Question

A professor's son, having made the wise decision to drop out of college, has been finding his way in life taking one job or another, leaving when his creativity is overly stifled or the employer tires of his creativity. The professor dutifully logs the duration of his son's last few careers and has determined that the average duration is normally distributed with a mean of sixty-six weeks and a standard deviation of ten weeks. The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?

Answer 1

88.88% probability that it endures for less than a year and a half

Explanation:

Problems of normally distributed 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 question, we have that:

`\mu = 66, \sigma = 10`

The next career begins on Monday; what is the likelihood that it endures for less than a year and a half?

One year has 52.14 weeks. So a year and a half has 1.5*52.14 = 78.21 weeks.

So this probability is the pvalue of Z when X = 78.21.

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

`Z = (78.21 - 66)/(10)`

`Z = 1.22`

`Z = 1.22` has a pvalue of 0.8888

88.88% probability that it endures for less than a year and a half