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Maths probability help please

On average, an engine part lasts 10 years. The length of time it lasts is exponentially distributed.
a) What is the probability that an engine part lasts more than 7 years?
b) On average, how long would 5 engine parts last if they are used one after another?

the formula is f(x) = lamda*e^(-lambda*x)
but which number is lambda? 10?
For part a, is x=7?
And I haven't got much idea how to start part b... Do I find the mean value of the length of time each part lasts, then times that by 5?

Thanks in advance
Reply 1
Original post by Bookworm524
On average, an engine part lasts 10 years. The length of time it lasts is exponentially distributed.
a) What is the probability that an engine part lasts more than 7 years?
b) On average, how long would 5 engine parts last if they are used one after another?

the formula is f(x) = lamda*e^(-lambda*x)
but which number is lambda? 10?
For part a, is x=7?
And I haven't got much idea how to start part b... Do I find the mean value of the length of time each part lasts, then times that by 5?

Thanks in advance

lambda = 1/mean
https://en.wikipedia.org/wiki/Exponential_distribution
If lambda was 10, then plugging the mean value of 10 years for x would give a tiny number for e^(-lambda*x) so e^(-100).

For part a) you want to find P(x>7) so the cumulative probability which requires you to integrate the pdf.
Reply 2
Original post by Bookworm524
On average, an engine part lasts 10 years. The length of time it lasts is exponentially distributed.
a) What is the probability that an engine part lasts more than 7 years?
b) On average, how long would 5 engine parts last if they are used one after another?

the formula is f(x) = lamda*e^(-lambda*x)
but which number is lambda? 10?
For part a, is x=7?
And I haven't got much idea how to start part b... Do I find the mean value of the length of time each part lasts, then times that by 5?

Thanks in advance

No lambda will be 1/10 post some working
Reply 3
Original post by mqb2766
lambda = 1/mean
https://en.wikipedia.org/wiki/Exponential_distribution
If lambda was 10, then plugging the mean value of 10 years for x would give a tiny number for e^(-lambda*x) so e^(-100).

For part a) you want to find P(x>7) so the cumulative probability which requires you to integrate the pdf.

For the integral is the limits going to be 0 and 7 then? Thanks!
Reply 4
Original post by Bookworm524
For the integral is the limits going to be 0 and 7 then? Thanks!


Its generally useful to sketch the distribution (decaying exponential) and area youre calculating.
What would the 0 -> 7 correspond to in terms of the cumulative probability?
Reply 5
Original post by Bookworm524
For the integral is the limits going to be 0 and 7 then? Thanks!


You want the prob of more than 7 so ...
(edited 6 months ago)

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