Question

A tire company makes tires that have a normal distribution with a mean of 65,000 miles with a standard deviation of 3000 miles prior to needing replacement. Find the probability that a tire lasts no more than 62,500 miles A tire company makes tires that have a normal distribution with a mean of 65,000 miles with a standard deviation of 3000 miles prior to needing replacement. Find the probability that a tire lasts at least 68,500 miles. A tire company makes tires that have a normal distribution with a mean of 65,000 miles with a standard deviation of 3000 miles prior to needing replacement. Find the probability that a tire lasts between 60,500 to 69,500 miles. A tire company makes tires that have a normal distribution with a mean of 65,000 miles with a standard deviation of 3000 miles prior to needing replacement. A tire that wears out in the bottom 3% of miles lasted before needing replacement is considered very poorly made. Find the highest number of miles that a tire would have to last to be considered very poorly made

Answer 1

Solution:- Given that mean 65000, sd = 3000,

=> P(X <= 62500) = P((X-mean)/sd < (62500-65000)/3000)
= P(Z < -0.8333)
= 1 - P(Z < 0.8333)
= 1 - 0.7967
= 0.2033

=> P(X > 68500) = P(Z > (68500-65000)/3000)
= P(Z > 1.1667)
= 0.1210

=> P(60500 < X < 69500) = P((60500-65000)/3000 < Z < (69500-65000)/3000 )
= P(-1.5 < Z < 1.5)
= 0.8664

=> for P(X < x) = 0.03, for Z = -1.88

X = mean + Z*Sd = 65000 - (1.88*3000) = 59360