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

0.8664