We can use the central limit theorem to approximate the sampling distribution of the sample mean, which will also have a normal distribution with mean μ = 71.8 and standard deviation σ/√n = 62.4/√238 ≈ 4.041.
Then, we can standardize the sample mean using the z-score formula:
z = (x - μ) / (σ/√n)
For a sample mean between 64.5 and 71, we have:
z1 = (64.5 - 71.8) / (62.4/√238) ≈ -1.846
z2 = (71 - 71.8) / (62.4/√238) ≈ -0.188
Using a standard normal distribution table or calculator, we can find the probabilities associated with these z-scores:
P(z < -1.846) ≈ 0.0325
P(z < -0.188) ≈ 0.4251
Then, we can subtract these probabilities to find the probability of the sample mean being between 64.5 and 71:
P(64.5 < M < 71) = P(-1.846 < z < -0.188) ≈ 0.4251 - 0.0325 ≈ 0.3926
Therefore, the probability that a sample of size n = 238 is randomly selected with a mean between 64.5 and 71 is approximately 0.3926.