Science:MATH105 Probability/Lesson 2 CRV/2.12 Example

Problem

The length of time X, needed by students in a particular course to complete a 1 hour exam is a random variable with PDF given by

$f(x)=\{\begin{array}{ll}k(x^2+x)& \text{if }0\le x\le 1,\\ 0& \text{else}\end{array}$

For the random variable X,

1. Find the value k that makes f(x) a probability density function (PDF)
2. Find the cumulative distribution function (CDF)
3. Graph the PDF and the CDF
4. Use the CDF to find
   1. Pr(X ≤ 0)
   2. Pr(X ≤ 1)
   3. Pr(X ≤ 2)
5. find the probability that that a randomly selected student will finish the exam in less than half an hour
6. Find the mean time needed to complete a 1 hour exam
7. Find the variance and standard deviation of X

Solution

Part 1

The given PDF must integrate to 1. Thus, we calculate

$\begin{array}{rl}1& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx\\ & =k\underset{0}{\overset{1}{\int }}(x^2+x)dx\\ & =k(\frac{x^3}{3}+\frac{x^2}{2}){|}_0^1\\ & =k\left(\frac{5}{6}\right)\end{array}$

Therefore, k = 6/5.

Part 2

The CDF, F(x), is area function of the PDF, obtained by integrating the PDF from negative infinity to an arbitrary value x.

If x is in the interval (-∞, 0), then

$\begin{array}{rl}F(x)& =\underset{-\mathrm{\infty }}{\overset{x}{\int }}f(t)dt\\ & =\underset{-\mathrm{\infty }}{\overset{x}{\int }}0dt\\ & =0\end{array}$

If x is in the interval [0, 1], then

$\begin{array}{rl}F(x)& =\underset{-\mathrm{\infty }}{\overset{x}{\int }}f(t)dt\\ & =\underset{-\mathrm{\infty }}{\overset{0}{\int }}f(t)dt+\underset{0}{\overset{x}{\int }}f(t)dt\\ & =0+\frac{6}{5}(\frac{x^3}{3}+\frac{x^2}{2})\\ & =\frac{6}{5}(\frac{x^3}{3}+\frac{x^2}{2})\\ \end{array}$

If x is in the interval (1, ∞) then

$\begin{array}{rl}F(x)& =\underset{-\mathrm{\infty }}{\overset{x}{\int }}f(t)dt\\ & =\underset{-\mathrm{\infty }}{\overset{0}{\int }}f(t)dt+\underset{0}{\overset{1}{\int }}f(t)dt+\underset{1}{\overset{x}{\int }}f(t)dt\\ & =0+\frac{6}{5}(\frac{x^3}{3}+\frac{x^2}{2}){|}_0^1+0\\ & =\frac{6}{5}\cdot \frac{5}{6}\\ & =1\end{array}$

Note that the PDF f is equal to zero for x > 1. The CDF is therefore given by

$F(x)=\{\begin{array}{ll}0& \text{if }x<0,\\ \frac{6}{5}(\frac{x^3}{3}+\frac{x^2}{2})& \text{if }0\le x\le 1,\\ 1& \text{if }x>1.\end{array}$

Part 3

The PDF and CDF of X are shown below.

Part 4

These probabilities can be calculated using the CDF:

$\begin{array}{rl}\text{Pr}(X\le 0)& =F(0)=\frac{6}{5}(\frac{0^3}{3}+\frac{0^2}{2})=0\\ \text{Pr}(X\le 1)& =F(1)=\frac{6}{5}(\frac{1^3}{3}+\frac{1^2}{2})=\frac{6}{5}\frac{5}{6}=1\\ \text{Pr}(X\le 2)& =1\\ \end{array}$

Note that we could have evaluated these probabilities by using the PDF only, integrating the PDF over the desired event.

Part 5

The probability that a student will complete the exam in less than half an hour is Pr(X < 0.5). Note that since Pr(X = 0.5) = 0, since X is a continuous random variable, we an equivalently calculate Pr(x ≤ 0.5). This is now precisely F(0.5):

$F(0.5)=\frac{6}{5}(\frac{0.5^3}{3}+\frac{0.5^2}{2})=\frac{6}{5}(\frac{1}{24}+\frac{1}{8})=\frac{6}{5}\cdot \frac{1}{6}=\frac{1}{5}$

Part 6

The mean time to complete a 1 hour exam is the expected value of the random variable X. Consequently, we calculate

$\begin{array}{rl}\mathbb{𝔼}(X)& =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}xf(x)dx\\ & =\frac{6}{5}\underset{0}{\overset{1}{\int }}x(x^2+x)dx\\ & =\frac{6}{5}\underset{0}{\overset{1}{\int }}{x}^3+x^{2}dx\\ & =\frac{6}{5}(\frac{x^4}{4}+\frac{x^3}{3}){|}_0^1\\ & =\frac{6}{5}(\frac{1}{4}+\frac{1}{3})\\ & =\frac{7}{10}\end{array}$

Part 7

To find the variance of X, we use our alternate formula to calculate

$\begin{array}{rl}\text{Var}(X)& =\mathbb{𝔼}(X^2)-[\mathbb{𝔼}(X){]}^2\\ & =\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2}f(x)dx-{\left(\frac{7}{10}\right)}^2\\ & =\frac{6}{5}\underset{0}{\overset{1}{\int }}{x}^2(x^2+x)dx-\frac{49}{100}\\ & =\frac{6}{5}\underset{0}{\overset{1}{\int }}{x}^4+x^{3}dx-\frac{49}{100}\\ & =\frac{6}{5}(\frac{x^5}{5}+\frac{x^4}{4}){|}_0^1-\frac{49}{100}\\ & =\frac{6}{5}(\frac{1}{5}+\frac{1}{4})-\frac{49}{100}\\ & =\frac{54}{100}-\frac{49}{100}\\ & =\frac{1}{20}\end{array}$

Finally, we see that the standard deviation of X is

$\text{StdDev}(X)=\sqrt{\frac{1}{20}}=\frac{1}{2\sqrt{5}}$