Question

The meann output of a certain type of amplifier is 384 whites with a variance of 100. if 41 amplifiers Are sampled what is the probability that the mean of the sample would differ from the population mean by greater than 1.8

that is the value

Answer 1

Given алм ) 384 Variance (m2) = 100 Stameland .eleviation () = 10 sample size na ul P(x > 41.8) we know Zscove - Free p( x2 M +18) - Plzy feet1-8) - 4 0155 PC x 4+18) = P(Z > 181 My p(x > 4+18) = P(Z > 1.15) Plze > M+18) - P- P(221.5) using a table PCZ31.165) = 0.8749 PC x> 4+18) – 1-0.8749 = 0.1251 P(x> 4+18) - TO.(251 Probability that the mean of sample would be greater than population mean by 18 i l 10.12511