toyota sibrogen tires why Go x Help Save &Exit Submit Chapter 4s Assignment 5 Pr

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toyota sibrogen tires why Go x Help Save &Exit Submit Chapter 4s Assignment 5 Problem 4S-23 Auto batteries have an average life of 3.0 years. Battery life is normally distributed with a mean of 3.0 years and a standard deviation of 0.39 year. The batteries are warranted to operate for a minimum of 2 years If a bettery fals within the warranty period, it will be replaced with a new battery at no charge. The company sells and installs the batteries. Also, the usual $7 instalation charge will be walved points Use table and TableB. a. What percentage of batteries would you expect to fal before the warranty period expires? (Round your z value to 2 declmal places. Round probabilities to 4 decimal places. Round your final answer to 2 declmal places Omit the "sign in your response Percentage % b. A compettor is offering a warranty of 28 months on its premium battery. The manager of this company is toying with the idea of using the same bantery with a different exterlor lebeling it as a premium batery, and offlering a 28-month warranty on n Whet percentage of the batteries would you expect to fail before this new warranty period expirest Round your z value to 2 decimal places. Round probabilities to 4 decimal places. Round your final answer to 2 decimal places. Omit thesign in your response) Percentage

Explanation / Answer

1.

Mean = 3 years

S.D =0.39 years

We have to find the Probability that a battery will last for less than warranty period of 2 years

Step 1

First find Z value [i.e. we find how many standard deviations it (the warranty years)   lies away from mean]

Z=(Given value- mean )/ s.d = ( 2 - 3 ) /0.39 =    - 2.5641

       

Step 2

Now we find the probability corresponding to the Z value from the table give [table 2]

Go to row Z -2.5 and then 0.06 in the column (Z= -2.56 -2.5+0.06)

P(X<2)

= P (Z< - 2.5641)

=0.0052

Probability=0.0052 *100 = 0.52 %

2.

Step 1

New warranty period = 28 months = 28/12 = 2.3333 years

We have to find the Probability that a battery will last for less than warranty period of 2.3333 years

Step 2

Find Z value

Z= ( 2.3333-3 ) /0.39 =    -1.0795

    

Step 3

       Now we find the probability corresponding to the Z value from the table give [table 2]

Z= -1.07 Go to row Z -1.0 and then 0.07 in the column

P(X<2.33)

= P (Z< -1.0795)

=0.1423

Probability=0.1423*100 = 14.23 %

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