Suppose that the life span of a certain automobile tire is normally distributed with mu equals...
Suppose that the life span of a certain automobile tire is normally distributed with mu equals 23,000 miles and sigmaequals2500 miles.
(a) Find the probability that a tire will last between 28,000 and 30,500 miles.
(b) Find the probability that a tire will last more than 29,000 miles.
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We have the parameter of normal distribution :
Mu = 23000
Stdev = 2500
a. P(28000<X<30500) = ?
= P(28000-23000/2500 <Z< 30500-23000/2500)
= P(2<Z<3)
= .0214
b.
P(X>29000)
= P(Z> 29000-23000 / 2500)
= P(Z> 6000/2500)
= 0.0082
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