Question

a certain brand of automobile tire has a mean life span of 36,000 miles and a standard deviation of 2,350 miles. (assume the life spans of the tires have a bell-shaped distribution.)

(a) the life spans of three randomly selected tires are 33,000 miles, 37,000 miles, and 31,000 miles. find the z-score that corresponds to each life span.

for the life span of 33,000 miles, z-score is .
(round to the nearest hundredth as needed.)

for the life span of 37,000 miles, z-score is .
(round to the nearest hundredth as needed.)

for the life span of 31,000 miles, z-score is .
(round to the nearest hundredth as needed.)

according to the z-scores, would the life spans of any of these tires be considered unusual?
no
yes

(b) the life spans of three randomly selected tires are 31,300 miles, 40,700 miles, and 36,000 miles. using the empirical rule, find the percentile that corresponds to each life span.

the life span 31,300 miles corresponds to a percentile of .
the life span 40,700 miles corresponds to a percentile of .
the life span 36,000 miles corresponds to a percentile of .
(round to the nearest whole number as needed.)

Explanation:

Calculate the z-scores for part (a)

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Determine if any z-scores are unusual

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Calculate percentiles using the empirical rule for part (b)

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Answer:

Question a

For the life span of 33,000 miles, z-score is <blank>-1.28</blank>.

For the life span of 37,000 miles, z-score is <blank>0.43</blank>.

For the life span of 31,000 miles, z-score is <blank>-2.13</blank>.

According to the z-scores, would the life spans of any of these tires be considered unusual?

• No
• Yes (Correct answer)

Question b

The life span 31,300 miles corresponds to a percentile of <blank>3</blank>.

The life span 40,700 miles corresponds to a percentile of <blank>98</blank>.

The life span 36,000 miles corresponds to a percentile of <blank>50</blank>.