Question 1

Provide an appropriate response.
-The standard error of the mean is given by
A) $\frac { \sigma } { \sqrt { n } }$
B) $\mu - \bar { x }$
C) $| \mu - \bar { x } |$
D) $\mu \pm \sigma$

Accepted Answer

The standard error of the mean is calculated as the population standard deviation ($\sigma$) divided by the square root of the sample size ($n$), which is represented by the formula $\frac { \sigma } { \sqrt { n } }$.

Question 2

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-The owner of a computer repair shop has determined that their daily revenue has mean $\$ 7200$ and standard deviation $\$ 1200$. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $\$ 7500$ ?
A) $0.0853$
B) $0.9131$
C) $0.0869$
D) $0.9147$

Accepted Answer

The correct answer is A.The distribution of the sample mean is approximately normal, even if the population distribution is not normal, according to the central limit theorem. The mean of the sample mean is equal to the mean of the population, and the standard deviation of the sample mean is equal to the standard deviation of the population divided by the square root of the sample size. In this case, the mean of the sample mean is $7200, and the standard deviation of the sample mean is $1200 / sqrt(30) = $219.09.The probability that the mean daily revenue for the next 30 days will exceed $7500 is equal to the area under the normal distribution curve to the right of $7500. This area is equal to 1 - P(Z < (7500 - 7200) / 219.09) = 1 - P(Z < 1.37) = 0.0853.

Question 3

Provide an appropriate response.
-The amount of time it takes a student to walk from her home to class has a skewed right distribution with a mean of 17 minutes and a standard deviation of $1.2$ minutes. If data were collected from 50 randomly selected walks, describe the sampling distribution of $\bar { x }$, the sample mean time.

Accepted Answer

By the Central Limit Theorem, the sampling distribution of $\bar { x }$ is approximately normal with $\mu _ { x } ^ { - } = \mu = 17$ minutes and $\sigma _ { x } ^ { - } = \frac { \sigma } { \sqrt { n } } = \frac { 1.2 } { \sqrt { 50 } } = 0.1697$ minutes.

Question 4

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-The owner of a computer repair shop has determined that their daily revenue has mean $\$ 7200$ and standard deviation $\$ 1200$. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $\$ 7000$ ?
A) $0.1814$
B) $0.8186$
C) $0.5675$
D) $0.4325$

Accepted Answer

The distribution of the sample mean will be approximately normal, even if the population distribution is not normal, according to the Central Limit Theorem. The mean of the sample mean will be the same as the mean of the population, which is $7200. The standard deviation of the sample mean will be the standard deviation of the population divided by the square root of the sample size, which is $1200 / sqrt(30) = $219.09.The probability that the mean daily revenue for the next 30 days will exceed $7000 is the area under the normal distribution curve to the right of $7000. This area can be calculated using a standard normal distribution table or a calculator. The z-score for $7000 is (7000 - 7200) / 219.09 = -0.91. The area to the right of -0.91 is 0.1814.Therefore, the probability that the mean daily revenue for the next 30 days will exceed $7000 is 0.1814.

Question 5

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-A farmer was interested in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows he figures the mean number of flights to be 57 with a standard deviation of 12 . What is the probability of the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor?
A) $0.0571$
B) $0.4429$
C) $0.9429$
D) $0.5710$

Accepted Answer

The Central Limit Theorem (CLT) states that the sampling distribution of the sample mean will be approximately normally distributed, regardless of the population distribution, if the sample size is large enough (usually n > 30). Here, the sample size is 40, which is greater than 30, so the CLT applies. To find the probability of observing a mean of 60 or more, we calculate the z-score for 60 using the formula $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is 60, $\mu$ is 57, $\sigma$ is 12, and $n$ is 40. The z-score tells us how many standard deviations 60 is from the mean. We then look up this z-score in a standard normal distribution table or use a calculator to find the corresponding probability. The correct answer reflects the probability of observing a sample mean of 60 or more under these conditions.

Question 6

Provide an appropriate response.
-According to the law of large numbers, as more observations are added to the sample, the difference between the sample mean and the population mean
A) Tends to become smaller
B) Tends to become larger
C) Remains about the same
D) Is inversely affected by the data added

Accepted Answer

The law of large numbers states that as the size of the sample increases, the sample mean will get closer to the population mean, hence the difference between them tends to become smaller.

