Question 1

Suppose that,on average,electricians earn approximately µ = $54,000 per year in the United States.Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is σ = $12,000.What is the probability that the average salary of four randomly selected electricians is less than $50,000?
A) 0.2514
B) 0.3707
C) 0.6293
D) 0.7486

Accepted Answer

The standard error for the sample mean is σ/√n = $12,000/√4 = $6,000. The z-score for $50,000 is (50,000 - 54,000) / 6,000 = -0.67. Using the standard normal distribution table, the probability of a z-score less than -0.67 is approximately 0.2514.

Question 2

Over the entire six years that students attend an Ohio elementary school,they are absent,on average,28 days due to influenza.Assume that the standard deviation over this time period is σ = 9 days.Upon graduation from elementary school,a random sample of 36 students is taken and asked how many days of school they missed due to influenza.The probability that the sample mean is between 25 and 30 school days is _______.
A) 0.0228
B) 0.0918
C) 0.8854
D) 0.9082

Accepted Answer

We are given the following information:
- Population mean (μ) = 28 days
- Standard deviation (σ) = 9 days
- Sample size (n) = 36
We need to find the probability that the sample mean (x̄) is between 25 and 30 school days.
We can use the central limit theorem to approximate the distribution of sample means as a normal distribution with mean (μx̄) = μ = 28 days and standard deviation (σx̄) = σ/√n = 9/√36 = 1.5 days.

P(25 ≤ x̄ ≤ 30) can be standardized to Z-score

z1= (25-28)/1.5 = -2

z2= (30-28)/1.5 = 1.33

The probability of z= -2 upto z = 1.33 will give the probability that the sample mean is between 25 and 30 days

Using standard normal table, we can find:

P(-2 ≤ Z ≤ 1.33) = P(Z ≤ 1.33) - P(Z ≤ -2)
= 0.9082 - 0.0228
= 0.8854

Therefore, the probability that the sample mean is between 25 and 30 school days is 0.8854. The best choice is (C).

Question 3

Susan has been on a bowling team for 14 years.After examining all of her scores over that period of time,she finds that they follow a normal distribution.Her average score is 225,with a standard deviation of 13.If during a typical month Susan bowls 64 games,what is the probability that her average score in this month is above 227?
A) 0.1093
B) 0.4404
C) 0.5596
D) 0.8907

Accepted Answer

To solve this problem, we need to first calculate the standard error of the mean (SEM) which is given by:
SEM = standard deviation / square root of sample size
SEM = 13 / square root of 64
SEM = 1.625
Next, we need to convert the given average score of 227 into a z-score using the formula:
z = (x - mu) / SEM
z = (227 - 225) / 1.625
z = 1.23
Using a standard normal distribution table, we can find the probability of getting a z-score of 1.23 or higher, which is 0.1093. Therefore, the probability that Susan's average score in this month is above 227 is 0.1093 or approximately 11%. Option A is the best choice.

Question 4

Suppose that,on average,electricians earn approximately µ = $54,000 per year in the United States.Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is σ = $12,000.Given a sample of four electricians,what is the standard deviation for the sampling distribution of the sample mean?
A) 6,000
B) 12,000
C) 36,000
D) 54,000

Accepted Answer

The standard deviation of the sampling distribution of the sample mean (σx̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n). Here, σ = $12,000 and n = 4, so σx̄ = $12,000 / √4 = $12,000 / 2 = $6,000.

Question 5

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is less than 15 is _______.
A) 0.0228
B) 0.3707
C) 0.6293
D) 0.9772

Accepted Answer

The sample mean ($\bar{x}$) distribution is normally distributed (by the Central Limit Theorem) with mean $\mu = 17$ and standard error $SE = \frac{\sigma}{\sqrt{n}} = \frac{6}{\sqrt{36}} = 1$. To find the probability that the sample mean is less than 15, we calculate the Z-score: $Z = \frac{15 - 17}{1} = -2$. Looking up -2 in the Z-table gives us a probability of 0.0228.

Question 6

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years.After examining his records from the past 10 years,he finds that the scores have a mean of 76 and a standard deviation of 12.What is the probability Professor Elderman's class of 36 has a class average below 78?
A) 0.1587
B) 0.5675
C) 0.8413
D) Cannot be determined.

Accepted Answer

To find the probability that the class average is below 78, we use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is large enough (n ≥ 30 is a common rule of thumb), regardless of the shape of the population distribution. Given that the sample size is 36 (which is greater than 30), we can proceed with this assumption.The mean of the sample means (μx̄) is equal to the population mean (μ), which is 76. The standard deviation of the sample means (σx̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n), which is 12 / sqrt(36) = 12 / 6 = 2.We are looking for the probability that the sample mean is less than 78. We first calculate the z-score for a sample mean of 78:z = (X - μx̄) / σx̄ = (78 - 76) / 2 = 2 / 2 = 1.Looking up the z-score of 1 in the standard normal distribution table, or using a calculator, we find that the area to the left of z = 1 (which represents the probability that the sample mean is less than 78) is approximately 0.8413. This means there is an 84.13% chance that the class average will be below 78.

Question 7

The central limit theorem states that,for any distribution,as n gets larger,the sampling distribution of the sample mean _______.
A) becomes larger
B) becomes smaller
C) is closer to a normal distribution
D) is closer to the standard deviation

Accepted Answer

The central limit theorem states that, regardless of the distribution of the population, as the sample size increases, the sampling distribution of the sample mean will approach a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. Therefore, option C is the correct choice.

Question 8

Over the entire six years that students attend an Ohio elementary school,they are absent,on average,28 days due to influenza.Assume that the standard deviation over this time period is σ = 9 days.Upon graduation from elementary school,a random sample of 36 students is taken and asked how many days of school they missed due to influenza.The probability that the sample mean is less than 30 school days is _______.
A) 0.0918
B) 0.4129
C) 0.5871
D) 0.9082

Accepted Answer

The sample mean follows a normal distribution with mean 28 and standard deviation 9/sqrt(36) = 1.5. The z-score for 30 is (30-28)/1.5 = 1.33. The probability of a z-score less than 1.33 is approximately 0.9082.

Question 9

Susan has been on a bowling team for 14 years.After examining all of her scores over that period of time,she finds that they follow a normal distribution.Her average score is 225,with a standard deviation of 13.If during a typical week Susan bowls 16 games,what is the probability that her average score for the week is between 220 and 228?
A) 0.0618
B) 0.2390
C) 0.7594
D) 0.8212

Accepted Answer

First, we need to find the z-scores for both the lower and upper bounds:

Lower z-score = (220 - 225) / (13 / sqrt(16)) = -2.06
Upper z-score = (228 - 225) / (13 / sqrt(16)) = 1.23

Next, we need to find the area between these two z-scores using a standard normal distribution table or calculator. The area between -2.06 and 1.23 is approximately 0.7594.

Therefore, the probability that Susan's average score for the week is between 220 and 228 is 0.7594 or approximately 76%.

Question 10

Susan has been on a bowling team for 14 years.After examining all of her scores over that period of time,she finds that they follow a normal distribution.Her average score is 225,with a standard deviation of 13.If during a typical week Susan bowls 16 games,what is the probability that her average score is more than 230?
A) 0.0618
B) 0.3520
C) 0.6480
D) 0.9382

Accepted Answer

The sampling distribution of the sample mean will also be normally distributed with mean 225 and standard deviation (standard error) of 13/sqrt(16) = 3.25. We calculate the z-score for 230: (230-225)/3.25 = 1.54. Using the standard normal distribution table, the probability of a z-score greater than 1.54 is approximately 0.0618.

Question 11

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years.After examining his records from the past 10 years,he finds that the scores have a mean of 76 and a standard deviation of 12.What is the probability that a class of 36 students will have an average greater than 70 on Professor Elderman's final exam?
A) 0.0014
B) 0.3085
C) 0.6915
D) 0.9986

Accepted Answer

To find the probability that a class of 36 students will have an average score greater than 70, 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 = 12 $ and $ n = 36 $, so $ \sigma_{\bar{x}} = \frac{12}{\sqrt{36}} = 2 $. We then calculate the z-score for a mean of 70: $ z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} = \frac{70 - 76}{2} = -3 $. Using the z-table, a z-score of -3 corresponds to a probability of 0.00135, which is the probability of having a mean of 70 or lower. Therefore, the probability of having a mean greater than 70 is 1 - 0.00135 = 0.99865, which rounds to 0.9986.

