Given a population with a mean of m = 100 and a variance of s2 = 81, the central limit theorem applies when the sample size is n Ú 25. A random sample of size n = 25 is obtained. a. What are the mean and variance of the sampling distribution for the sample means? b. What is the probability that x 7 102? c. What is the probability that 98 … x … 101? d. What is the probability that x … 101.5?

a) Sample mean = 100

Variance of the sampling distribution = 3.24

b) P(x > 102) = 0.1335

c) P(98 ≤ x ≤ 101) = 0.57876

d) P(x ≤ 101.5) = 0.79673

Step-by-step explanation:

For a given sample, the sample mean and standard deviation of the distribution of sample means are related to the mean and standard deviation of the population through

μₓ = μ = 100

σₓ = (σ/√n)

σ = population standard deviation = √(population variance) = √81 = 9

n = sample size = 25

σₓ = (σ/√n) = (9/√25) = 1.8

Variance of the sampling distribution = 1.8² = 3.24

b) Probability that the sample mean is greater than 102 P(x > 102)

This is a normal distribution problem

With mean = xbar = 100

And standard deviation = σ = 1.8

we need to first obtain the z-score of 102.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ = (102 - 100)/1.8 = 1.11

To determine the probability Probability that the sample mean is greater than 102

P(x > 102) = P(z > 1.11)

We'll use data from the normal probability table for these probabilities

P(x > 102) = P(z > 1.11) = 1 - P(z ≤ 1.11) = 1 - 0.86650 = 0.1335.

c) P(98 ≤ x ≤ 101)

We first normalize/standardize/obtain the z-scores of 98 and 101

For 98

z = (x - xbar)/σ = (98 - 100)/1.8 = -1.11

For 101

z = (x - xbar)/σ = (101 - 100)/1.8 = 0.56

P(98 ≤ x ≤ 101) = P(-1.11 ≤ z ≤ 0.56)

= P(z ≤ 0.56) - P(z ≤ -1.11)

= 0.71226 - 0.13350 = 0.57876

d) P(x ≤ 101.5)

Normalizing 101.5

z = (x - xbar)/σ = (101.5 - 100)/1.8 = 0.83

P(x ≤ 101.5) = P(z ≤ 0.83) = 0.79673

Hope this Helps!!!

the answer is b.)

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