Question

The mean output of a certain type of amplifier is 110 watts with a variance of 100.

If 44 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 3.4 watts? Round your answer to four decimal places. 0.9762 is answer

Answer 1

Mean, = 110 W

Standard deviation, = = 10 W

Sample size, n = 44

For sampling distribution of mean, mean = = 110 W

Standard error, =

=

= 1.507

P( < A) = P(Z < (A - )/)

P(sample would differ from the population mean by less than 3.4 watts) = P(110 - 3.4 < < 110 + 3.4)

= P(106.6 < < 113.4)

= P( < 113.4) - P( < 106.6)

= P(Z < (113.4 - 110)/1.507) - P(Z < (106.6 - 110)/1.507)

= P(Z < 2.26) - P(Z < -2.26) (Take values corresponding to z from standard normal distribution table)

= 0.9881 - 0.0119

= 0.9762