Question

A test instrument needs to be calibrated periodically to prevent measurement errors. After some time of use without calibration, it is known that the probability density function of the measurement error is f(x)=1-0.5x for 0 < x < 2 millimeters. Determine the mean, variance and standard deviation of the random variable.

Answer 1

Answer:

$E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}$

For the variance we need to calculate first the second moment given by:

$E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx$

And after solve the integral we got:

$E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}$

And for this case the variance would be:

$Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}$

And the deviation would be:

$Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}$

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Solution to the problem

For this case w ehave the following density function:

$f(x) = 1 -0.5x , 0 < x< 2$

We can determine the mean with the following integral:

$E(X) = \int_{0}^2 x(1-0.5x) dx$

And if we solve the integral we got:

$E(X) =\frac{x^2}{2} -\frac{0.5}{3}x^3 \Big|_0^2 = (\frac{2^2}{2}- \frac{0.5}{3}*(2^2)) -0 = 0.667 = \frac{2}{3}$

For the variance we need to calculate first the second moment given by:

$E(X^2) = \int_{0}^2 x^2 (1-0.5x) dx=\int_{0}^2 x^2-0.5x^3 dx$

And after solve the integral we got:

$E(X^2) = \frac{x^3}{3} -\frac{0.5}{4} x^4 \Big|_0^2 = \frac{2^3}{3} -\frac{0.5}{4} 2^4 = \frac{8}{3} -2 = \frac{2}{3}$

And for this case the variance would be:

$Var(X) = E(X^2) -[E(X)]^2 = \frac{2}{3} -(2/3)^2 = \frac{2}{9}$

And the deviation would be:

$Sd(X) = \sqrt{\frac{2}{9}}= \frac{\sqrt{2}}{3}$