Jawab:

We are given:

The population variance, σ^2 = $525

The sample size, n = 32

The sample mean, x̄ = $110

We can use the central limit theorem to approximate the sampling distribution of the sample mean as a normal distribution, since the sample size is large enough. The mean of this sampling distribution is the population mean, μ, and the standard deviation is σ/√n.

The population mean is μ = $125.

The standard deviation of the sampling distribution is σ/√n = sqrt(525/32) = $4.30 (rounded to two decimal places).

Now, we can answer the questions:

What is the probability of getting a sample mean of $110 or larger?

To find the probability of getting a sample mean of $110 or larger, we need to find the area to the right of $110 on the sampling distribution. We can standardize the sample mean using the formula z = (x̄ - μ) / (σ/√n), where x̄ = $110, μ = $125, σ/√n = $4.30. Thus,

z = (110 - 125) / 4.30 = -3.49

Using a standard normal table or calculator, we can find that the area to the right of z = -3.49 is 0.9995. Therefore, the probability of getting a sample mean of $110 or larger is 1 - 0.9995 = 0.0005 or 0.05%.

What is the probability of getting a sample mean larger than $135 per hour?

To find the probability of getting a sample mean larger than $135 per hour, we need to find the area to the right of $135 on the sampling distribution. We can standardize the sample mean using the same formula as above, where x̄ = $135, μ = $125, σ/√n = $4.30. Thus,

z = (135 - 125) / 4.30 = 2.32

Using a standard normal table or calculator, we can find that the area to the right of z = 2.32 is 0.0107. Therefore, the probability of getting a sample mean larger than $135 per hour is 0.0107 or 1.07%.

What is the probability of getting a sample mean of between $120 and $130 per hour?

To find the probability of getting a sample mean of between $120 and $130 per hour, we need to find the area between $120 and $130 on the sampling distribution. We can standardize both sample means using the same formula as above, where μ = $125 and σ/√n = $4.30. Thus,

z1 = (120 - 125) / 4.30 = -1.16

z2 = (130 - 125) / 4.30 = 1.16

Using a standard normal table or calculator, we can find that the area to the left of z = -1.16 is 0.1230, and the area to the left of z = 1.16 is 0.8769. Therefore, the area between z1 and z2 is 0.8769 - 0.1230 = 0.7539. Thus, the probability of getting a sample mean of between $120 and $130 per hour is 0.7539 or 75.39%.