The scores of students on the ACT college entrance examination in a recent year had a normal distribution with a mean of 18.6 and a standard deviation of 5.9. A simple random sample of 50 students who took the exam is selected for study:

A) What is the shape, mean, and standard deviation of the sampling distribution

of the sample mean for samples of size 50?

B) What is the probability that the sample mean is 21 or higher?

C) What is the probability that the sample mean falls within 2 points of the

population mean?

D) What is the probability that the sample mean is 17 or less?

Last year, a national opinion poll found that 44% of all American adults agree that parents should be given vouchers good for education at any public or private school of their choice. Assume that in fact the population proportion is 0.44. A sample of 300 is to be selected and asked the same question:

A) What is the shape, mean, and standard deviation of the sampling distribution

of the sample proportion for samples of size 300?

B) What is the probability that the sample proportion will fall within 3% of the

population proportion?

C) What is the probability that the sample proportion will exceed 50%?

D) If the sample size were increased to 900, what is the probability that the

sample proportion will fall within 3% of the population proportion

13.2555

7.62

A) For samples of size 50, the shape of the sampling distribution of the sample mean is approximately normal. The mean of the sampling distribution is equal to the population mean, which is 18.6. The standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size. Therefore,

Mean of the sampling distribution = 18.6
Standard deviation of the sampling distribution = 5.9 / √50 ≈ 0.835

B) To find the probability that the sample mean is 21 or higher, we need to convert the value to a standard Z-score and then use the standard normal distribution table. The formula for the Z-score is:

Z = (x - μ) / (σ / √n)

Where:
x = desired sample mean (21)
μ = population mean (18.6)
σ = population standard deviation (5.9)
n = sample size (50)

Calculating the Z-score,

Z = (21 - 18.6) / (5.9 / √50) ≈ 2.6

Using the standard normal distribution table, the probability of obtaining a Z-score of 2.6 or higher is approximately 0.0047.

C) To find the probability that the sample mean falls within 2 points of the population mean, we need to find the area under the normal curve between (18.6 - 2) and (18.6 + 2). Again, we convert the values to Z-scores:

Z1 = (18.6 - (18.6 - 2)) / (5.9 / √50) ≈ -0.68
Z2 = (18.6 + 2 - 18.6) / (5.9 / √50) ≈ 0.68

Using the standard normal distribution table, the probability of obtaining a Z-score between -0.68 and 0.68 is approximately 0.4764.

D) To find the probability that the sample mean is 17 or less, we again convert the value to a Z-score:

Z = (17 - 18.6) / (5.9 / √50) ≈ -2.03

Using the standard normal distribution table, the probability of obtaining a Z-score of -2.03 or lower is approximately 0.0219.

Now let's move on to the questions about the sample proportion:

A) For samples of size 300, the shape of the sampling distribution of the sample proportion is approximately normal, following the central limit theorem. The mean of the sampling distribution of the sample proportion is equal to the population proportion, which is 0.44. The standard deviation of the sampling distribution of the sample proportion is calculated using the formula:

Standard deviation of the sample proportion = √((p * (1 - p)) / n)

Where:
p = population proportion (0.44)
n = sample size (300)

Calculating the standard deviation,

Standard deviation of the sample proportion = √((0.44 * (1 - 0.44)) / 300) ≈ 0.024

B) To find the probability that the sample proportion will fall within 3% of the population proportion, we need to find the area under the normal curve between (0.44 - 0.03) and (0.44 + 0.03). We convert the values to Z-scores:

Z1 = (0.44 - (0.44 - 0.03)) / 0.024 ≈ 1.25
Z2 = (0.44 + 0.03 - 0.44) / 0.024 ≈ 1.25

Using the standard normal distribution table, the probability of obtaining a Z-score between -1.25 and 1.25 is approximately 0.7887.

C) To find the probability that the sample proportion will exceed 50%, we convert the value to a Z-score:

Z = (0.5 - 0.44) / 0.024 ≈ 2.5

Using the standard normal distribution table, the probability of obtaining a Z-score of 2.5 or higher is approximately 0.0062.

D) If the sample size were increased to 900, the standard deviation of the sampling distribution of the sample proportion would be:

Standard deviation of the sample proportion = √((0.44 * (1 - 0.44)) / 900) ≈ 0.014

To find the probability that the sample proportion will fall within 3% of the population proportion, we use the same formula as in part B:

Z1 = (0.44 - (0.44 - 0.03)) / 0.014 ≈ 1.07
Z2 = (0.44 + 0.03 - 0.44) / 0.014 ≈ 1.07

Using the standard normal distribution table, the probability of obtaining a Z-score between -1.07 and 1.07 is approximately 0.7162.

A) For samples of size 50, the shape of the sampling distribution of the sample mean is approximately normal (assuming the sample is random and independent). The mean of the sampling distribution is the same as the population mean, which is 18.6 in this case. The standard deviation of the sampling distribution, known as the standard error, is the population standard deviation divided by the square root of the sample size. So in this case, it would be 5.9 divided by the square root of 50.

B) To find the probability that the sample mean is 21 or higher, you can standardize the sample mean using the formula z = (x - µ) / σ, where x is the sample mean, µ is the population mean, and σ is the standard deviation of the sampling distribution. In this case, you want to find P(x ≥ 21). First, standardize the value of 21 using the formula: z = (21 - 18.6) / (5.9 / √50). Then, you can look up the probability corresponding to the standardized value using a standard normal distribution table or calculator.

C) To find the probability that the sample mean falls within 2 points of the population mean, you would calculate P(18.6 - 2 ≤ x ≤ 18.6 + 2). This can be done using the same standardization process as in part B, with z values corresponding to the lower and upper bounds.

D) To find the probability that the sample mean is 17 or less, you can calculate P(x ≤ 17) using the standardization process described in part B.

A) For samples of size 300, the shape of the sampling distribution of the sample proportion is approximately normal (assuming the sample is random and independent). The mean of the sampling distribution is the same as the population proportion, which is 0.44 in this case. The standard deviation of the sampling distribution, known as the standard error, is the square root of (p * (1 - p) / n), where p is the population proportion and n is the sample size.

B) To find the probability that the sample proportion falls within 3% of the population proportion (0.44), you would calculate P(0.44 - 0.03 ≤ p ≤ 0.44 + 0.03). This can be done using the same standardization process as in part B, with z values corresponding to the lower and upper bounds.

C) To find the probability that the sample proportion exceeds 50% (0.50), you would calculate P(p > 0.50) using the standardization process described in part B.

D) If the sample size were increased to 900, the standard deviation of the sampling distribution (standard error) would decrease. To find the probability that the sample proportion falls within 3% of the population proportion, you would calculate P(0.44 - 0.03 ≤ p ≤ 0.44 + 0.03) using the updated standard error for a sample size of 900.