celesteh12

2022-10-13

usiness Weekly conducted a survey of graduates from several top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is \$172,000. Assume the standard deviation is \$36,000. Suppose you take a simple random sample of 16 graduates. Round all answers to four decimal places if necessary.

• What is the standard deviation of distribution of $X$
• What is the standard deviation of $\overline{X}$
• For a single randomly selected graduate, find the probability that her salary is between \$164,800 and \$177,900.
• For a simple random sample of 16 graduates, find the probability that the average salary is between \$164,800 and \$177,900.
• For part d), is the assumption of normal necessary? No or Yes

Don Sumner

To solve this problem, let's first define the variables:
$\mu$ = Mean annual salary for graduates 10 years after graduation = 172,000
$\sigma$ = Standard deviation of the population = $36,000$
$n$ = Sample size = 16
(a) Standard deviation of the distribution of X:
The standard deviation of the distribution of X, denoted as ${\sigma }_{X}$, can be calculated using the formula:
${\sigma }_{X}=\frac{\sigma }{\sqrt{n}}$
Substituting the given values, we have:
${\sigma }_{X}=\frac{36,000}{\sqrt{16}}$
${\sigma }_{X}=\frac{36,000}{4}$
${\sigma }_{X}=9,000$
Therefore, the standard deviation of the distribution of X is 9,000.
(b) Standard deviation of X:
The standard deviation of X, denoted as ${s}_{X}$, is an estimate of the population standard deviation based on the sample. It can be calculated using the formula:
${s}_{X}=\frac{\sigma }{\sqrt{n}}$
Substituting the given values, we have:
${s}_{X}=\frac{36,000}{\sqrt{16}}$
${s}_{X}=\frac{36,000}{4}$
${s}_{X}=9,000$
Therefore, the standard deviation of X is also 9,000.

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