Question

9. The probability density function for a diameter of a drilled hole in millimeters is f(x) =

10e^−10(x−5)

for x > 5 mm. Although the target diameter is 5 mm, vibrations, tool wear,

and other nuisances produce diameters larger than 5 mm.

a) Determine the mean and variance of the diameter of the holes. [Hint: Use

integration by parts.]

b) Determine the probability that the hole exceeds 5.1 mm.

Answer 1

$EX=\int_5^{+\infty}{x\cdot 10e^{-10\left( x-5 \right)}dx}=\\=-xe^{-10\left( x-5 \right)}|_{5}^{+\infty}+\int_5^{+\infty}{e^{-10\left( x-5 \right)}dx}=\\=5+\frac{1}{10}=5.1\\EX^2=\int_5^{+\infty}{x^2\cdot 10e^{-10\left( x-5 \right)}dx}=\\=-x^2e^{-10\left( x-5 \right)}|_{5}^{+\infty}+\int_5^{+\infty}{2xe^{-10\left( x-5 \right)}dx}=\\=25+\frac{1}{5}EX=25+\frac{1}{5}\cdot 5.1=26.02\\DX=EX^2-\left( EX \right) ^2=26.02-5.1^2=0.01\\b:\\P\left( X>5.1 \right) =\int_{5.1}^{+\infty}{10e^{-10\left( x-5 \right)}dx}=\\=-e^{-10\left( x-5 \right)}|_{5.1}^{+\infty}=e^{-1}$