Answer to Question #127568 in Statistics and Probability for Jessica

Question

Integration by parts is required. The probability density function for the diameter of a drilled hole in millimeters is

10e^(-10(x-60)) for x > 6.0 mm


Although the target diameter is 6.0 millimeters, vibrations, tool wear, and other nuisances produce diameters larger than 6.0 millimeters.



a) Determine the mean and variance of the diameter of the holes.

Mean = _______ (Round your answer to 1 decimal place.)
Variance = ______ (Round your answer to 2 decimal places.)

b) Determine the probability that a diameter exceeds 6.1 millimeters. (Round your answer to 3 decimal places.)

Answer 1

Note

The property of a probability density function: integral over all space is equal to 1.

$f(x)=10e^{-10(x-60)},\;x>6$

is not a probability density function, because the property fails.

Let's try to fix it. Let the probability density function be

$f(x)=10e^{-10(x-6)},\;x>6$.

Then the property succeeds:

$\int\limits_{-\infty}^{\infty} f(x) \;dx = \int\limits_{6}^{\infty}10e^{-10(x - 6)} dx=-e^{-10(x-6)}|_6^{\infty}=0-(-1)=1$

a)

Solution

$M(X)=\int\limits_{-\infty}^{\infty} x\cdot f(x) \;dx = \int\limits_{6}^{\infty}10x\cdot e^{-10(x - 6)} dx=\\ =x\cdot(-e^{-10(x-6)})|_6^{\infty}-\int\limits_6^{\infty}-e^{-10(x-6)}dx=\\ =0-(-6)-0.1(e^{-10(x-6)})|_6^{\infty}=6-(0-0.1)=6.1\\[0.3cm]$

$M(X^2)=\int\limits_{-\infty}^{\infty} x^2\cdot f(x) \;dx = \int\limits_{6}^{\infty}10x^2\cdot e^{-10(x - 6)} dx=\\ = x^2\cdot(-e^{-10(x-6)})|_6^{\infty}-\int\limits_6^{\infty}2x(-e^{-10(x-6)}) dx =\\ = 0-(-36)-\left(\left(-0.2\right)\cdot\int\limits_{6}^{\infty}10x\cdot e^{-10(x - 6)} dx\right)=36+0.2\cdot6.1=37.22$

$D(X)=M(X^2)-M^2(X)=37.22-37.21=0.01$

Answer

Mean = 6.1

Variance = 0.01

b)

$\int\limits_{6.1}^{\infty} f(x) \;dx = \int\limits_{6.1}^{\infty}10e^{-10(x - 6)} dx=-e^{-10(x-6)}|_{6.1}^{\infty}=0-(-e^{-1})=1/e\approx0.368$