Question 7

Provide an appropriate response.
-Suppose a population has a mean of 7 for some characteristic of interest and a standard deviation of $9.6$. A sample is drawn from this population of size 64. What is the standard error of the mean?
A) $1.2$
B) $0.7$
C) $0.15$
D) $3.3$

Accepted Answer

The standard error of the mean is calculated as the population standard deviation divided by the square root of the sample size: $ \frac{9.6}{\sqrt{64}} = \frac{9.6}{8} = 1.2 $.

Question 8

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-A farmer was interested in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows he figures the mean number of flights to be 57 with a standard deviation of 12 . What is the probability of the farmer will count 52 or fewer flights or 60 or more flights on average in the next 40 rows down which he drives his tractor?
A) $0.0612$
B) $0.4959$
C) $0.9388$
D) $0.0530$

Accepted Answer

The Central Limit Theorem (CLT) applies here, stating that the distribution of the sample mean will approach a normal distribution as the sample size increases, regardless of the population's initial distribution. Given a sample size of 40, which is sufficiently large, we can use the normal distribution to calculate probabilities. The probability of observing a mean outside the range of 52 to 60 is found by calculating the Z-scores for these values and using the standard normal distribution. The correct probability is $0.0612$, indicating a relatively low chance of observing such extreme average counts in the next 40 rows.

Question 9

Provide an appropriate response.
-The amount of corn chips dispensed into a 12 -ounce bag by the dispensing machine has been identified at possessing a normal distribution with a mean of $12.5$ ounces and a standard deviation of $0.1$ ounce. Suppose 400 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 400 bags exceeded $12.6$ ounces.
A) approximately 0
B) $0.1915$
C) $0.3085$
D) $0.6915$

Accepted Answer

Given the large sample size (400 bags) and the small standard deviation (0.1 ounce), the probability of the sample mean exceeding 12.6 ounces is extremely low, given the distribution's mean is 12.5 ounces. This is because the standard error of the mean decreases as the sample size increases, making the distribution of sample means tighter around the population mean.

Question 10

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-The owner of a computer repair shop has determined that their daily revenue has mean $\$ 7200$ and standard deviation $\$ 1200$. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will be between $\$ 7000$ and $\$ 7500$ ?
A) $0.7333$
B) $0.2667$
C) $0.9147$
D) $0.8186$

Accepted Answer

The central limit theorem states that the distribution of sample means will be approximately normal, regardless of the shape of the population distribution, if the sample size is large enough. In this case, the sample size is 30, which is large enough to invoke the central limit theorem. Therefore, the distribution of the sample mean will be approximately normal with mean $\$ 7200$ and standard deviation $\$ 1200/\sqrt{30} = \$219.09$.To find the probability that the mean daily revenue for the next 30 days will be between $\$ 7000$ and $\$ 7500$, we can use the standard normal distribution (z-distribution). The z-score for $\$ 7000$ is $(7000 - 7200)/219.09 = -0.91$, and the z-score for $\$ 7500$ is $(7500 - 7200)/219.09 = 1.37$. Using a standard normal distribution table, we find that the probability of a z-score between -0.91 and 1.37 is 0.7333.Therefore, the probability that the mean daily revenue for the next 30 days will be between $\$ 7000$ and $\$ 7500$ is 0.7333.

Question 11

Provide an appropriate response.
-Assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of 9.6. If 144 people are randomly selected, find the probability that their mean blood pressure will be less than $124 .$
A) $0.9938$
B) $0.0062$
C) $0.9998$
D) $0.8615$

Accepted Answer

To find the probability that the mean blood pressure of 144 randomly selected people is less than 124, we use the formula for the standard error of the mean: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Here, $\sigma = 9.6$ and $n = 144$, so $\sigma_{\bar{x}} = \frac{9.6}{\sqrt{144}} = \frac{9.6}{12} = 0.8$. We then calculate the z-score for a mean of 124: $z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{124 - 122}{0.8} = 2.5$. Looking up the z-score of 2.5 in the standard normal distribution table gives a probability of 0.9938 that the mean blood pressure of 144 people will be less than 124.

Question 12

Provide an appropriate response.
-Assume that the heights of men are normally distributed with a mean of $67.9$ inches and a standard deviation of $2.1$ inches. If 36 men are randomly selected, find the probability that they have a mean height greater than $68.9$ inches.
A) $0.0021$
B) $0.0210$
C) $0.9005$
D) $0.9979$

Accepted Answer

The standard error of the mean is $ \frac{2.1}{\sqrt{36}} = 0.35 $. The z-score for a mean height of 68.9 inches is $ \frac{68.9 - 67.9}{0.35} = 2.857 $. Using the standard normal distribution table, the probability of a z-score greater than 2.857 is approximately 0.0021.

