Please help! : A simple random sample of size n-36 is obtained from a population that is skewed right with mean = 72 and SD = 12. (a) Describe the sampling distribution of #

Question

A simple random sample of size n-36 is obtained from a population that is skewed right with mean = 72 and Standard Deviation = 12.

1. Describe the sampling distribution of #
2. What is P (× > 75.1)?
3. What is P (X = 67.3) ?
4. What is P (70.3 < × < 75.1) ?

(round four decimal places as needed)

1. please walk through steps if possible, thank you.

Answer 1

Answer:

Describe the sampling distribution of X:

The sampling distribution of X will have a mean of 72 and a Standard Deviation of (12/√36) ≈ 2.60.

What is P (X > 75.1)?

P (X > 75.1) = 1 - P (X ≤ 75.1) = 1 - P(z ≤ (75.1 - 72) / 2.60) = 1 - 0.977 = 0.023

What is P (X = 67.3) ?

P (X = 67.3) = P(z = (67.3 - 72) / 2.60) = P (z = -2.12) = 0.017

What is P (70.3 < X < 75.1) ?

P (70.3 < X < 75.1) = P (z < (75.1 - 72) / 2.60) - P (z < (70.3 - 72) / 2.60) = 0.977 - 0.226 = 0.751