Assume that the population of heights of male college students is approximately normally distributed with mean of 72.15 inches and standard deviation of 6.39 inches. A random sample of 96 heights is obtained. Show all work.
(A) Find P (x > 73.25)
(B) Find the mean and standard error of the x distribution
(C) Find P (x > 73.25
(D) Why is the formula required to solve (A) different than (C)?
(A) To find P(x > 73.25), we need to find the z-score first. The z-score formula is given by:
z = (x - μ) / σ
where x is the value of interest, μ is the population mean, and σ is the population standard deviation.
In this case, x = 73.25, μ = 72.15, and σ = 6.39. Plugging these values into the formula, we get:
z = (73.25 - 72.15) / 6.39
Simplifying,
z = 1.1 / 6.39
z ≈ 0.172
Now we need to find the probability corresponding to this z-score in the standard normal distribution table or using a calculator.
P(z > 0.172) ≈ 0.4306
Therefore, P(x > 73.25) ≈ 0.4306.
(B) The mean (μ) of the x distribution is the same as the population mean, which is 72.15 inches. The standard error (SE) of the x distribution can be calculated using the formula:
SE = σ / √n
where σ is the population standard deviation and n is the sample size.
In this case, σ = 6.39 and n = 96. Plugging these values into the formula, we get:
SE = 6.39 / √96
Simplifying,
SE ≈ 0.653
Therefore, the mean of the x distribution is 72.15 inches and the standard error is approximately 0.653 inches.
(C) The calculation for P(x > 73.25) is the same as in part (A), which is approximately 0.4306.
(D) The formula used to solve (A) is different from (C) because in (A), we are dealing with an individual value (x = 73.25) and finding its probability in the distribution. In (C), we are finding the cumulative probability of x being greater than a certain value (x > 73.25). The formula for (A) involves calculating the z-score for an individual value, while the formula for (C) involves finding the cumulative probability using the z-score table or a calculator.