Question

13) 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

Given: The data representing the Duration of the Professor' son's career can be represented by the normal distribution curve with the following details:

Mean of Duration of his Career, = 66 Weeks

Standard deviation of Duration of his career, = 10 Weeks

To Find: Probability that duration of career will be lesser than year and a half (i.e. 52 Weeks + 26 Weeks = 78 Weeks)

[Assuming 1 Year Consists of 52 Weeks]

So, We need to find P(X<78), where X represents the random variable of the original normal distribution representing the duration of career.

We can now plot the given normal distribution data to a standard normal distribution with the variable, Z that has mean 0 and standard deviation as 1.

The Value of Z is given by:

Z = (X - )/

For X = 78,

Z = (78-66)/10

= 1.2

So, P(X<78) = P(Z<1.2)

Refer to Z-Table, Area under normal distribution curve to the left of Z=1.2 is 0.8849

Hence, P(Z<1.2) = 0.8849

There is 88.49% Likelihood that his next career will endure or last lesser than 1.5 Years