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

Answer:

$\lambda=10000$ miles

Step-by-step explanation:

We are given that

$\beta=10000 miles$

Exponential distribution function

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

             0, otherwise

Using the formula

$f(x)=\frac{1}{10000}e^{-\frac{x}{10000}},x\geq0$

              0, otherwise

$E(x)=\int_{-\infty}^{\infty}xf(x) dx$

Using the formula

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

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

Using formula

$\int u\cdot vdx=u\int vdx-\int (\frac{du}{dx}\int vdx)dx$

$\lambda=\frac{1}{10000}(0-(10000)^2(0-1))=10000$

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