Given:
Normally distributed = $40,000
a standard deviation= $5,000.
Required:
Find the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000.
Explanation:
The probability formula when the mean and standard deviation is known:
[tex]P(x)=P(\frac{x-mean}{standard\text{ deviation}})[/tex][tex]\begin{gathered} P(x\ge30k)=P(\frac{30k-40k}{5k}) \\ \begin{equation*} P(x\ge30k)= \end{equation*} \end{gathered}[/tex]