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()
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
where, = population mean
= standard deviation
So, Percentage of salaries that are at least $74,000 is given by = P(X $74,000)
P(X 74,000) = P( ) = P(Z -1) = P(Z 1)
= 0.84134 {using z table}
Therefore, 84.13% of salaries that are at least $74,000.