Question

A population is normally distributed with a mean of 30 and a standard deviation of 4. a. What is the mean of the sampling distribution (μM) for this population? b. If a sample of n = 16 participants is selected from this population, what is the standard error of the mean (σM)? c. Let’s say that a sample mean is 32. 1) What is the z-score for a sample mean of 32? (calculate this) 2) What is the probability of a sample mean greater than 32? (use table).

Answer 1

Solution:

We are given

µ = 30

σ = 4

a. What is the mean of the sampling distribution (μM) for this population?

Mean of the sampling distribution = μM = µ = 30

Answer: 30

b. If a sample of n = 16 participants is selected from this population, what is the standard error of the mean (σM)?

Standard error = σ/sqrt(n) = 4/sqrt(16) = 4/4 = 1

Answer: 1

c. Let’s say that a sample mean is 32. 1) What is the z-score for a sample mean of 32? (calculate this)

Z = (Xbar - µ)/[ σ/sqrt(n)]

Z = (32 – 30)/[4/sqrt(16)]

Z = 2/1

Z = 2

Answer: 2

2) What is the probability of a sample mean greater than 32? (use table).

P(Xbar>32) = 1 – P(Xbar<32)

P(Xbar<32) = P(Z<2) = 0.97725

(by using z-table)

P(Xbar>32) = 1 – P(Xbar<32)

P(Xbar>32) = 1 – 0.97725

P(Xbar>32) = 0.02275

Required probability = 0.02275