Question 1

Your friend took an introductory statistics class last year and learned all about density curves. He draws a smooth curve with a long left tail. For such a curve, which of the following is true for the mean and median?

Accepted Answer

For a distribution with a long left tail (negatively skewed), the mean is less than the median because the long tail pulls the mean toward the lower values.

Question 2

Your friend took an introductory statistics class last year and learned all about density curves. She tells you that two requirements of a density curve are:

Accepted Answer

Density curves must always be above the horizontal axis, and the total area under the curve must equal one. This represents the entire probability distribution, where the total probability is 1.

Question 3

The duration (in days) of human pregnancies follows approximately the N(266,16) distribution. What proportion of pregnancy durations is more than 270 days?

Accepted Answer

We need to find P(X > 270), where X follows N(266,16).
Using the standard normal distribution, we can calculate the z-score as:
z = (270-266)/4 = 1
P(Z > 1) = 0.159
Therefore, the proportion of pregnancy durations that is more than 270 days is:
1 - P(Z < 1) = 1 - 0.841 = 0.159
So, the answer is B)0.401.

Question 4

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The median of the distribution is:

Accepted Answer

Since the distribution is known to be normal and we have the mean and standard deviation, we can use the formula:
Median = Mean = F1F1F1S1F1F1F10 = 150 minutes
Therefore, the median of the distribution is 150 minutes, which is answer choice B.

Question 5

The graph below shows a distribution that is called a chi-squared distribution.  For this distribution, the median:

Accepted Answer

The chi-squared distribution is typically skewed to the right, meaning that the median is smaller than the mean. Therefore, the best choice is B.

Question 6

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. Thus, 99.7% of running times are approximately between:

Accepted Answer

The 99.7% range for a normal distribution is defined by the mean plus or minus three standard deviations. Here, with a mean of 93 seconds and a standard deviation of 16 seconds, the range is 93 ± (3*16) = 93 ± 48, which gives us 45 seconds to 141 seconds.

Question 7

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. What would a man's running time have to be to put him in the top 1% of runners?

Accepted Answer

To find the time that puts a runner in the top 1%, we need to find the z-score that corresponds to the 99th percentile of a standard normal distribution, which is approximately 2.33. Using the formula: time = mean + (z-score * standard deviation), we get: 93 + (2.33 * 16) = 55.8 seconds.

Question 8

Suppose you received a score of 91 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. What score do you need on exam 2 to do equally well, if the mean is 60 and the standard deviation is 12?

Accepted Answer

To do equally well relative to the other scores, you need to achieve the same z-score. The z-score for exam 1 is (91 - 79) / 8 = 1.5. To achieve the same z-score on exam 2, where the mean is 60 and the standard deviation is 12, you solve for the score: 1.5 = (X - 60) / 12. Solving for X gives X = 1.5 * 12 + 60 = 18 + 60 = 78.

Question 9

Suppose you received a score of 95 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. If your score on exam 2 is 90 out of 100, and the mean was 60 with a standard deviation of 15, then you did:

Accepted Answer

The answer of Suppose you received a score of 95...

Question 10

The typical college freshman spends an average of $\mu$= 150 minutes per day, with a standard deviation of $\sigma$= 50 minutes, on social media. The distribution of time on social media is known to be Normal. The proportion of students spending at least 2 hours on social media equals:

Accepted Answer

To find the proportion of students spending at least 2 hours (120 minutes) on social media, we calculate the z-score: z = (120 - 150) / 50 = -0.6. Using the standard normal distribution table, the proportion of students spending less than 120 minutes is approximately 0.274. Therefore, the proportion spending at least 120 minutes is 1 - 0.274 = 0.726.

Question 11

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of$\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The third quartile is:

Accepted Answer

The third quartile is the 75th percentile which can be found using the Z-score formula:
Z = (x - μ)/σ 
where x is the value at the 75th percentile, μ is the mean, and σ is the standard deviation.
Using the values given, we have:
Z = (x - 150) / 50
From a standard Normal distribution table, the Z-score for the 75th percentile is approximately 0.674.
Therefore, 0.674 = (x - 150) / 50
Solving for x, we get:
x = 0.674 * 50 + 150 = 183.72
So the answer is A) 183.72 minutes.

Question 12

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percent of the time can you expect to wait longer than a full (8-hour) business day for a resolution?

Accepted Answer

To find the percentage of time you can expect to wait longer than 8 hours, you calculate the Z-score for 8 hours using the formula Z = (X - μ) / σ, where X is 8 hours, μ is the mean (4.3 hours), and σ is the standard deviation (3.1 hours). The Z-score is (8 - 4.3) / 3.1 = 1.19. Looking up this Z-score in a standard normal distribution table or using a calculator, you find that the area to the left of this Z-score is approximately 88.3%, meaning the area to the right (the percentage of time you wait longer than 8 hours) is 100% - 88.3% = 11.7%.

Question 13

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percentage of the tickets are resolved in less than half an hour?

Accepted Answer

We need to find the percentage of tickets resolved in less than half an hour. 
Using the Z-score formula, 
Z = (X - μ) / σ
where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation. 
Plugging in the values, we get 
Z = (0.5 - 4.3) / 3.1 = -1.23 
Using a standard normal distribution table or calculator, we can find that the percentage of values to the left of Z = -1.23 is approximately 11.0%. Therefore, choice A is the best answer.

Question 14

You recently took a statistics exam in a large class. The instructor tells the class that the scores were Normally distributed, with a mean of 72 (out of 100) and a standard deviation of 12. Your score was 90. Your friend had a time conflict and took a course with another instructor. Your friend had a score of 75 on a test, with a mean of 60 and a standard deviation of 10. What can you conclude?

