(b) The life spans of three randomly selected tires are 31 comma 80031,800 miles

Question

(b) The life spans of three randomly selected tires are

31 comma 80031,800

miles,

40 comma 20040,200

miles, and

36 comma 00036,000

miles. Using the empirical rule, find the percentile that corresponds to each life span.

A certain brand of automobile tire has a mean life span of 36,000 miles and a standard deviation of 2,100 miles. (Assume the life spans of the tires have a bell-shaped distribution.) (a)The life spans of three randomly selected tires are 35 00 miles, 38.000 mies and 32.0 miles. Find the z-score that corresponds to ea life span. For the life span of 35,000 miles, z-score is 0.48 Round to the nearest hundredth as needed.) For the life span of 38,000 miles, z-score is 0.95 Round to the nearest hundredth as needed.) For the life span of 32,000 miles, z-score is -1.90 . (Round to the nearest hundredth as needed.) According to the z-sccres, would the life spans of any of these tires be considered unusual? No Yes (b) The life spans of three randomly selected tires are 31,800 miles, 40,200 miles, and 36,000 miles. Using the empirical rule, find the percenti e that corresponds to each life span The life span 31,800 miles corresponds to the 25 th percentile. The life span 40,200 miles corresponds to the 75 th percentile. The life span 36,000 miles corresponds to the 50 th percentile. Enter your answer in the edit fields and then click Check Answer. All parts showing Clear All Final Check

Explanation / Answer

Ans:

Given that

mean=36000

standard deviation =2100

a)

z=(35000-36000)/2100=-0.48

z=(38000-36000)/2100=0.95

z=(32000-36000)/2100=-1.90

No,as none of the z score is less than -2 or greater than 2

b)

For 31800

z=(31800-36000)/2100=-2

P(z<-2)=0.02275

2.28th percentie

For 40200

Z=(40200-36000)/2100=2

P(z<2)=0.9772

97.72th percentile

For 36000

z=0

P(z<0=0.5

50th percentile

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