Question

The HR department of the companies found that the time required for randomly selected staff members to perform a certain computer task can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. Those who complete the task in 145 seconds or more must attend specialized training. What is the probability that a staffer takes less than 145 seconds and thus will not require additional training?

1. Complete inequality in terms of X: P(X )

2. Corresponding Z-score:

3. Probability that staffer will not require additional training:

Answer 1

(A) We have to find probability that that a staffer takes less than 145 seconds

So, inequality term is

(B) Using the z score formula

z = (x-mean)/(sd)

we have x = 145, mean= 120 and standard deviation (sd) = 20

this gives us

=(145-120)/20

= 1.25

(C) We have to find the value of P(z<1.25)

using z distribution, find 1.2 in the left most column and 0.05 in the top most row, select the intersecting cell, we get

probability = 0.8944

Therefore,  Probability that staffer will not require additional training is 0.8944