
Lifetime of Tires
Here is the question:
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
is normal,
(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 : )

$\displaystyle Z={X\mu\over \sigma}$, so $\displaystyle X=\mu + \sigma Z$
Now match up percentile points.
When X=35 we have the lower 7 percentile of a Z, known as $\displaystyle Z_{.07}\approx$ 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 $\displaystyle Z_{.11}\approx$ 1.22 or 1.23.
You now have two equations with two unknowns. Find $\displaystyle \mu$ and $\displaystyle \sigma$.
IN part (b) you have a binomial question, where n=1000 and a success is for a tire to fail before 40 months.
So $\displaystyle p=P(X<40)=P\biggl(Z<{40\mu\over\sigma}\biggr)$.
HOWEVER, the question should ask you to APPROXIMATE this via the Central Limit Theorem.
Otherwise you need to obtain via a summation
$\displaystyle P(BINOMIAL >50)=\sum_{x=51}^{1000}{1000\choose x}p^x(1p)^{1000x}$
or you can use the complement.
BUT I bet they want you to approximate via the normal distribution.

I was just surfing and I found...
http://bayes.bgsu.edu/nsf_web/jscrip...ormal_icdf.htm
Plug in mean 0, st dev 1, prob .07 and I got....1.4749
Plug in mean 0, st dev 1, prob .89 and I got....1.2249