Accepted Answer

The answer of You recently took a statistics exam in...

Question 15

You recently took a statistics exam in a large class with n = 500 students. The instructor tells the class that the scores were Normally distributed, with a mean of 72 (out of 100) and a standard deviation of 8, but when you talk to other students in the class, you find out that more than 30 students have scores below 45. That violates which rule for the Normal distribution?

Accepted Answer

The 68-95-99.7 rule, also known as the empirical rule, states that for a normal distribution, about 68% of the data falls within one standard deviation of the mean, about 95% falls within two standard deviations, and about 99.7% falls within three standard deviations. Given the mean score is 72 and the standard deviation is 8, a score of 45 is more than three standard deviations away from the mean (72 - 3*8 = 48). Therefore, having more than 30 students score below 45 is highly unlikely in a normal distribution, indicating a violation of the 68-95-99.7 rule.

Question 16

You recently took a statistics exam in a large class. The instructor tells the class that the scores were Normally distributed, with a mean of 72 (out of 100) and a standard deviation of 12. Your score was 96. You would like to know what proportion of students did better than you did. Unfortunately, your calculator batteries are dead and you must now rely on Normal tables. You need to find which of the following?

Accepted Answer

The answer of You recently took a statistics exam in...

Question 17

A cappuccino vending machine is designed to dispense an average of $\mu$ oz per cup. If the ounces per fill are Normally distributed, with a standard deviation of 0.4 oz, at what value should $\mu$ be set so that 6 oz cups will overflow only 2% of the time?

Accepted Answer

The answer of A cappuccino vending machine is designed to...

Quiz 4: The Normal Distributions

Your friend took an introductory statistics class last year and learned all about density curves. He draws a smooth curve with a long left tail. For such a curve, which of the following is true for the mean and median?

Your friend took an introductory statistics class last year and learned all about density curves. She tells you that two requirements of a density curve are:

The duration (in days) of human pregnancies follows approximately the N(266,16) distribution. What proportion of pregnancy durations is more than 270 days?

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The median of the distribution is:

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. Thus, 99.7% of running times are approximately between:

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. What would a man's running time have to be to put him in the top 1% of runners?

Suppose you received a score of 91 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. What score do you need on exam 2 to do equally well, if the mean is 60 and the standard deviation is 12?

Suppose you received a score of 95 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. If your score on exam 2 is 90 out of 100, and the mean was 60 with a standard deviation of 15, then you did:

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The proportion of students spending at least 2 hours on social media equals:

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The third quartile is:

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percent of the time can you expect to wait longer than a full (8-hour) business day for a resolution?

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percentage of the tickets are resolved in less than half an hour?

A cappuccino vending machine is designed to dispense an average of $\mu$ oz per cup. If the ounces per fill are Normally distributed, with a standard deviation of 0.4 oz, at what value should $\mu$ be set so that 6 oz cups will overflow only 2% of the time?

Quiz 4: The Normal Distributions

Your friend took an introductory statistics class last year and learned all about density curves. He draws a smooth curve with a long left tail. For such a curve, which of the following is true for the mean and median?

Your friend took an introductory statistics class last year and learned all about density curves. She tells you that two requirements of a density curve are:

The duration (in days) of human pregnancies follows approximately the N(266,16) distribution. What proportion of pregnancy durations is more than 270 days?

The typical college freshman spends an average of μ\muμ = 150 minutes per day, with a standard deviation of σ\sigmaσ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The median of the distribution is:

The graph below shows a distribution that is called a chi-squared distribution. For this distribution, the median:

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. Thus, 99.7% of running times are approximately between:

Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age, with a mean of 93 seconds and a standard deviation of 16 seconds. What would a man's running time have to be to put him in the top 1% of runners?

Suppose you received a score of 91 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. What score do you need on exam 2 to do equally well, if the mean is 60 and the standard deviation is 12?

Suppose you received a score of 95 out of 100 on exam 1. The mean was 79 and the standard deviation was 8. If your score on exam 2 is 90 out of 100, and the mean was 60 with a standard deviation of 15, then you did:

The typical college freshman spends an average of μ\muμ = 150 minutes per day, with a standard deviation of σ\sigmaσ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The proportion of students spending at least 2 hours on social media equals:

The typical college freshman spends an average of μ\muμ = 150 minutes per day, with a standard deviation of σ\sigmaσ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The third quartile is:

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percent of the time can you expect to wait longer than a full (8-hour) business day for a resolution?

The average time taken for your Internet service provider to remotely resolve a trouble ticket has a Normal distribution, with a mean of 4.3 hours and a standard deviation of 3.1 hours. What percentage of the tickets are resolved in less than half an hour?

A cappuccino vending machine is designed to dispense an average of μ\muμ oz per cup. If the ounces per fill are Normally distributed, with a standard deviation of 0.4 oz, at what value should μ\muμ be set so that 6 oz cups will overflow only 2% of the time?

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The median of the distribution is:

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The proportion of students spending at least 2 hours on social media equals:

The typical college freshman spends an average of $\mu$ = 150 minutes per day, with a standard deviation of $\sigma$ = 50 minutes, on social media. The distribution of time on social media is known to be Normal. The third quartile is:

A cappuccino vending machine is designed to dispense an average of $\mu$ oz per cup. If the ounces per fill are Normally distributed, with a standard deviation of 0.4 oz, at what value should $\mu$ be set so that 6 oz cups will overflow only 2% of the time?