An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the warranty code is the free-replacement period; the second number is the prorated-credit period. Under this warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charge to the consumer. If the battery fails after 31 months but within 54 months, the manufacturer provides a prorated credit toward the purchase of a new battery. The manufacturer assumes that X, the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months.

1. If the manufacturer's assumptions are correct, it would reed to replace of its batteries free.2. The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect. A standard deviation of results in a 1.07% replacement rate.3. Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

1) If the manufacturer's assumptions are correct, it would reed to replace 0.621% of its batteries free

2) The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect. A standard deviation of 6.084 results in a 1.07% replacement rate

3) Using the revised standard deviation for battery life, the percentage of the manufacturer's batteries that don't qualify for free replacement but do qualify for the prorated credit = 92%

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 45 months

Standard deviation = σ = 5.6 months.

Batteries that fail within the first 31 months get a free replacement and batteries that fail after 31 months but within 54 months get a prorated credit toward the purchase of a new battery.

1) Percentage of batteries that'll qualify for a replacement are the batteries that fail within 31 months. That is, P(X ≤ 31)

To obtain this, we first normalize or standardize 31 months.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (31 - 45)/5.6 = - 2.50

The required probability

P(X ≤ 31) = P(z ≤ -2.50)

We'll use data from the normal probability table for these probabilities

P(X ≤ 31) = P(z ≤ -2.50) = 0.00621 = 0.621%

2) The company finds that the percentage of free batteries replaced is actually 1.07%

P(X ≤ 31) = 1.07% = 0.0107

Let the z value of 31 months under this new normal distribution be z'

P(X ≤ 31) = P(z ≤ z') = 0.0107

From the normal distribution table,

z' = -2.301

z' = (x - μ)/σ

-2.301 = (31 - 45)/σ

σ = (-14) ÷ (-2.301) = 6.084 months

3) Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

That is, P(31 < X ≤ 54)

We first normalize 31 and 54 using the revised standard deviation

For 31

z = (x - μ)/σ = (31 - 45)/6.084 = - 2.30

For 54

z = (x - μ)/σ = (54 - 45)/6.084 = 1.48

The required probability = P(31 < X ≤ 54)

= P(-2.30 < z ≤ 1.48)

Using the normal distribution tables

P(31 < X ≤ 54) = P(-2.30 < z ≤ 1.48)

= P(z ≤ 1.48) - P(z < -2.30)

= 0.93056 - 0.0107

= 0.91986 = 91.986% = 92% to nearest whole number.

Hope this Helps

it 2

step-by-step explanation:

8^1/3=2

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