In a given year,the average annual salary of a NFL football player was $189,000 with a standard deviation of $20,500.If a sample of 50 players was taken,then probability that the sample mean will be $192,000 or more is:
A) 0.6970.
B) 0.3485.
C) 0.1515.
D) 0.0398.
E) 0.4602.

Question

Quizplus · Accepted Answer

The probability that the sample mean will be $192,000 or more can be found using the Z-score formula for 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: $Z = \frac{192,000 - 189,000}{20,500/\sqrt{50}} \approx 1.036$. Looking up 1.036 on the Z-table gives the area to the left, so we subtract this value from 1 to find the area to the right, which corresponds to the probability of the sample mean being $192,000 or more. This gives approximately 0.1515.

In a Given Year,the Average Annual Salary of a NFL