Question 13

Provide an appropriate response.
-One year, professional sports players salaries averaged $\$ 1.6$ million with a standard deviation of $\$ 0.7$ million. Suppose a sample of 100 major league players was taken. Find the approximate probability that the average salary of the 100 players exceeded $\$ 1.1$ million.
A) approximately 1
B) approximately 0
C) $0.2357$
D) $0.7357$

Accepted Answer

Given the large sample size (n=100), we can use the Central Limit Theorem to approximate the distribution of the sample mean. The mean of the sample means is \$1.6 million, and the standard deviation of the sample means (standard error) is $\frac{\$0.7\ million}{\sqrt{100}} = \$0.07\ million$. To find the probability that the average salary exceeds \$1.1 million, we calculate the z-score for \$1.1 million and find the corresponding probability. However, since \$1.1 million is significantly below the mean (\$1.6 million), the probability that the sample mean exceeds \$1.1 million is very high, approaching 1.

Question 14

Provide an appropriate response.
-The average score of all golfers for a particular course has a mean of 66 and a standard deviation of 5 . Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded $67 .$
A) $0.0228$
B) $0.1293$
C) $0.4772$
D) $0.3707$

Accepted Answer

To find the probability that the average score of the 100 golfers exceeded 67, we use the formula for the z-score: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the values, we get $z = \frac{67 - 66}{5/\sqrt{100}} = \frac{1}{0.5} = 2$. Looking up 2 in the z-table gives us the area to the left, which is 0.9772. To find the probability of exceeding 67, we subtract this from 1, giving $1 - 0.9772 = 0.0228$.

Question 15

Provide an appropriate response.
-A random sample of size $\mathrm { n }$ is to be drawn from a population with $\mu = 500$ and $\sigma = 100$. What size sample would be necessary in order to ensure a standard error of 25 ?

Accepted Answer

The answer of Provide an appropriate response.
-A random sample of...

Question 16

Describe the Distribution of the Sample Mean-Samples from a Population That Is Not Normal
-A farmer was interest in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away. After several rows he figures the mean number of flights to be 57 with a standard deviation of 12 . What is the probability of the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor?
A) $0.0041$
B) $0.4959$
C) $0.9959$
D) $0.0410$

Accepted Answer

The Central Limit Theorem states that the distribution of the sample mean will be approximately normal if the sample size is large enough (usually n > 30 is considered sufficient). Here, the sample size is 40, which is large enough. The standard error of the mean is $ \frac{12}{\sqrt{40}} \approx 1.897 $. The z-score for 52 is $ \frac{52 - 57}{1.897} \approx -2.637 $. Using the standard normal distribution table, the probability of a z-score being less than -2.637 is approximately 0.0041.

Question 17

Provide an appropriate response.
-Furnace repair bills are normally distributed with a mean of 271 dollars and a standard deviation of 15 dollars. If 36 of these repair bills are randomly selected, find the probability that they have a mean cost between 271 dollars and 273 dollars.
A) $0.2881$
B) $0.7881$
C) $0.2119$
D) $0.5517$

Accepted Answer

Using the Central Limit Theorem, we calculate the probability of the sample mean being between 271 and 273 dollars. The z-score for 271 is (271-271)/15/√36 = 0, and for 273 it is (273-271)/15/√36 = 2/15*6 = 0.8. Looking up these z-scores in the standard normal distribution table, the probability of being between these two values is approximately 0.2881.

Question 18

Provide an appropriate response.
-Suppose a population has a mean weight of 180 pounds and a standard deviation of 25 pounds. A sample of 100 items is drawn from this population. What is the standard error of the mean?
A) $2.5$
B) $1.8$
C) $18.0$
D) $0.25$

Accepted Answer

The standard error of the mean is calculated as the population standard deviation divided by the square root of the sample size. Here, it is $25 / \sqrt{100} = 25 / 10 = 2.5$.

