SOLUTION: A large flashlight is powered by 5 batteries. Suppose that the life of a battery is normally distributed with mean 150 and standard deviation 15. The flashlight will cease function

Question 1166037: A large flashlight is powered by 5 batteries. Suppose that the life of a battery is normally distributed with mean 150 and standard deviation 15. The flashlight will cease functioning if one or more of its batteries go dead. Assuming the lives of batteries are independent. What is the probability that flashlight will operate more than 130 hours?
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
So what is the probability all 5 batteries will last more than 130 hours?
the probability of one is z=(x-mean)/sd or z>(130-150)/15 or z> -1.33
this has probability of 0.9082
for all 5 batteries to do this would be 0.9082^5=0.6179.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
this looks like it's a binomial distribution probability type of problem.

the formula for binomial distribution probability is:

p(x) = a ^ x * b ^ (n-x) * c(n,x)

c(n,x) is equal to n! / (x! * (n-x)!)

in your problem:
n is equal to 5.
x is equal to any integer from 0 to 5 inclusive.
a is equal to the probability that the life of a randomly selected battery is less than 130 hours.
b is equal to the probability that the life of a randomly selected battery is greater than 130 hours.
the probability of b would be equal to 1 minus the probability of a.

using a normal distribution calculator such as the one found at https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html, the probability that any randomly selected battery will have a life less than 130 hours is equal to 0.09121.
the probability of a randomly selected battery having a life greater than 130 hours is therefore equal to 1 - .09121 = .90879.

here's a display of the calculator results for the probability of a random selected battery having a life of less than 30.

the binomial distribution probability formula becomes:

p(x) = .09121 ^ x * .90879 ^ (5-x) * c(5,x)

the flashlight will not operate for more than 130 hours if 1 or more of its batteries dies in less than 130 hours.

this means that the flashlight will operate for more than 130 hours if 0 of its batteries dies in less than 130 hours.

what you are then looking for is x equal to 0 in the formula of p(x) = .09121 ^ x * .90879 ^ (5 - x) * c(5,x)

the formula becomes p(0) = .09121 ^ 0 * .90879 ^ (5 - 0) * c(5,0).

c(5,0) is equal to 5! / (0! * 5!) which is equal to 5! / 5! which is equal to 1.

the formula becomes p(0) = .09121 ^ 0 * .90879 ^ 5 which is equal to .6198943783.

that means that the probability that 0 batteries will fail in less than 130 hours is .6198943783.

if 0 batteries fail in less than 130 hours, then the flashlight will be operating for more than 130 hours.

i believe that .6198943783 is the answer they are looking for.

it is the probability that 0 of the flashlight batteries dies in less than 130 hours.

you can round that answer to whatever number of decimal points they require.

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