Question

Quick Start Company makes 12-volt car batteries. From historical data, the company knows that the life of such a battery is a normally distributed random variable with a mean life of 44 months and a standard deviation of 1010 months.
(a) What percentage of Quick Start 12-volt batteries will last between 32 months and 53 months?

(b) If Quick Start does not want to make refunds for more than 10% of its batteries under a full-guarantee policy, how long should the company guarantee the 12-volt batteries?

months
(c) Seventy-five 12-volt batteries are randomly selected n=75. What is the probability that the mean lifetime of the batteries in this sample will be between 42 and 43 months?

(Input answer to four decimal places)

Answer 1

Mean life of battery = 44 months

standard deviation of battery = 10 months

(a) Here if x is the lifespan of a random battery

Pr(32 month < x < 53 month) = Pr(x < 53 months ; 44 months ; 10 months) - Pr(x < 32 months ; 44 months ; 10 months)

Z₂ = (53 - 44)/10 = 0.9

Z₁ = (32 - 44)/10 = -1.2

Pr(32 month < x < 53 month) = Pr(x < 53 months ; 44 months ; 10 months) - Pr(x < 32 months ; 44 months ; 10 months)

= Pr(Z < 0.9) - Pr(Z < -1.2)

= 0.8159 - 0.1151 = 0.7009

(b) Here the battery lifespan of top 10% of battery. So we have to find 10% percentile heree

Pr(x < x₀ ; 44 ; 10) = 0.10

so here Z value would be

Z = -1.28155

(x₀ - 44)/10 = -1.28155

x₀ = 31.1845 months

(C) Here n = 75

standard error of sample mean = 10/sqrt(75) = 1.1547 months

Here the sample mean of lifetime of the batteries are

Pr(42 months < x < 43 months) = Pr( < 43 months ; 44 months ; 1.1547 month) - Pr( <44 months ; 44 months ; 1.1547 months)

Z₂ = (43 - 44)/1.1547 = -0.866

Z₁ = (42 - 44)/1.1547 = -1.732

Pr(42 months < x < 43 months) = Pr( < 43 months ; 44 months ; 1.1547 month) - Pr( <44 months ; 44 months ; 1.1547 months)

= Pr(Z < -0.866) - Pr(Z < -1.732)

= 0.1932 - 0.0416

= 0.1517