A population has a mean of 50 and a standard deviation of 14. If a random sample of 49 is taken, what is the probability that the sample mean is each of the following?

a. Greater than 54
b. Less than 53
c. Less than 48
d. Between 45.5 and 52.5
e. Between 50.9 and 51.7

(Round the values of z to 2 decimal places, e.g., 15.25. Round your answer to 4 decimal places, e.g., 0.2513.)

The probability for the sample means of 49 given a population with a mean of 50 and standard deviation of 14 is calculated using the z-score method. The results are: for greater than 54, 0.0228; less than 53, 0.9332; less than 48, 0.1587; between 45.5 and 52.5, 0.67; and between 50.9 and 51.7, 0.1151.

Explanation:

To answer your question, one can use the z-score formula: z = (X - μ) / (σ / sqrt(n)), where X is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.

Greater than 54: Here, the z-score is (54 - 50) / (14 / sqrt(49)) = 2. So, we are looking for P(Z > 2) which is approximately 0.0228.
Less than 53: The z-score is (53 - 50) / (14 / sqrt(49)) = 1.5. Then, P(Z < 1.5) is approximately 0.9332.
Less than 48: The z-score is (48 - 50) / (14 / sqrt(49)) = -1, so P(Z < -1) is approximately 0.1587.
Between 45.5 and 52.5: Here we calculate P(45.5 < X < 52.5) = P(-1 < Z < 1) = 0.67.
Between 50.9 and 51.7: This is P(50.9 < X < 51.7) = P(0.5 < Z < 0.8) = 0.1151.

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