A population has a mean of 53 and a standard deviation of 21. A sample of 49 observations will be taken. The probability that the sample mean will be greater than 57.95 is
A)0
B).0495
C).4505
D)None of the alternative answers is correct.

Question

Quizplus · Accepted Answer

We can use the central limit theorem and the z-score formula to solve the problem.

The formula for the z-score of a sample mean is:

z = (x̄ - μ) / (σ / sqrt(n))

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we get:

z = (57.95 - 53) / (21 / sqrt(49))
z = 1.43

To find the probability of a z-score greater than 1.43, we can look up the area to the right of 1.43 in a standard normal distribution table. This area is approximately 0.0495.

Therefore, the probability that the sample mean will be greater than 57.95 is 0.0495 or 4.95%. The correct answer is B.

A Population Has a Mean of 53 and a Standard