Question 12

According to the central limit theorem,the distribution of the sample means is normal if _______.
A) the underlying population is normal
B) the sample size n ≥ 30
C) the standard deviation of the population is known
D) both the underlying population is normal and the sample size n ≥ 30 are correct

Accepted Answer

The central limit theorem states that the distribution of the sample means approaches a normal distribution as the sample size increases, typically considered to be n ≥ 30, regardless of the shape of the population distribution.

Question 13

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years.After examining his records from the past 10 years,he finds that the scores have a mean of 76 and a standard deviation of 12.Professor Elderman offers his class of 36 a pizza party if the class average is above 80.What is the probability that he will have to deliver on his promise?
A) 0.0228
B) 0.3707
C) 0.6293
D) 0.9772

Accepted Answer

The answer of Professor Elderman has given the same multiple-choice...

Question 14

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years.After examining his records from the past 10 years,he finds that the scores have a mean of 76 and a standard deviation of 12.What is the probability that a class of 15 students will have a class average greater than 70 on Professor Elderman's final exam?
A) 0.0262
B) 0.6915
C) 0.9738
D) Cannot be determined.

Accepted Answer

The answer of Professor Elderman has given the same multiple-choice...

Question 15

Suppose that,on average,electricians earn approximately µ = $54,000 per year in the United States.Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is σ = $12,000.What is the probability that the average salary of four randomly selected electricians exceeds $60,000?
A) 0.1587
B) 0.3085
C) 0.6915
D) 0.8413

Accepted Answer

The answer of Suppose that,on average,electricians earn approximately µ =...

Question 16

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.What are the expected value and the standard deviation for the sampling distribution of the sample mean?
A) 0.425 and 1.00
B) 0.425 and 2.83
C) 17 and 1.00
D) 17 and 2.83

Accepted Answer

The answer of A random sample of size 36 is...

Question 17

Suppose that,on average,electricians earn approximately µ = $54,000 per year in the United States.Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is σ = $12,000 dollars.What is the probability that the average salary of four randomly selected electricians is more than $50,000 but less than $60,000?
A) 0.5899
B) 0.7486
C) 0.8413
D) 0.9048

Accepted Answer

The probability that the average salary of four randomly selected electricians is more than $50,000 but less than $60,000 is 0.5899. This can be calculated using the normal distribution with mean µ = $54,000 and standard deviation σ = $12,000. The z-score for $50,000 is (50000 - 54000) / 12000 = -0.33, and the z-score for $60,000 is (60000 - 54000) / 12000 = 0.5. The probability that the average salary is between these two values is the area under the normal curve between z = -0.33 and z = 0.5, which is 0.5899.

Question 18

Susan has been on a bowling team for 14 years.After examining all of her scores over that period of time,she finds that they follow a normal distribution.Her average score is 225,with a standard deviation of 13.What is the probability that in a one-game playoff,her score is more than 227?
A) 0.2676
B) 0.4404
C) 0.5596
D) 0.7324

Accepted Answer

The answer of Susan has been on a bowling team...

Question 19

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is greater than 18 is _______.
A) 0.1587
B) 0.4325
C) 0.5675
D) 0.8413

Accepted Answer

The answer of A random sample of size 36 is...

Question 20

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is between 15 and 18 is _______.
A) 0.0228
B) 0.8185
C) 0.8413
D) 0.8641

Accepted Answer

The sample mean follows a normal distribution with mean µ = 17 and standard deviation σ/√n = 6/√36 = 1. The z-scores for 15 and 18 are (15-17)/1 = -2 and (18-17)/1 = 1, respectively. The probability of a z-score between -2 and 1 is approximately 0.8185.

Quiz 7: Sampling and Sampling Distributions

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is less than 15 is _______.

The central limit theorem states that,for any distribution,as n gets larger,the sampling distribution of the sample mean _______.

According to the central limit theorem,the distribution of the sample means is normal if _______.

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.What are the expected value and the standard deviation for the sampling distribution of the sample mean?

Susan has been on a bowling team for 14 years.After examining all of her scores over that period of time,she finds that they follow a normal distribution.Her average score is 225,with a standard deviation of 13.What is the probability that in a one-game playoff,her score is more than 227?

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is greater than 18 is _______.

A random sample of size 36 is taken from a population with mean µ = 17 and standard deviation σ = 6.The probability that the sample mean is between 15 and 18 is _______.