According to comScore, Facebook users spent an average of 432 minutes on the site during a recent month. Assume that this population has a standard deviation of 127 minutes. A random sample of 50 users was selected from this population. What is the probability that the average number of minutes on the site in the recent month was less than 410 minutes?
A)0.1112
B)0.2806
C)0.0643
D)0.3384

Question

Quizplus · Accepted Answer

To find the probability that the average number of minutes on the site in the recent month was less than 410 minutes, we use the formula for the z-score of a sample mean: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $z = \frac{410 - 432}{127/\sqrt{50}}$, we find the z-score and then use the standard normal distribution to find the probability corresponding to this z-score. The correct probability, based on the calculation, is 0.0643.

According to ComScore, Facebook Users Spent an Average of 432