Question

The mean output of a certain type of amplifier is 496 watts with a variance of 144 . If 40 amplifiers are sampled, what is the probability that the mean of the sample would be less than 492.3 watts? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 496

standard deviation = = 144

n = 40

= 496

₌ / n = 144 40 = 22.7684

P( < 492.3 )

P ( - /) < (492.3 - 496 /22.7684)

P( z < - 3.7 / 22.7684 )

P ( z < -0.16 )   

Using z table

= 0.4364

Probability = 0.4364