Question

The time, in 100 hours, that a student uses her game console over a year is a random variable X with probability density function
f(x) = { x if 0 2-x if 1<=x<2
0 otherwise}

The power (in number of kilowatt hours) expended by the student's game console each year is 48X^2+26. For these problems, please ensure your answers are accurate to within 3 decimals.


Part a)

Find the mean amount of power expended by the student's game console per year.

Part b)

Find the variance of power expended by the student's game console per year.

Answer 1

Answer:

Step-by-step explanation:

The concepts of probability density function, mean, and variance are used to solve the problem.

The probability density function can be used for the continuous random variable. The value of the probability density function gives the value at a particular sample point.

The expected value of the random variable is the mean value of the random variable. It can be obtained by summing the multiplied value of the observation with their corresponding probability.

The mean value of a random variable is the long run average value of repetitions of the experiment it represents, or the expected average outcome for many observations. It is denoted by

The variance of the distribution indicates that how far the values are spread from its mean.

That is, it measures the spread or variability.

(a)

According to the question, the random variable X has probability density function. The probability density function is,

$f\left( X \right) = \left\{ \begin{array}{l}\\X{\rm{ if }}0 < X < 1\\\\2 - X{\rm{ if }}1 \le X < 2\\\\0{\rm{ Otherwise}}\\\end{array} \right$

​  

According to the question, the power expanded by the student’s game console each year is

The mean amount of power expended by the student’s game console per year is calculated as:

$E\left( {48{X^2} + 26} \right) = 48E\left( {{X^2}} \right) + 26$

The is calculated as:

$\begin{array}{c}\\E\left( {{X^2}} \right) = \int\limits_0^1 {\left( {{x^2} \times x} \right)} \,\,dx + \int\limits_1^2 {{x^2}\left( {2 - x} \right)} \,\,dx\\\\ = \int\limits_0^1 {{x^3}} \,\,dx + \int\limits_1^2 {\left( {2{x^2} - {x^3}} \right)} \,\,dx\\\\ = \frac{1}{4}\left[ {{x^4}} \right] + \left[ {2\left[ {\frac{{{x^3}}}{3}} \right]_1^2 - \left[ {\frac{{{x^4}}}{4}} \right]_1^2} \right]\\\\ = \frac{1}{4} + \left[ {\frac{2}{3}\left( {{2^3} - {1^3}} \right) - \frac{1}{4}\left( {{2^4} - {1^4}} \right)} \right]=1.167\\\end{array}$

The mean amount of power expended by the student’s game console per year is calculated as:

$\begin{array}{c}\\E\left( {48{X^2} + 26} \right) = 48E\left( {{X^2}} \right) + 26\\\\ = 48 \times \left( {1.167} \right) + 26\\\\ = 56.112 + 26\\\\ = 82.016\\\end{array}$

​(b)

The variance of power expended by the student’s game console per year is calculated as,

$\begin{array}{c}\\V\left( {48{X^2} + 26} \right) = {\left( {48} \right)^2}V\left( {{X^2}} \right)\\\\ = {48^2}\left( {E\left( {{X^4}} \right) - {{\left[ {E\left( {{X^2}} \right)} \right]}^2}} \right)\\\end{array}$

The value of is calculated as:

$\begin{array}{c}\\E\left( {{X^4}} \right) = \int\limits_0^1 {\left( {{x^4} \times x} \right)} \,\,dx + \int\limits_1^2 {{x^4}\left( {2 - x} \right)} \,\,dx\\\\ = \int\limits_0^1 {{x^4}} \,\,dx + \int\limits_1^2 {\left( {2{x^4} - {x^5}} \right)} \,\,dx\\\\ = \frac{1}{5}\left[ {{x^5}} \right]_0^1 + \left[ {2\left[ {\frac{{{x^5}}}{5}} \right]_1^2 - \left[ {\frac{{{x^6}}}{6}} \right]_1^2} \right]\\\\ = \frac{1}{5} + \left[ {\frac{2}{5}\left( {{2^5} - {1^5}} \right) - \frac{1}{6}\left( {{2^6} - {1^6}} \right)} \right]=2.1\\\end{array}$

The variance of is calculated as:

$\begin{array}{c}\\V\left( {{X^2}} \right) = E\left( {{X^4}} \right) - {\left[ {E\left( {{X^2}} \right)} \right]^2}\\\\ = 2.1 - {\left( {1.167} \right)^2}\\\\ = 0.738\\\end{array}$

Therefore, the required value of variance is calculated as: