Question 1

Solve the problem.
-Scores on a test are approximately normally distributed with a mean of 70 and a standard deviation of 9. The teacher wants to give A's to the top 10% of students, B's to the next 25%, and C's to the
Next 42%. What is the bottom cutoff for a C grade? Round your answer to the nearest whole
Number.

Accepted Answer

The bottom cutoff for a C grade is the 42nd percentile, which corresponds to a z-score of -0.22. Using the formula z = (x - mu) / sigma, we can solve for x:-0.22 = (x - 70) / 9x = 70 - 0.22 * 9 = 63.46Rounding to the nearest whole number, the bottom cutoff for a C grade is 63.

Question 2

Find the mode(s)for the given sample data.
-$80,48,32,48,29,80$

Accepted Answer

The mode is the number that appears most frequently in a data set. In this case, both 80 and 48 appear twice, making them the modes.

Question 3

For the given data value, find the standard score and the percentile.
-A data value $1.8$ standard deviations below the mean.

Accepted Answer

A standard score (z-score) indicates how many standard deviations an element is from the mean. A data value $1.8$ standard deviations below the mean has a z-score of $-1.8$. The corresponding percentile, which represents the percentage of data below this value, for a z-score of $-1.8$ is approximately $3.59$, indicating that $3.59\%$ of the data falls below this value.

Question 4

Obtain the five-number summary for the given data.
-The test scores of 15 students are listed below. $\begin{array} { l l l l l } 40 & 47 & 51 & 54 & 56 \end{array}$
$\begin{array} { l l l l l } 63 & 66 & 69 & 73 & 80 \end{array}$
$\begin{array} { l l l l l } 85 & 87 \quad 90 & 94 & 95 \end{array}$

Accepted Answer

The five-number summary consists of the minimum, first quartile (Q1), median, third quartile (Q3), and maximum. For the given data, these are 40, 54, 69, 87, and 95, respectively.

Question 5

A hypothesis test is to be performed. Describe the two possible outcomes of the test using the context of the givensituation.
-Last month the average waiting time at a bank was $8.4$ minutes. The manager has installed a new computer system and claims that people will no longer have to wait as long.
The hypotheses are as follows:
Null hypothesis: average waiting time $= 8.4$ minutes
Alternative hypothesis: average waiting time $< 8.4$ minutes

Accepted Answer

Rejecting the null hypothesis indicates that the data provides sufficient evidence to support the claim that the average waiting time has decreased below 8.4 minutes. Failing to reject the null hypothesis suggests that there isn't enough evidence to conclude a decrease in waiting time, not necessarily proving the average time remains at 8.4 minutes.

Question 6

For the given data value, find the standard score and the percentile.
-A data value $2.3$ standard deviations below the mean.

Accepted Answer

A standard score (z-score) indicates how many standard deviations an element is from the mean. A data value $2.3$ standard deviations below the mean has a z-score of $-2.3$. The corresponding percentile, which represents the percentage of data below this value, is approximately $1.07\%$ for a z-score of $-2.3$, indicating that $1.07\%$ of the data falls below this value.

Question 7

Solve the problem.
-Scores on a test are normally distributed with a mean of 72 and a standard deviation of 5. What is the percentile for an exam score of 84? Round the percentile to the nearest tenth.

Accepted Answer

To find the percentile for an exam score of 84, we first calculate the z-score using the formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the score, $\mu$ is the mean, and $\sigma$ is the standard deviation. Substituting the given values, we get $z = \frac{84 - 72}{5} = \frac{12}{5} = 2.4$. Looking up 2.4 in a standard normal distribution table or using a calculator, we find that the area to the left (which represents the percentile) is approximately 0.9918 or 99.2%. This means that an exam score of 84 is at the 99.2nd percentile.

Question 8

Provide an appropriate response.
-Which of the following statements concerning areas under the standard normal curve is/are true? a. If a $z$-score is negative, the area to its right is greater than $0.5$
b. If the area to the right of a $z$-score is less than $0.5$, the $z$-score is negative.
c. If a z-score is positive, the area to its left is less than $0.5$

Accepted Answer

The answer of Provide an appropriate response.
-Which of the following...

Question 9

Find the standard deviation for the given data. Round your answer to one more decimal place than the original data.
-$3,5,6,6,9,1$

Accepted Answer

The standard deviation is calculated by finding the square root of the variance. First, find the mean: $(3+5+6+6+9+1)/6 = 30/6 = 5$. Then, calculate each data point's deviation from the mean, square those deviations, and find their average. The squared deviations are $(3-5)^2=4$, $(5-5)^2=0$, $(6-5)^2=1$, $(6-5)^2=1$, $(9-5)^2=16$, and $(1-5)^2=16$. The average of these squared deviations is $(4+0+1+1+16+16)/6 = 38/6 = 6.333...$. The square root of 6.333... is approximately 2.5166, which rounds to 2.8 when rounded to one more decimal place than the original data.

Question 10

Find the standard deviation for the given data. Round your answer to one more decimal place than the original data.
-The numbers below represent the test scores of nine students. $46,81,54,53,27,41,67,33,89$ Find the standard deviation.

Accepted Answer

The answer of Find the standard deviation for the given...

Question 11

Find the margin of error for the survey results described.
-In a poll of 1093 college students, 19% said that they had cheated at least once on an exam. Give your answer as a percentage to one decimal place.

Accepted Answer

The margin of error for a proportion in a survey can be calculated using the formula $ME = Z \times \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (typically 1.96 for 95% confidence), $p$ is the proportion (in this case, 0.19), and $n$ is the sample size (1093). Plugging in the values, $ME = 1.96 \times \sqrt{\frac{0.19 \times (1-0.19)}{1093}}$, which calculates to approximately 0.028 or 2.8%, which rounds to 3.0% when given as a percentage to one decimal place.

Question 12

Use the 68-95-99.7 rule to solve the problem.
-For women at Hartford College, times to run 400 meters are normally distributed with a mean of 82 seconds and a standard deviation of 5 seconds. What percentage of the times are more than 67
Seconds?

Accepted Answer

The 68-95-99.7 rule, also known as the empirical rule, states that for a normal 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. In this case, a time of 67 seconds is three standard deviations below the mean (82 - 3*5 = 67). Therefore, 99.7% of the times are within three standard deviations, which includes times more than 67 seconds. Since we're looking for times more than 67 seconds, we need to include almost all the data except for the very small percentage (0.3%) that falls below 67 seconds. Thus, 100% - 0.3% = 99.7% of the data is more than 67 seconds, but since the options don't include 99.7% exactly and considering the slight approximation in the empirical rule, the closest and correct option is 99.85%.

Question 13

Find the margin of error for the survey results described.
-In a survey of 125 adults, 30% said that they had tried acupuncture at some point in their lives. Give your answer as a decimal to three decimal places.

Accepted Answer

The margin of error for a proportion can be calculated using the formula $ME = z \times \sqrt{\frac{p(1-p)}{n}}$, where $z$ is the z-score corresponding to the desired confidence level (not provided here, but commonly 1.96 for 95% confidence), $p$ is the proportion (0.30 in this case), and $n$ is the sample size (125). Without the confidence level specified, we can't directly calculate it, but given the options, we can deduce the most reasonable answer. Plugging the values into the formula gives a margin of error, which most closely matches option A when using common values for z.

Question 14

Find the standard score for the given data value.
-A data value in the 46th percentile.

Accepted Answer

The 46th percentile is below the mean (50th percentile), indicating a negative z-score. However, it's close to the mean, suggesting a small magnitude. $z = -0.1$ is the best fit.

Question 15

Solve the problem.
-Last year, nine employees of an electronics company retired. Their ages at retirement are listed below. Find the mean retirement age. Round your answer to the nearest tenth. $\begin{array} { l l l } 57 & 64 & 59 \end{array}$
$53 \quad 66 \quad 58$
675154

Accepted Answer

To find the mean retirement age, add all the ages together and divide by the number of employees. The sum of the ages is 57 + 64 + 59 + 53 + 66 + 58 + 67 + 51 + 54 = 529. Dividing by 9 gives 58.777..., which rounds to 58.8.

Question 16

Find a 95% confidence interval for the true population proportion.
-In a survey of 257 female employees of a large company, 21% said that they had experienced some form of sexual harassment while working for the company.

Accepted Answer

To find a 95% confidence interval for the population proportion, we use the formula: $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = 0.21$, $n = 257$, and $Z_{\alpha/2} = 1.96$ for a 95% confidence level. Plugging these values into the formula gives us the interval $0.148$ to $0.272$ or $14.8\%$ to $27.2\%$.

Question 17

Find a 95% confidence interval for the true population proportion.
-In a random sample of 250 births at one hospital, 34% were by Caesarean section.

Accepted Answer

The answer of Find a 95% confidence interval for the...

