Math / Data & Statistics

QuestionThe incomes in a certain large population of college teachers have a Normal distribution, with mean \ $75,000 and standard deviation$ \ $10,000$ . Sixteen teachers are selected at random from this population to serve on a committee. What is the probability that their average salary is more than $\$ 77,500$ ? 0.8413 essentially 0 0.0228 0.1587

Studdy Solution

STEP 1

1. The distribution of incomes is Normal with mean $\mu = 75,000$ and standard deviation $\sigma = 10,000$ .
2. We are selecting a sample of $n = 16$ teachers.
3. We need to find the probability that the sample mean $\bar{X}$ is greater than \$77,500.

STEP 2

1. Determine the distribution of the sample mean.
2. Calculate the z-score for the sample mean.
3. Find the probability using the standard normal distribution.

STEP 3

The distribution of the sample mean $\bar{X}$ is Normal with mean $\mu_{\bar{X}} = \mu = 75,000$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{10,000}{\sqrt{16}}$ .
$\sigma_{\bar{X}} = \frac{10,000}{4} = 2,500$

STEP 4

Calculate the z-score for $\bar{X} = 77,500$ .
$z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{77,500 - 75,000}{2,500}$
$z = \frac{2,500}{2,500} = 1$

STEP 5

Find the probability that the z-score is greater than 1 using the standard normal distribution table.
The probability that $Z > 1$ is $1 - P(Z \leq 1)$ .
From the standard normal distribution table, $P(Z \leq 1) = 0.8413$ .
Therefore, $P(Z > 1) = 1 - 0.8413 = 0.1587$ .
The probability that the average salary of the selected teachers is more than \$77,500 is:
$\boxed{0.1587}$

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