data. $\begin{array} { c c c c c c } 865 & 1010 & 655 & 565 & 1465 & 1110 \\710 & 765 & 820 & 1310 & 555 & 1065\end{array}$

If your score on your next statistics test is converted to a z score, which of these z scores would you prefer?

Find the standard deviation for the given sample data. Round your answer to one more_ decimal place than the original data. $\begin{array} { l l l l l l l l l } 18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15\end{array}$

Scores on the SAT test have a mean of 1518 and a standard deviation of 325. Scores on the ACT test have a mean of 21.1 and a standard deviation of 4.8. Which of the following Choices is not true?

Find the variance for the given sample data. Round your answer to one more decimal place_ than the original data. $\begin{array} { l l l l l } 7 & 7 & 2 & 5 & 1\end{array}$

Use the given sample data to find $Q _ { 2 } .$ ._ $\begin{array} { l l l l l l l l } 49 & 52 & 52 & 52 & 74 & 67 & 55 & 55\end{array}$

Human body temperatures have a mean of 98.20° F and a standard deviation of 0.62° F. Sally's temperature can be described by z = -1.5. What is her temperature? Round your Answer to the nearest hundredth.

Approximately what percentage of apple diameters is greater than the 66th percentile?

Quiz 3: Describing, Exploring, and Comparing Data

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data. 22.616.136.136.023.820.3\begin{array} { l l l l l l } 22.6 & 16.1 & 36.1 & 36.0 & 23.8 & 20.3\end{array}22.6​16.1​36.1​36.0​23.8​20.3​

If the standard deviation for a set of data is 0, which of the following must be true?_

If your score on your next statistics test is converted to a z score, which of these z scores would you prefer?

Find the standard deviation for the given sample data. Round your answer to one more_ decimal place than the original data. 181818915510515\begin{array} { l l l l l l l l l } 18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15\end{array}18​18​18​9​15​5​10​5​15​

Scores on the SAT test have a mean of 1518 and a standard deviation of 325. Scores on the ACT test have a mean of 21.1 and a standard deviation of 4.8. Which of the following Choices is not true?

Find the variance for the given sample data. Round your answer to one more decimal place_ than the original data. 77251\begin{array} { l l l l l } 7 & 7 & 2 & 5 & 1\end{array}7​7​2​5​1​

Use the given sample data to find Q2.Q _ { 2 } .Q2​. ._ 4952525274675555\begin{array} { l l l l l l l l } 49 & 52 & 52 & 52 & 74 & 67 & 55 & 55\end{array}49​52​52​52​74​67​55​55​

Human body temperatures have a mean of 98.20° F and a standard deviation of 0.62° F. Sally's temperature can be described by z = -1.5. What is her temperature? Round your Answer to the nearest hundredth.

Approximately what percentage of apple diameters is greater than the 66th percentile?

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in the original data. $\begin{array} { l l l l l l } 22.6 & 16.1 & 36.1 & 36.0 & 23.8 & 20.3\end{array}$

Find the standard deviation for the given sample data. Round your answer to one more_ decimal place than the original data. $\begin{array} { l l l l l l l l l } 18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15\end{array}$

Find the variance for the given sample data. Round your answer to one more decimal place_ than the original data. $\begin{array} { l l l l l } 7 & 7 & 2 & 5 & 1\end{array}$

Use the given sample data to find $Q _ { 2 } .$ ._ $\begin{array} { l l l l l l l l } 49 & 52 & 52 & 52 & 74 & 67 & 55 & 55\end{array}$