A normally distributed population has a mean of 20 and a standard deviation of 5. Suppose you draw a sample of size n=16 from this population. How likely are we to see a sample …

Question

A normally distributed population has a mean of 20 and a standard deviation of 5. Suppose you draw a sample of size
`n=16` from this population. How likely are we to see a sample average greater than 22, that is
`P(\ \textgreater \ 22)`?

Answer 1

The probability of observing a sample average greater than 22 is 0.8997, or 89.97%.

Step-by-step explanation:

To determine the probability of observing a sample average greater than 22, we need to calculate the Z-score for 22. Z = (X - μ) / (σ / √n), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, X = 22, μ = 20, σ = 5, and n = 16.

Plugging these values into the formula gives us Z = (22 - 20) / (5 / √16) = 1.6 / 1.25 = 1.28.

Next, we need to find the probability of Z > 1.28 using the standard normal distribution table or a calculator. The probability of Z > 1.28 is approximately 0.1003, or 10.03%.

Therefore, P(> 22) = 1 - P(≤ 22) = 1 - 0.1003 = 0.8997, or 89.97%.