Question

a certain brand of automobile tire has a mean life - span of 34,000 miles and a standard deviation of 2,400 miles. (assume the life - spans of the tires have a bell - shaped distribution.)
for the life - span of 38,000 miles, z - score is 1.67
(round to the nearest hundredth as needed.)
for the life - span of 31,000 miles, z - score is 1.25
(round to the nearest hundredth as needed.)
according to the z - scores, would the life - span of any of these tires be considered unusual?
no
yes
(b) the life - spans of three randomly selected tires are 29,200 miles, 38,800 miles, and 34,000 miles. using the empirical rule, find the percentile that corresponds to each life - span.
the life - span 29,200 miles corresponds to a percentile of 2
the life - span 38,800 miles corresponds to a percentile of
the life - span 34,000 miles corresponds to a percentile of 50
(round to the nearest whole number as needed.)

Explanation:

Step1: Recall z - score and percentile relationship

The empirical rule for a normal distribution: approximately 68% of the data lies within 1 standard - deviation of the mean ($z=\pm1$), 95% within 2 standard - deviations ($z = \pm2$), and 99.7% within 3 standard - deviations ($z=\pm3$). A z - score of 0 corresponds to the 50th percentile (the mean).

Step2: For a life - span of 31000 miles with $z = 1.25$

We know that the area to the left of $z = 0$ is 0.5. Using the standard normal distribution table or z - score properties, the area to the left of $z=1.25$ can be found. Looking up the value in the standard normal table, the area to the left of $z = 1.25$ is approximately 0.8944. So the percentile is approximately 89.

Answer:

89