Question

A simple random sample of size n = 39 is obtained from a population with u = 61 and sigma = 16

(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample meanthis condition is true, describe the sampling distribution of X.
(b) Assuming the normal model can be used, determine P( tilde x < 64.2 ).
(c) Assuming the normal model can be used, determine P(x >= 62.2)
(a) What must be true regarding the distribution of the population?
A. The population must be normally distributed and the sample size must be large.
B. Since the sample size is large enough, the population distribution does not need to be normal.
C. The population must be normally distributed.
D. The sampling distribution must be assumed to be normal.

Answer 1

In order to use the normal model to compute probabilities involving the sample mean, the distribution of the population must be approximately normal and the sample size must be large. The z-score for P(tilde x < 64.2) is approximately 0.783 and P(x >= 62.2) is approximately 0.42.

Step-by-step explanation:

In order to use the normal model to compute probabilities involving the sample mean, the distribution of the population must be approximately normal and the sample size must be large. Therefore, option A is correct - the population must be normally distributed and the sample size must be large.

Assuming the normal model can be used, to determine P( tilde x < 64.2 ), we need to calculate the z-score and find the corresponding probability. The formula for z-score is: z = (x - μ) / (σ / √n) where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values, the z-score is z = (64.2 - 61) / (16 / √39) ≈ 0.782. Using a standard normal distribution table or calculator, we can find that P(z < 0.782) is approximately 0.783.

Similarly, to determine P(x >= 62.2), we can calculate the z-score as z = (62.2 - 61) / (16 / √39) ≈ 0.195. Again, using a standard normal distribution table or calculator, we find that P(z >= 0.195) is approximately 0.42.