Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. what is the probability that a tire lasts between 54,000 and 66,000 miles?
To solve for the P(54,000≤x≤66000) we proceed as follows: z-score=(x-μ)/σ μ-60000 σ-4000 thus: when x=66,000 z-score=(66000-60000)/4000=1.5 P(z≤1.5)=0.9332 when x=54000 z=(54000-60000)/4000 z=-1.5 P(z≤-1.5)=0.0668 thus P(54,000≤x≤66000) =P(z≤1.5)-P(z≤-1.5) =0.9332-0.0668 =0.8664
