# SOLUTION: The manager of the Portland Electronics store is concerned that his suppliers have been giving him TV sets with lower than average quality. His research shows that replacement tim

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: The manager of the Portland Electronics store is concerned that his suppliers have been giving him TV sets with lower than average quality. His research shows that replacement tim      Log On

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 Click here to see ALL problems on Probability-and-statistics Question 179216This question is from textbook : The manager of the Portland Electronics store is concerned that his suppliers have been giving him TV sets with lower than average quality. His research shows that replacement times for TV sets have a mean of 8.2 years and a standard deviation of 1.1 years (based on data from 'Getting Things Fixed,' Consumer Reports). He then randomly selects 50 TV sets sold in the past and finds that the mean replacement time is 7.8 years. a. Assuming that TV replacement times have a mean of 8.2 years and a standard deviation of 1.1 years, find the probability that 50 randomly selected TV sets will have a mean replacement time of 7.8 years or less. b. Based on the result from part (a), does it appear that the Portland Electronics store has been given TV sets with lower than average quality? This question is from textbook Answer by stanbon(57410)   (Show Source): You can put this solution on YOUR website!a. Assuming that TV replacement times have a mean of 8.2 years and a standard deviation of 1.1 years, find the probability that 50 randomly selected TV sets will have a mean replacement time of 7.8 years or less. ------ Find the z-value of 7.8 z(7.8) = (7.8-8.2)/[1.1/sqrt(50)] = -2.57129 P(x <= 7.8 yrs) = P(z <= -2.5713) = 0.00507 -------------------------------------------- b. Based on the result from part (a), does it appear that the Portland Electronics store has been given TV sets with lower than average quality? Yes. Only (1/2) of 1% of TV's would be expected to have a lower life expectancy. ============================== Cheers, Stan H.