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

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

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 =
`(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(
`(X-\mu)/(\sigma)`
`\geq`
`(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.

Answer 2

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

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 = (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 = (X - \mu)/(\sigma)`

`Z = (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