The average life of a brand of automobile tires is 30,000 miles, with a standard deviation of 2,000 miles. if a tire is selected and tested, find the probability that it will have a lifetime between 25,000 and 31,000 miles. assume the variable is normally distributed. give your answer rounded to four decimal places (to the nearest ten-thousandth).

Find the z scores for 25000 and 28000.

z1=( x – average)/ standard deviation =( 25000-30000)/ 2000 = -2.5

z2=( x – average)/ standard deviation =( 31000-30000)/ 2000 = 0.5

Find the corresponding P(z).

P (z1) = 0.00621

P (z2) = 0.15866

Subtract the probabilities to find the probability for the range,

(25000<= x <=28000)

P(x) = 0.15866-0.00621 = 0.1524

astronomers is your final answer! : )

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