Question

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 8383 months with a variance of 100100.

If the claim is true, what is the probability that the mean monitor life would differ from the population mean by less than 1.71.7 months in a sample of 8989 monitors? Round your answer to four decimal places.

Answer 1

Given: Mean() 83 Noriance Co2 =100 Standard deviation () Vanonce Sample mean CX) Sample sizecn) Differ by mean by lessthn 17 re 81-3 < B4 89 81-3 Fixst take : X=81-3 By using z SCODE fomala 81-3-3 X-4 60 calculate P(Z<-60 Using z toble Hence we get O 00 -O.OS48 G1.3 Nert take X847 By using z SCore foimut XIA 34.7-83 ०१.। USing ztable, calculote P(ZI66 Hen ce e get फन्पपः B4.7 I6 O-g452

Hence awe get PZ <I.60) & p(z-160) = 0-9452 = 00S48 Now Tule PCa<z <b) P Z<6)-PZ <a) p-1-60z<I'60)= P(z160-P(2G 60) O,O548 0-9452 O-8904 O&g0 4 60 OOS48 60 O.9452 O.3904 ADS