11. Assume that the population of heights of male college students is approximately normally distributed with mean m of 72.15 inches and standard deviation s of 6.39 inches. A random sample of 96 heights is obtained. Show all work.

(A) Find
(B) Find the mean and standard error of the distribution
(C) Find
(D) Why is the formula required to solve (A) different than (C)?
(Points : 12)

A and C are not complete. Copy and paste may not work here.

B. mean = 72.15

SEm = SD/√n

(A) To find P(X > 74.2), we need to convert the given value 74.2 to a z-score using the formula:

z = (x - m) / s

where x is the given value, m is the mean, and s is the standard deviation.

In this case, x = 74.2, m = 72.15, and s = 6.39.

Plugging in the values, we get:

z = (74.2 - 72.15) / 6.39 = 0.3208

Next, we look up the z-score in the z-table to find the corresponding probability. The z-table provides the area under the standard normal distribution curve to the left of a given z-score.

Looking up the z-score 0.3208 in the z-table, we find the corresponding probability to be 0.6255.

Therefore, P(X > 74.2) = 1 - P(X ≤ 74.2) = 1 - 0.6255 = 0.3745.

(B) The mean of the distribution is the same as the population mean, which is 72.15 inches.

The standard error of the distribution can be calculated using the formula:

SE = s / sqrt(n)

where SE represents the standard error, s is the standard deviation, and n is the sample size.

In this case, s = 6.39 and n = 96.

Plugging in the values, we get:

SE = 6.39 / sqrt(96) ≈ 0.6521.

Therefore, the mean of the distribution is 72.15 inches and the standard error is approximately 0.6521 inches.

(C) To find the sample size needed to estimate the population mean with a specific margin of error, we use the formula:

n = (z * s / E)^2

where n represents the sample size, z is the z-score corresponding to the desired confidence level (usually found in a z-table), s is the standard deviation, and E is the margin of error.

In this case, we want to find the sample size needed to estimate the mean within a margin of error of 0.5 inches, and we want to use a 95% confidence level, which corresponds to a z-score of approximately 1.96.

Plugging in the values, we get:

n = (1.96 * 6.39 / 0.5)^2 ≈ 617.79

Therefore, the sample size needed is approximately 618.

(D) The formula used to solve (A) is different from the formula used to solve (C) because they are addressing different questions.

In (A), we are given a specific value (74.2 inches) and asked to find the probability of a random variable being greater than that value. To do this, we convert the value to a z-score and use the z-table to find the probability.

In (C), we are given a desired margin of error and asked to find the sample size needed to estimate the population mean within that margin of error. To do this, we use a formula that takes into account the standard deviation, desired margin of error, and desired confidence level.

While both (A) and (C) involve calculations with z-scores, they have different goals and require different formulas to obtain the desired information.