Question 862165: 3. The lifetimes of a colour TV picture tube is normally distributed, with a mean of 8 yrs and a standard deviation of 2 years.
(i) What is the probability that a picture tube will last more than 10 yrs?
(ii) If the firm guarantees the picture tube for 4 yrs, what percentage of the tubes sold will have to be replaced?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! m = mean
s = standard deviation.
x = actual value
z = (x-m) / s
p(z) = area under the normal distribution curve taken from the normal distribution tables.
in your problem:
m = 8
s = 2
for part i, the formula becomes:
x = 10
z = (10-8)/2 = 2/2 = 1
p(z > 1) = .1587
if you look up a z of 1 in the following z-score table, it will tell you that the probability of getting a z-score less than 1 would be equal to .8413.
you would then take 1 - .8413 to get the probability of getting a z-score greater than 1.
http://lilt.ilstu.edu/dasacke/eco148/ztable.htm
that takes care of part i.
part ii is taken care of as follows:
x = 4
z = (4-10)/2 = -6/2 = -3
p(z < -3) = .0013
this means that the percentage of tubes that would have to be replaced is .13%.
in this case you take the probability straight off the table, since you want the probability of less than the z-score which is what this particular table is designed to give you.
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