The life expectancy of a particular brand of tire is normally distributed with a

Question

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. 17. Refer to Exhibit 4. What is the random variable in this experiment? he life ex he normal distribution of this brand of tire ,000 miles None of the alternative answers is correct ANS A PTS: 18. Refer to Exhibit 4. What is the probability that a randomly selected tire will have a life of at least 30,000 miles? 4772 9772 None of the altemative answers is ANS: B PTS: 19. Refer to Exhibit 4. What is the probability that a randomly selected tire will have a life of at least 47,500 miles? ,0668 None of the altemative answers is correct ANS: C PTS:1 Refer to Exhibit 4. What percentage of tires will have a life 20 of 34,000 to 46,000 miles 38.49% None of the altenative answers is correct ANS B PTS:1

Explanation / Answer

for these problems we are requried to find z score from formula a score=(X-mean)/std deviation

and than corresponding z score probability from table; calculator or any software. PLease let me know if you want any help in calculating z value

17)

as distriution is about life expectancy of tire therefore option A is correct

18)

P(X>30000)=P(Z>(30000-40000)/2)=P(Z>-2) =1-P(Z<-2)=1-0.02275=0.97725

option B

19)P(X>47500)=P(Z>(47500-40000)/2)=P(Z>1.5) =1-P(Z<1.5)=1-0.9332=0.0668

option C

20)P(34000<X<46000)=P((34000-40000)/5000<Z<(46000-40000)/5000)=P(-1.2<Z<1.2)=0.8849-0.1151 =0.7698

option B

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