Question

Suppose that the probability density function ofthe length of computer cables is f (x) = 0.1 from 1200 to1210 millimeters.a. Determine the mean and standard deviation of the cablelength.b. If the length specifications are 1195 < x < 1205 millimeters,what proportion of cables is within specifications?

Answer 1

Notice that $f(x)=0.1$ for $1200\le x\le1210$ implies that $f(x)=0$ elsewhere, since

$\displaystyle\int_{1200}^{1210}f(x)\,\mathrm dx=1$

where $X=x$ is a random variable representing cable lengths according to the PDF $f(x)$.

a. By definition of expectation, the mean is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x\,\mathrm dx=1205$

The variance is

$\operatorname{Var}[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=0.1\int_{1200}^{1210}x^2\,\mathrm dx=\frac{4,356,100}3$

so that the variance is $\frac{25}3$, making the standard deviation $\frac5{\sqrt3}$.

b. The proportion of cables within specs is

$P(1195<X<1205)=\displaystyle\int_{1195}^{1205}f(x)\,\mathrm dx=0.1\int_{1200}^{1205}\mathrm dx=0.5$