Question

The monthly overtime for employees of​ Fred’s Manufacturing Company is normal with a mean of

27

hours and a standard deviation of

4.4

hours. If

16

members of this population are selected at​ random, what is the probability the sample mean will be between

25.79

and

29.42

​hours? (Round to the nearest tenth of a​ percent.)

Answer 1

Solution :

= / n = 4.4 / 16 = 1.1

= P[(25.79 - 27) / 1.1 < ( - ) / < (29.42 - 27) / 1.1)]

= P(-1.1 < Z < 2.2)

= P(Z < 2.2) - P(Z < -1.1)

= 0.9861 - 0.1357

= 0.8504

probability = 85.0%