The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?1) 0.97722) 0.50003) 0.99874) 0.0228

1) 0.9772

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

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

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:

What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

This is 1 subtracted by the pvalue of Z when X = 30000. So

has a pvalue of 0.0228.

1 - 0.0228 = 0.9772

0.9772 that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.