QUESTION 1 1 points Save Answer Use the following scenario and data for all ques

Question

QUESTION 1 1 points Save Answer Use the following scenario and data for all questions The useful life of one type of truck tire has a normal distribution with a mean of -529 thousand miles and a standard deviation of . 7.50 thousand miles of a tire selected at random. thousand miles. Use x to denote the life in The probability that the useful life of a tire selected at random is shorter than or equal to 47.82 thousand miles is closest to 0 2324 0 2676 0.6240 0.7324 O none of the above QUESTION 2 1 points Save Answer The probability that the useful life of a tire selected at random is shorter than or equal to 58.64 thousand miles is closest to 0.2081 02939 0.7939 0.8187 none of th above QUESTION 3 1 points Save Answer The probability that the useful life of a tire selected at random is longer than 60.25 thousand miles is closest to O0.1514 .3485 08485 1.0333 none of the above

Explanation / Answer

µ = 52.5

= 7.5

Since, we have population mean and standard deviation we can use Z-test.

Z = (X - µ) /

1.) X = 47.82

Z = (47.82 - 52.5) / 7.5

= -0.624

P(useful life < X) can be found out using Z-test table.

P = 0.2676

2.) Similarly as previous one.

X = 58.64

Z = (58.64 - 52.5) / 7.5

= 0.8186

P(useful life < X) can be found out using Z-test table.

P = 0.7939

3.) X = 60.25

Z = (60.25 - 52.5) / 7.5

= 1.033

P(useful life > X) can be found out using Z-test table.

P(useful life > X) = 1 - P(useful life < X )

= 0.1515

4.) Similar as previous one

X = 55.85

Z = (55.85 - 52.5) / 7.5

= 0.4466

P(useful life > X) can be found out using Z-test table.

P(useful life > X) = 1 - P(useful life < X )

= 0.3264

5.) P(47.25<X<56.25) = P(X<56.25) - P(X<47.25)

We will do exactly the way we did prevoius part to find out these probabilities.

Z = (56.25 - 52.5) / 7.5

= 0.5

P(X<56.25) = 0.691

Z = (47.25-52.5) / 7.5

= -0.7

P(X<47.25) = 0.241

P(47.25<X<56.25) = 0.691-0.41

~ 0.4495

6.) Doing it exaclty in same way as we did previous one.

P(45.8<X<50.8) = 0.2223

7.) Similarly as previous 2 questions

P(54.75<X<58.75) = 0.1788

8.) Now we know that probability of Useful life less than warranted life (X) should be 0.1

Z = (X-µ) /

P(Useful life<X) = 0.1

From this we can calculate Z value using inverse normal distribution table.

Z = -1.281

Now, putting this value in equation of Z

-1.281 = (X-52.5)/7.5

X ~ 43

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