Question

Professors at a local university earn an average salary of $80,000 with a standard deviation of $6,000. The salary distribution is approximately bell-shaped. What can be said about the percentage of salaries that are at least $74,000?

Answer 1

Answer:

84.13% of of salaries that are at least $74,000

Step-by-step explanation:

Problems of normally distributed(bell-shaped) samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 80000, \sigma = 6000$

What can be said about the percentage of salaries that are at least $74,000?

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{74000 - 80000}{6000}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587.

1 - 0.1587 = 0.8413

84.13% of of salaries that are at least $74,000

Answer 2

Answer:

84.13% of salaries are at least $74,000.

Step-by-step explanation:

We are given that Professors at a local university earn an average salary of $80,000 with a standard deviation of $6,000. The salary distribution is approximately bell-shaped.

Let X = Percentage of Salary

So, X ~ N($\mu=80000,\sigma^{2}=6000^{2})$)

Now, the z score probability distribution is given by;

        Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean

           $\sigma$ = standard deviation

So, Percentage of salaries that are at least $74,000 is given by = P(X $\geq$ $74,000)

    P(X $\geq$ 74,000) = P( $\frac{X-\mu}{\sigma}$ $\geq$  $\frac{74,000-80,000}{6,000}$ ) = P(Z $\geq$ -1) = P(Z $\leq$ 1)

                                                                     = 0.84134 {using z table}

Therefore, 84.13% of salaries that are at least $74,000.