Question

The number of miles that a particular car can run before itsbattery wears out is exponentially distributed with an averageof 10,000 miles.Problems1. The mean of a probability distribution, , is given by . Find given that theaverage of this exponential distribution is 10,000.

Answer 1

`\lambda=10000` miles

Explanation:

We are given that

`\beta=10000 miles`

Exponential distribution function

`f(x)=(1)/(\beta)e^{-(x)/(\beta)}, x\geq 0`

0, otherwise

Using the formula

`f(x)=(1)/(10000)e^{-(x)/(10000)},x\geq0`

0, otherwise

`E(x)=\int_(-\infty)^(\infty)xf(x) dx`

Using the formula

`\lambda=E(x)=(1)/(10000)\int_(0)^(\infty)xe^{-(x)/(10000)}dx`

`\lambda=(1)/(10000)([-10000xe^{-(x)/(10000)}]^(\infty)_(0)+10000[-10000e^{-(x)/(10000)}]^(\infty)_(0))`

Using formula

`\int u\cdot vdx=u\int vdx-\int ((du)/(dx)\int vdx)dx`

`\lambda=(1)/(10000)(0-(10000)^2(0-1))=10000`

Hence, the value of
`\lambda=10000` miles