Fuel Consumption and Cars The battery manufacturer varta sells a car battery wit
ID: 3170254 • Letter: F
Question
Fuel Consumption and Cars The battery manufacturer varta sells a car battery with 800 cold-cranking amps and advertises great performance even in biting cold weather. Varta claims that after sitting on a frozen Minnesota lake for 10 days at temperatures below 32 degree F, this battery will still have enough power to start a car. Suppose the actual probability of starting a car following this experiment is 0.75, and 15 randomly selected cars (equipped with this battery) are subjected to these grueling conditions. a. Find the probability that fewer than 10 cars will start. b. Find the probability that more than 12 cars will start.Explanation / Answer
Probability that car will start P= 0.75
Probability that car will not start Q= 1- P = 1-0.75 =0.25
(A)
Probability that fewer than 10 cars will start P(X>=10) =15C10 P10Q5 + 15C11 P11Q4 + 15C12 P12Q3 + 15C13 P13Q2 + 15C14 P14Q1 + 15C15 P15Q0
= [15!/{10!(15-10)!}](0.75)10(0.25)5 + [15!/{11!(15-11)!}](0.75)11(0.25)4 + [15!/{12!(15-12)!}](0.75)12(0.25)3 + [15!/{13!(15-13)!}](0.75)13(0.25)2 + [15!/{14!(15-14)!}](0.75)14(0.25)1+ [15!/{15!(15-15)!}](0.75)15(0.25)0
= 0.85
(B)
Probability that fewer than 13 cars will start =15C13 P13Q2 + 15C14 P14Q1 + 15C15 P15Q0
= [15!/{13!(15-13)!}](0.75)13(0.25)2 + [15!/{14!(15-14)!}](0.75)14(0.25)1+ [15!/{15!(15-15)!}](0.75)15(0.25)0
= 0.24
So, Probability that more than 12 cars will start P(X<=12) = 1- P(X>=13)
= 1-0.24
=0.76
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