Question

A random sample of size n = 122 is taken from a population of size N = 2,700 with mean μ = −56 and variance σ2 = 90. [You may find it useful to reference the z table.]



a-1. Is it necessary to apply the finite population correction factor?

multiple choice
Yes
No


a-2. Calculate the expected value and the standard error of the sample mean. (Negative values should be indicated by a minus sign. Round “expected value” to a whole number and "standard error" to 2 decimal places.)



b. What is the probability that the sample mean is between −58 and −54? (Round final answer to 4 decimal places.)



c. What is the probability that the sample mean is greater than −55? (Round final answer to 4 decimal places.)

Answer 1

a-1. No

a-2. The expected value of the sample mean E(x̄) is equal to the population mean μ:

E(x̄) = μ = -56

The standard error of the sample mean σ(x̄) is calculated using the formula:

σ(x̄) = σ / √n

σ(x̄) = √90 / √122

σ(x̄) ≈ 2.43 (rounded to 2 decimal places)

b. To find the probability that the sample mean is between -58 and -54, we need to calculate the z-scores for these values and use the z-table.

Find the z-scores:

z1 = (x1 - μ) / σ(x̄) = (-58 - (-56)) / 2.43
z2 = (x2 - μ) / σ(x̄) = (-54 - (-56)) / 2.43

P(-58 < x < -54) = P(z1 < Z < z2)

Using the z-table, find the probabilities corresponding to the z-scores z1 and z2, and subtract the smaller probability from the larger one:

P(-58 < x < -54) = P(z1 < Z < z2) = P(Z < z2) - P(Z < z1)

Round the final answer to 4 decimal places.

c. To find the probability that the sample mean is greater than -55, we need to find the z-score for -55 and use the z-table.

Find the z-score:

z = (x - μ) / σ(x̄) = (-55 - (-56)) / 2.43

P(x > -55) = P(Z > z)

Using the z-table, find the probability corresponding to the z-score z.

Round the final answer to 4 decimal places.