Question

A large sleep study involving over 5000 American teenagers examined how many hours the participants slept on the weekend. The nighty sleep times were distinctly non-normal with a mean of 10 hours and standard deviation of 2 hours. Suppose we take a random sample of 100 nightly sleep times from this population. What is the probability that the mean of these 100 sleep times is farther than 0.4 hours away from the population mean?

a. 0.02
b. 0.05
c. 0.42
d. 0.84
e. We cannot calculate this probability because the sampling distribution is not normal.

Answer 1

The probability that the mean of 100 sleep times is farther than 0.4 hours away from the population mean is approximately 0.0228. The correct answer is option (a).

Step-by-step explanation:

To find the probability that the mean of the 100 sleep times is farther than 0.4 hours away from the population mean, we need to calculate the z-score for this difference and then find the corresponding probability from the standard normal distribution.

The z-score is calculated as (0.4 - 0) / (2 / sqrt(100)), which simplifies to 0.4 / 0.2 = 2.

Using a standard normal table or a calculator, we can find that the probability of a z-score being greater than 2 is approximately 0.0228.

Therefore, the probability that the mean of the 100 sleep times is farther than 0.4 hours away from the population mean is approximately 0.0228, which is closest to option a. 0.02.