Question

8. Suppose the probability density function of the length of computer cables is f(x) = 0.1

from 1200 to 1210 millimeters.

a) Determine the mean and standard deviation of the cable length.

b) If the length specifications are 1195 < x < 1205, what proportions of cables are

within specifications?

Answer 1

$a:\\EX=\int_{1200}^{1210}{xf\left( x \right) dx}=\int_{1200}^{1210}{x\cdot 0.1dx}=0.05\left( 1210^2-1200^2 \right) =1205\\EX^2=\int_{1200}^{1210}{x^2f\left( x \right) dx}=\int_{1200}^{1210}{x^2\cdot 0.1dx}=\frac{0.1}{3}\left( 1210^3-1200^3 \right) =1.45203\times 10^6\\DX=EX^2-\left( EX \right) ^2=1.45203\times 10^6-1205^2=8.33333\\\sigma X=\sqrt{DX}=\sqrt{8.33333}=2.88675\\b:\\P\left( 1195<X<1205 \right) =P\left( 1200<X<1205 \right) =\int_{1200}^{1205}{0.1dx}=0.5\\$