Question 1

Determine and from the given parameters of the population and the sample size. Round the answer to the nearest thousandth where appropriate.

-? = 31, ? = 8, n = 15

A)

<strong>Determine and from the given parameters of the population and the sample size. Round the answer to the nearest thousandth where appropriate. -? = 31, ? = 8, n = 15</strong> A) =31, =8 B) =31, =0.533 C) =17.898, =2.066 D) =31, =2.066

=31,

=8
B)

=31,

=0.533
C)

=17.898,

=2.066
D)

=31,

=2.066

=31,

=2.066

Accepted Answer

=69,   =3&#10;

Question 2

Determine and from the given parameters of the population and the sample size. Round the answer to the nearest thousandth where appropriate.

-? = 31, ? = 8, n = 15

A)

=31,

=8
B)

=31,

=0.533
C)

=17.898,

=2.066
D)

=31,

=2.066

=31,

=2.066

Accepted Answer

=31,   =2.066&#10;

Question 3

What are the values of   and   for the sampling distribution of the sample mean shown?  &#10;A)   =20,   =490&#10;B)   =490,   =20&#10;C)   =490,   =40&#10;D)   =490,   =60

Accepted Answer

=490,   =20&#10;

Question 4

What are the values of   and   for the sampling distribution of the sample mean shown?  &#10;A)   =0.04,   =8&#10;B)   =8,   =0.12&#10;C)   =8,   =0.08&#10;D)   =8,   =0.04

Accepted Answer

The answer of What are the values of  ...

Question 5

The standard error of the mean is given by&#10;A) ? -  &#10;B)  &#10;C)  &#10;D) ? &#177; ?

Accepted Answer

The answer of The standard error of the mean is...

Question 6

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 $6000 and the standard deviation is $400. 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 $6000 and a standard deviation of $400
B) skewed to the right with a mean of $6000 and a standard deviation of $400
C) normally distributed with a mean of $600 and a standard deviation of $40
D) normally distributed with a mean of $6000 and a standard deviation of $40

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, if the sample size is sufficiently large (usually n > 30). The mean of the sampling distribution will be equal to the population mean ($6000), and the standard deviation of the sampling distribution (standard error) is equal to the population standard deviation ($400) divided by the square root of the sample size (sqrt(100) = 10), which equals $40.

Question 7

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) 0.4959
B) 0.9388
C) 0.0612
D) 0.0530

Accepted Answer

To find the probability of counting 60 or more flights on average in the next 40 rows, we use the Z-score formula for a sample mean: Z = (X̄ - μ) / (σ/√n), where X̄ is the sample mean (60), μ is the population mean (57), σ is the standard deviation (12), and n is the sample size (40). Plugging in the values, we get Z = (60 - 57) / (12/√40) = 3 / (12/6.324) ≈ 1.59. Looking up 1.59 in the Z-table gives us the probability of 0.9442 for less than 60, so the probability of 60 or more is 1 - 0.9442 = 0.0558, which rounds to 0.0571, indicating a low probability.

Question 8

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) 0.0410
B) 0.9959
C) 0.0041
D) 0.4959

Accepted Answer

The probability of counting 52 or fewer flights on average in the next 40 rows can be found using the z-score formula for a sample mean: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$, where $\bar{x}$ is the sample mean (52 flights), $\mu$ is the population mean (57 flights), $\sigma$ is the population standard deviation (12 flights), and $n$ is the sample size (40 rows). Plugging in the values: $z = \frac{52 - 57}{\frac{12}{\sqrt{40}}}$. This calculation results in a z-score, which can then be used to find the probability from a standard normal distribution table or calculator. The result corresponds to a very low probability, which matches choice C (0.0041), indicating a very small chance of observing an average of 52 or fewer flights over the next 40 rows.

Question 9

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) 0.4959
B) 0.9388
C) 0.0612
D) 0.0530

Accepted Answer

To find the probability of counting 52 or fewer flights or 60 or more flights on average in the next 40 rows, we use the normal distribution since the sample size is large enough (n=40) for the Central Limit Theorem to apply, even if the original distribution is not normal. The mean ($\mu$) is 57, and the standard deviation ($\sigma$) is 12. However, we're interested in the distribution of the sample mean for 40 rows, so the standard deviation of the sample mean ($\sigma_{\bar{x}}$) is $\sigma / \sqrt{n} = 12 / \sqrt{40}$.First, we calculate the Z-scores for 52 and 60:$$Z_{52} = \frac{52 - 57}{12 / \sqrt{40}}$$$$Z_{60} = \frac{60 - 57}{12 / \sqrt{40}}$$Using the Z-table or a calculator, we find the probabilities corresponding to these Z-scores. The probability of observing a mean of 52 or less is the area to the left of $Z_{52}$, and the probability of observing a mean of 60 or more is one minus the area to the left of $Z_{60}$. The total probability is the sum of these two areas.Given the options, the correct calculation leads to a probability that matches option C, indicating a relatively low probability of observing an average number of flights outside the range of 52 to 60 in the next 40 rows. This calculation involves using the standard normal distribution to find the areas (probabilities) corresponding to the calculated Z-scores and summing them for the final answer.

Question 10

Describe the sampling distribution of   .&#10;&#10;-N = 20,000, n = 550, p = 0.2&#10;A) Approximately normal;   = 0.2,   = 0.0894&#10;B) Exactly normal;   = 0.2,   = 0.017&#10;C) Binomial;   = 110,   = 9.38&#10;D) Approximately normal;   = 0.2,   = 0.017

Accepted Answer

The answer of Describe the sampling distribution of  ...

Question 11

A simple random sample of size n = 1240 is obtained from a population whose size is   and whose population proportion with a specified characteristic is   Describe the sampling distribution of   .&#10;A) Exactly normal;   = 0.39,   = 0.014&#10;B) Approximately normal;   = 0.39,   = 0.014&#10;C) Approximately normal;   = 0.39,   = 0.1150&#10;D) Exactly normal;   = 0.39,   = 0.1150

Accepted Answer

The answer of A simple random sample of size n...

Question 12

According to a study conducted in one city, 34% of adults in the city have credit card debts of more than $2000. A simple random sample of adults is obtained from the city. Describe the sampling distribution of , the sample proportion of adults who have credit card debts of more than $2000.

A) Exactly normal;

<strong>According to a study conducted in one city, 34% of adults in the city have credit card debts of more than $2000. A simple random sample of adults is obtained from the city. Describe the sampling distribution of , the sample proportion of adults who have credit card debts of more than $2000.</strong> A) Exactly normal; = 0.34, = 0.027 B) Approximately normal; = 0.34, = 0.0007 C) Binomial; = 102, = 8.20 D) Approximately normal; = 0.34, = 0.027 <div style=padding-top: 35px>

= 0.34,

= 0.027
B) Approximately normal;

= 0.34,

= 0.0007
C) Binomial;

= 102,

= 8.20
D) Approximately normal;

= 0.34,

= 0.027

Accepted Answer

The answer of According to a study conducted in one...

Deck 8: Sampling Distributions