A certain brand of automobile tire has a mean life span of 39,000 miles and a st

Question

A certain brand of automobile tire has a mean life span of 39,000 miles and a standard deviation of 2,350 miles. (Assume the life spans of the tires have a bell-shapeo distribution.) (a) The life spans of three randomly selected tires are 35,000 miles, 37,000 miles, and 31,000 miles. Find the z-score that corresponds to each life span For the life span of 35,000 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? O No Yes (b) The life spans of three randomly selected tires are 34,300 miles, 43,700 miles, and 39,000 miles. Using the empirical rule, find the percentile that corresponds to each life span. The life span 34,300 miles corresponds to the th percentile The life span 43,700 miles corresponds to the th percentilie The life span 39,000 miles corresponds to the Dh percentile

Explanation / Answer

solutiona:

z=x-mean/stddev

for 35000 miles

z=35000-39000/2350

z=-1.70

For 37000 miles

Z=37000-39000/2350

Z=-0.85

For 31000 miles

Z=31000-39000/2350

Z=-3.40

as all z scores are negative

they are unusual

ANSWER:

YES

Solutionb:

for 34300

z=34300-39000/2350

Z=-2

2nd percentile

for 43700

z=43700-39000/2350

Z=2

98 percentile

For 39000

z=39000-39000/2350

z=0

50 th percentile

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