The mean annual income for people in a certain city (in thousands of dollars) is 46, with a standard deviation of 33. A pollster draws a sample of 55 people to interview. What is the probability that the Sample mean income is between 42 and 50 (thousands of dollars)?

Question

Quizplus · Accepted Answer

To solve this problem, we'll use the central limit theorem, which tells us that the distribution of sample means will be approximately normal as sample size increases (in this case, n=55).

We can standardize our sample mean using the formula:

z = (x̄ - μ) / (σ / sqrt(n))

Where x̄ is our sample mean (which we don't know yet), μ is the population mean (given as 46), σ is the population standard deviation (given as 33), and n is our sample size (given as 55).

Plugging in our values, we get:

z = (50 - 46) / (33/sqrt(55)) = 1.29
z = (42 - 46) / (33/sqrt(55)) = -1.29

We can now find the probability that z falls between these values using a standard normal distribution table or a calculator.

P(-1.29 < z < 1.29) = 0.6319

Therefore, the probability that the sample mean income is between 42 and 50 (thousands of dollars) is 0.6319, or answer choice B.

The Mean Annual Income for People in a Certain City