Question

1. Glenn Howell, Vice President of Standard Insurance staff, has developed a new training program fully adaptable to the pace of users. New employees work in several stages at their own pace of work; The training term is given when the material is learned. The Howell program has been especially effective in accelerating the training process, since the salary of an employee during training is only 67% of what he would earn when completing the program. In recent years, the average term of the program has been 44 days, with a standard deviation of 12 days.

a) Find the probability that an employee will finish the program between 46 and 54 days.
b) Find the probability that an employee ends the program between 41 and 50 days.
c) What is the probability of not finishing the program in 47 days?

Answer 1

Please keep the z tables ready for this

we know that the z score is given as

z = x-mu/sd

here mu = 44

sd = 12

a) Find the probability that an employee will finish the program between 46 and 54 days.

here lets calculate the z score as

z46 = (46-44)/12 = 0.166

z=54 = (54-44)/12 = 0.833

To find the probability of P (0.166<Z<0.833), we use the following formula:

P (0.166<Z<0.833 )=P ( Z<0.833 )−P (Z<0.166 )

We see that P ( Z<0.833 )=0.7967.

We see that  P ( Z<0.166 )=0.5675.

At the end we have:

P (0.166<Z<0.833 )=0.2292

b) Find the probability that an employee ends the program between 41 and 50 days.

here lets calculate the z score as

z41 = (41-44)/12 = -0.25

z=50 = (50-44)/12 = 0.5

To find the probability of P (−0.25<Z<0.5), we use the following formula:

P (−0.25<Z<0.5 )=P ( Z<0.5 )−P (Z<−0.25 )

We see that P ( Z<0.5 )=0.6915.

We see that P ( Z<0.25 )=0.5987 so,

P ( Z<−0.25)=1−P ( Z<0.25 )=1−0.5987=0.4013

At the end we have:

P (−0.25<Z<0.5 )=0.2902

c) What is the probability of not finishing the program in 47 days?

z47 = (47-44)/12 = 0.25

P ( Z<0.25 )=0.5987

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