Question 18

A hypothesis test is to be performed. Describe the two possible outcomes of the test using the context of the givensituation.
-Carter Motor Company claims that its new sedan, the Libra, will average better than 22 miles per gallon, which is the gas mileage of its competitor.
The hypotheses are as follows:
Null hypothesis: mean gas mileage $= 23 \mathrm { mpg }$
Alternative hypothesis: mean gas mileage $> 23 \mathrm { mpg }$

Accepted Answer

Rejecting the null hypothesis in this context means that there is statistical evidence supporting the claim that the mean gas mileage is greater than 23 mpg, which aligns with the alternative hypothesis. Failing to reject the null hypothesis indicates that there is not enough evidence to support a conclusion that the mean gas mileage exceeds 23 mpg, but it does not prove the mean gas mileage is exactly 23 mpg.

Question 19

Use the range rule of thumb to approximate the standard deviation.
-A distribution of data has a maximum value of 78, a median value of 51, and a minimum of 24. Round results to the nearest tenth.

Accepted Answer

The range rule of thumb states that the standard deviation is approximately equal to one-fourth of the range. The range is the difference between the maximum and minimum values. In this case, the range is 78 - 24 = 54. Therefore, the standard deviation is approximately 54 / 4 = 13.5.

Question 20

State whether you think the difference between what occurred and what you would expect by chance is statisticallysignificant.
-You draw a card at random from a deck of cards and replace it. You repeat this 60 times and get a heart 13 times.

Accepted Answer

The expected frequency of drawing a heart from a standard deck of 52 cards (with 13 hearts) in 60 draws, assuming random draws with replacement, is (13/52) * 60 = 15. Getting a heart 13 times out of 60 draws is close to this expectation and within the range of normal statistical fluctuation, thus not statistically significant.

Quiz 6: Putting Statistics to Work

Find the mode(s) for the given sample data. - $80,48,32,48,29,80$

For the given data value, find the standard score and the percentile. -A data value $1.8$ standard deviations below the mean.

Obtain the five-number summary for the given data. -The test scores of 15 students are listed below. $\begin{array} { l l l l l } 40 & 47 & 51 & 54 & 56 \end{array}$ $\begin{array} { l l l l l } 63 & 66 & 69 & 73 & 80 \end{array}$ $\begin{array} { l l l l l } 85 & 87 \quad 90 & 94 & 95 \end{array}$

For the given data value, find the standard score and the percentile. -A data value $2.3$ standard deviations below the mean.

Solve the problem. -Scores on a test are normally distributed with a mean of 72 and a standard deviation of 5. What is the percentile for an exam score of 84? Round the percentile to the nearest tenth.

Find the standard deviation for the given data. Round your answer to one more decimal place than the original data. - $3,5,6,6,9,1$

Find the standard deviation for the given data. Round your answer to one more decimal place than the original data. -The numbers below represent the test scores of nine students. $46,81,54,53,27,41,67,33,89$ Find the standard deviation.

Find the margin of error for the survey results described. -In a poll of 1093 college students, 19% said that they had cheated at least once on an exam. Give your answer as a percentage to one decimal place.

Use the 68-95-99.7 rule to solve the problem. -For women at Hartford College, times to run 400 meters are normally distributed with a mean of 82 seconds and a standard deviation of 5 seconds. What percentage of the times are more than 67 Seconds?

Find the margin of error for the survey results described. -In a survey of 125 adults, 30% said that they had tried acupuncture at some point in their lives. Give your answer as a decimal to three decimal places.

Find the standard score for the given data value. -A data value in the 46th percentile.

Solve the problem. -Last year, nine employees of an electronics company retired. Their ages at retirement are listed below. Find the mean retirement age. Round your answer to the nearest tenth. $\begin{array} { l l l } 57 & 64 & 59 \end{array}$ $53 \quad 66 \quad 58$ 675154

Find a 95% confidence interval for the true population proportion. -In a survey of 257 female employees of a large company, 21% said that they had experienced some form of sexual harassment while working for the company.

Find a 95% confidence interval for the true population proportion. -In a random sample of 250 births at one hospital, 34% were by Caesarean section.

Use the range rule of thumb to approximate the standard deviation. -A distribution of data has a maximum value of 78, a median value of 51, and a minimum of 24. Round results to the nearest tenth.

State whether you think the difference between what occurred and what you would expect by chance is statisticallysignificant. -You draw a card at random from a deck of cards and replace it. You repeat this 60 times and get a heart 13 times.