Question

The annual income for teachers is normally distributed with a mean of $38,000 and a standard deviation of $4000. Of a random sample of 36 teachers is selected, what is the probability that the mean income of this sample is less than $37,000?

A) 0. 4013


B)0. 5897


C) 0. 9332


D) 0. 669

Answer 1

Answer: D

Step-by-step explanation:

To calculate the probability, we can use the standard normal distribution and the concept of the sampling distribution of the sample mean.

Given that the annual income for teachers is normally distributed with a mean (μ) of $38,000 and a standard deviation (σ) of $4,000.

To find the probability that the mean income of a random sample of 36 teachers is less than $37,000, we need to calculate the z-score and then find the corresponding cumulative probability.

First, calculate the standard error of the mean (SE) using the formula:

SE = σ / sqrt(n)

SE = 4000 / sqrt(36) = 4000 / 6 = 666.67

Next, calculate the z-score using the formula:

z = (x - μ) / SE

z = (37000 - 38000) / 666.67 = -1.5

Now, we can look up the cumulative probability for the z-score -1.5 in the standard normal distribution table.

Looking up the z-score -1.5 in the table, we find that the cumulative probability is approximately 0.0668.

Therefore, the probability that the mean income of this sample is less than $37,000 is approximately 0.0668.