For a tire manufacturing company, it is known that only 7% of their tire
products last less than 35 months, whereas 11% of them last more tham 63
months. Assuming the distribution of the lifetime of the manufactured tires
(a) what is the expected lifetime for a tire manufactured by this company,
and its standard deviation?
(b) The manufacturer company gives a warranty that if the tire does not last
more than 40 months, they will give a replacement for it. The company
sold 1000 tires during the last year. What is the probability that they
will have to replace more than 50 tires?
So I have
a) P (X <= 35 months) = 0.07
P (X >= 63 months) = 0.11
and for b I've started with...
b) P ( X <= 40 months) = ...?
Where do I go from here, for each part?
Thank you so much in advance : )
March 22nd 2009, 09:23 PM
Now match up percentile points.
When X=35 we have the lower 7 percentile of a Z, known as -1.47 or -1.48.
That's via the tables, but you can do better via a link on line.
And when X=63 we have the upper 11 percentile of a Z, known as 1.22 or 1.23.
You now have two equations with two unknowns. Find and .
IN part (b) you have a binomial question, where n=1000 and a success is for a tire to fail before 40 months.
HOWEVER, the question should ask you to APPROXIMATE this via the Central Limit Theorem.
Otherwise you need to obtain via a summation
or you can use the complement.
BUT I bet they want you to approximate via the normal distribution.