Question

1.The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. What is the probability that a randomly selected tire will have a life of of less than 30,000 miles?

Answer 1

Let life expectancy of that particular brand of tire be denoted by random variable X.

Mean = 40,000, Std. dev = 5,000.

We need to calculate probability of X<= 30,000

Z score at (X=30,000) = (X - mean(X))/Std. Dev = (30,000 - 40,000) / 5,000 = -2

Since, standard normal distribution is symmetric, calculating probability( Z < -2 )= probability( Z > 2)

Probability ( Z > 2 ) = 0.5 - Probability( 0 < Z < 2 )

At Z = 2, According to Z-table (cumulative standard normal distribution table),

Probability( 0 < Z < 2 ) = 0.47725

Probability ( Z > 2 ) = 0.5 -  0.47725 = 0.02275

So, probability of a randomly selected tire having a life of less than 30,000 miles = 0.02275