Question

Assume you have a normal distribution representing the likelihood of project completion times. The mean of this distribution is 14, and the standard deviation is 4. The probability of completing the project in 15 or fewer days is:

a. 0.59

b. 0.27

c. 0.93

d. 0.75

Answer 1

X =15

Mean=14

Standard Deviation=4

Z=(X-Mean)/Standard Deviation

Z=(15-14)/4

Z=0.25

Considering the Z table, probability of completing the project in 15 or fewer days =59.87%

Hence, correct answer is 0.59