Question 1 of 2 15 Points
A report revealed that the average no. of months that an employee stays in a factory is 36
months. Assuming that the no. of months of an employee tenure in the factory is normally
distributed with a standard deviation of 6 months, find the probability that a certain employee
will stay
a. More than 30 months
b. Less than 24 months
c. Between 24 to 48 months

Step-by-step explanation:

Given the following :

Mean(m) = 36 months

Standard deviation (s) = 6 months

A) probability of staying more than 30 months :

Zscore = ( x - m) / s

x = 30

Zscore = (30 - 36) / 6

Zscore = - 6/6

= - 1

P(Z > - 1) = 1 - p(Z < - 1) ; p(Z < - 1) = 0.1587

1 - p(Z < - 1) = 1 - 0.1587 = 0.8413

2)less than 24 months :

Zscore = ( x - m) / s

x = 24

Zscore = (24 - 36) / 6

Zscore = - 12/6

= - 2

P(Z < - 2) = 0.0228

3.) between 24 to 48 months

For 24 months : from above = 0.0228

48 months :

Zscore = ( x - m) / s

x = 48

Zscore = (48 - 36) / 6

Zscore = 12/6

= 2

P(Z < 2) = 0.9772 ( from z table)

Hence,

0.9772 - 0.0228 = 0.9544

7

step-by-step explanation:

i am thinking c. sorry if im wrong .

step-by-step explanation: