Question

A population of values has a normal distribution with = 71.8 and o= 62.4. You intend to draw a

random sample of size n = = 238.
Find the probability that a sample of size n = 238 is randomly selected with a mean between 64.5 and 71.
P(64.5 M<71) =
Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-
scores rounded to 3 decimal places are accepted.

Answer 1

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.