Question 179216
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.