SOLUTION: I cannot figure out this problem:
The useful life of an electrical component is exponentially distributed with a mean of 4000 hours.
a. What is the probability the circuit will l
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Question 855493: I cannot figure out this problem:
The useful life of an electrical component is exponentially distributed with a mean of 4000 hours.
a. What is the probability the circuit will last more than 4500 hours?
b. What is probability the circuit will last between 4000 and 4750 hours?
c. What is probability the circuit will fail within the first 3250 hours?
I cannot figure out what the value of e is, or the value of lambda symbol.
Thanks for your help.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The expected value (E) is the mean of an exponential distribution
E[X] = 1 / lamda = 4000
lamda = 1 / 4000
f(x, lamda) = lamda * e^(-lamda*x)
note that e is equal to 2.71828
a) P(X>4500) = e^(-4500*lamda) = 2.71828^(-4500*1/4000)
P(X>4500) = 0.32
b)P(4000 < X < 4750) = P(X>4000) - P(X>4750)
P(4000 < X < 4750) = e^(-4000*lamda) - e^(-4750*lamda)
P(4000 < X < 4750) = .37 - .30 = 0.07
c)P(X<3250) = 1 - P(X>3250)
P(X<3250) = 1 - e^(-3250*lamda)= 1 - 0.44 = 0.56
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