A population has a mean of 125 and a standard deviation of 16.A sample of 64 observations will be taken.The probability that the sample mean will be between 122.4 and 126.1 is

A) 0.1821.
B) 0.0919.
C) 0.6120.
D) 0.3880.

First, we need to transform the sample mean to a z-score using the formula:

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.

Using the values given in the problem, we have:

z1 = (122.4 - 125) / (16 / sqrt(64)) = -1.5
z2 = (126.1 - 125) / (16 / sqrt(64)) = 0.3875

Next, we need to find the probability of getting a z-score between -1.5 and 0.3875. We can use a standard normal distribution table or calculator to find these probabilities.

P(-1.5 < z < 0.3875) = 0.6120

Therefore, the answer is C.