Question 19

Provide an appropriate response.
-The number of violent crimes committed in a day possesses a distribution with a mean of 3.8 crimes per day and a standard deviation of 2 crimes per day. A random sample of 100 days was observed, and the mean number of crimes for the sample was calculated. Describe the sampling distribution of the sample mean.
A) approximately normal with mean = 3.8 and standard deviation = 0.2
B) approximately normal with mean = 3.8 and standard deviation = 2
C) shape unknown with mean = 3.8 and standard deviation = 2
D) shape unknown with mean = 3.8 and standard deviation = 0.2

Accepted Answer

The sampling distribution of the sample mean will be approximately normal due to the Central Limit Theorem, with the same mean as the population (3.8) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (2/√100 = 0.2).

Question 20

Provide an appropriate response.
-The amount of money collected by a snack bar at a large university has been recorded daily for the past five years. Records indicate that the mean daily amount collected is $1950 and the standard deviation is $500. The distribution is skewed to the right due to several high volume days (including football game days). Suppose that 100 days were randomly selected from the five years and the average amount collected from those days was recorded. Which of the following describes the sampling distribution of the sample mean
A) normally distributed with a mean of $1950 and a standard deviation of $50
B) normally distributed with a mean of $1950 and a standard deviation of $500
C) normally distributed with a mean of $195 and a standard deviation of $50
D) skewed to the right with a mean of $1950 and a standard deviation of $500

Accepted Answer

The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normally distributed regardless of the shape of the population distribution, as long as the sample size is sufficiently large (usually n ≥ 30). The mean of the sampling distribution will be equal to the population mean ($1950), and the standard deviation of the sampling distribution (standard error) is equal to the population standard deviation divided by the square root of the sample size, which is $500/√100 = $50.

Quiz 8: Sampling Distributions

Provide an appropriate response. -The standard error of the mean is given by

Provide an appropriate response. -According to the law of large numbers, as more observations are added to the sample, the difference between the sample mean and the population mean

Provide an appropriate response. -Suppose a population has a mean of 7 for some characteristic of interest and a standard deviation of $9.6$ . A sample is drawn from this population of size 64. What is the standard error of the mean?

Provide an appropriate response. -Assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of 9.6. If 144 people are randomly selected, find the probability that their mean blood pressure will be less than $124 .$

Provide an appropriate response. -Assume that the heights of men are normally distributed with a mean of $67.9$ inches and a standard deviation of $2.1$ inches. If 36 men are randomly selected, find the probability that they have a mean height greater than $68.9$ inches.

Provide an appropriate response. -The average score of all golfers for a particular course has a mean of 66 and a standard deviation of 5 . Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded $67 .$

Provide an appropriate response. -A random sample of size $\mathrm { n }$ is to be drawn from a population with $\mu = 500$ and $\sigma = 100$ . What size sample would be necessary in order to ensure a standard error of 25 ?

Provide an appropriate response. -Furnace repair bills are normally distributed with a mean of 271 dollars and a standard deviation of 15 dollars. If 36 of these repair bills are randomly selected, find the probability that they have a mean cost between 271 dollars and 273 dollars.

Provide an appropriate response. -Suppose a population has a mean weight of 180 pounds and a standard deviation of 25 pounds. A sample of 100 items is drawn from this population. What is the standard error of the mean?

Quiz 8: Sampling Distributions

Provide an appropriate response. -The standard error of the mean is given by

Provide an appropriate response. -According to the law of large numbers, as more observations are added to the sample, the difference between the sample mean and the population mean

Provide an appropriate response. -Suppose a population has a mean of 7 for some characteristic of interest and a standard deviation of 9.69.69.6 . A sample is drawn from this population of size 64. What is the standard error of the mean?

Provide an appropriate response. -Assume that blood pressure readings are normally distributed with a mean of 122 and a standard deviation of 9.6. If 144 people are randomly selected, find the probability that their mean blood pressure will be less than 124.124 .124.

Provide an appropriate response. -Assume that the heights of men are normally distributed with a mean of 67.967.967.9 inches and a standard deviation of 2.12.12.1 inches. If 36 men are randomly selected, find the probability that they have a mean height greater than 68.968.968.9 inches.

Provide an appropriate response. -The average score of all golfers for a particular course has a mean of 66 and a standard deviation of 5 . Suppose 100 golfers played the course today. Find the probability that the average score of the 100 golfers exceeded 67.67 .67.

Provide an appropriate response. -A random sample of size n\mathrm { n }n is to be drawn from a population with μ=500\mu = 500μ=500 and σ=100\sigma = 100σ=100 . What size sample would be necessary in order to ensure a standard error of 25 ?