Question

A certain brand of automobile tire has a mean life span of 36 comma 000 miles and a standard deviation of 2 comma 450miles.​(Assume the life spans of the tires have a​ bell-shaped distribution.)(a) The life spans of three randomly selected tires are 34,000​miles, 38,000 ​miles, and 31,000 miles. Find the​ z-score that corresponds to each life span.*For the life span of 34,000 miles, z-score is___________? ​(Round to the nearest hundredth as​ needed.)*For the life span of 38,000 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? Yes or No​(b) The life spans of three randomly selected tires are 33,550​miles, 38,450 ​miles, and 36,000 miles. Using the empirical​ rule, find the percentile that corresponds to each life span.The life span 33,550 miles corresponds to the _________?th percentile.The life span 38,450 miles corresponds to the__________ ?th percentile.The life span 36,000 miles corresponds to the________--- ?th percentile.

Answer 1

Explanation:

Let X be the mean lifespan of automobile tire

X is N(36,000, 2450)

The Z scores are calculated as (x-36000)/2450

x x-36000 Z=(x-36000)/2450

34000 -2000 -0.816326531 =-0.82

38000 2000 0.816326531 =0.82

31000 -5000 -2.040816327 =-2.04

No only if beyond |3| z score is there we say unusual

For 33550 miles

Z = -1

P(Z<-1) = 0.3413

So percentile is 34.13

For 36000 , z=0 so 50th percentile