Learning Objectives:

• Understand relation between Probability Density Function (PDF) and Cumulative Distribution Function (CDF).

---

Solved Example: 9-1-01

Let X denotes the time a person waits for a bus to arrive as shown below. Calculate the probability that a person waits 90 seconds for the bus to arrive.

A. 37.5%

B. 57.5%

C. 71.9%

D. 78.8%

90 seconds is 1.5 minutes.

\[P(X \leq 1.5) = 0.5 \times 1 + \dfrac{1}{2} \times 0.5 \times (0.5 + 0.375)= 0.719\ \mathrm{or\ } 71.9\%\]

Correct Answer: C

Solved Example: 9-1-02

X is a random variable with a uniform probability density function in the interval [-2, 10]. For Y = 2X - 6, the conditional probability P(Y ≤ 7 | X ≥ 5):

A. 0.3

B. 0.4

C. 0.5

D. 0.6

The area under the curve must be 1, \[p=\dfrac{1}{12}\] \[{{f}_{x}}\left( x \right)=\left\{ \begin{matrix} \dfrac{1}{12}~~~-2\le x\le 10 \\ 0~~~~\mathrm{otherwise} \\ \end{matrix} \right\}\] Similarly, \[{{f}_{y}}\left( y \right)=\left\{ \begin{matrix} ~~~\frac{1}{24}~~~-10\le x\le 14 \\ 0~~~~~~~~\mathrm{otherwise} \\ \end{matrix} \right.\] \begin{align*} P\left( y\le 7\text{ }\!\!|\!\!\text{ }x\ge 5 \right)&=\dfrac{P\left( 4\le y\le 7 \right)}{P\left( 4\le y\le 14 \right)}\\ &=\dfrac{3}{10}\\ &= 0.3 \end{align*}

Correct Answer: A

Solved Example: 9-1-03

Probability density function of a random variable X is given below

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {0.25}&{\mathrm{if}\;1 \le x \le 5}\\ 0&{\mathrm{otherwise}} \end{array}} \right.\)

P (X ≤ 4) is:

A. $\dfrac{3}{4}$

B. $\dfrac{1}{2}$

C. $\dfrac{1}{4}$

D. $\dfrac{1}{8}$

Since the value of the function is constant, it is difficult to figure it out whether the function is discrete or continuous.

Correct Answer: A

Solved Example: 9-1-04

Consider a random variable X that takes values +1 and -1 with probability 0.5 each. The values of the cumulative distribution function F(x) at x = -1 and +1 are:

A. 0 and 0.5

B. 0 and 1

C. 0.5 and 1

D. 0.25 and 0.75

Correct Answer: C

Learning Objectives:

• Understand conditions for discrete vs. continuous probability distribution function.
• Calculate the expected value of a discrete function.
• Calculate the expected value of a continuous function.

---

Expected Value of a Discrete Function:

Expected Value of a Continuous Function:

---

Solved Example: 9-2-01

You pick 3 different numbers between 1 and 9. If you pick all the numbers correctly you win \$50. What are your expected earnings if it costs \$1 to play?

A. -\$0.241

B. -\$0.279

C. -\$0.334

D. -\$0.405

Let X = your earnings

• X = 50 - 1 = 49, when you are winning (Prize money minus cost of playing)

• X = -1, when you are losing (Cost of playing)

Since the order is not important, we will use combination.
Probability of winning, $P(\mathrm{Winning\ \$49}) = \dfrac{1}{({}^{9}C_{3})} = \dfrac{1}{84}$
Probability of losing, $P(\mathrm{Losing\ \$1}) = 1- \dfrac{1}{84} = \dfrac{83}{84}$
Expected Earnings, $E(X) = 49 * \dfrac{1}{84} + (-1)* \dfrac{83}{84} = - \dfrac{34}{84} = -\$0.405$

Correct Answer: D

Solved Example: 9-2-02

A fair die is rolled two times independently. Given that the outcome on the first roll is 1, the expected value of the sum of the two outcomes is:

A. 4

B. 4.5

C. 3

D. 5.5

The sample space for the outcome at second rolling is: $S_2 = \{1,2,3,4,5,6\}$ Given that we have already got 1 in the first rolling, \begin{align*} S &= \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)\} \\ &= \{2,3,4,5,6,7\} \end{align*} Each of the above outcome is equally likely, which is $\dfrac{1}{6}^{\mathrm{th}}$ times. \begin{align*} E &= \sum_{i=1}^{n} P_i x_i \\&= \left(\dfrac{1}{6} \times 2\right) + \left(\dfrac{1}{6} \times 3\right) + \left(\dfrac{1}{6} \times 4\right) \\ &+ \left(\dfrac{1}{6} \times 5\right) + \left(\dfrac{1}{6} \times 6\right) + \left(\dfrac{1}{6} \times 7\right)\\ &= \dfrac{1}{6} \left(2+3+4+5+6+7\right)\\ &= \dfrac{1}{6} (27) = 4.5 \end{align*}

Correct Answer: B

Solved Example: 9-3-01

Let the random variable X denotes the time a person waits for a bus to arrive. X varies according to the graph shown below. What is the expected value of waiting, in minutes, for a person on this bus stop?

A. 0.577

B. 1.083

C. 1.235

D. 1.502

Method I: Using fundamental definition of expected value. Let us find the equation of declining line.

$\mathrm{Slope} = m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{0-0.5}{3 - 1} = \dfrac{-1}{4}$ and the line passes through ($x_1$, $y_1$) = (3,0).

Using, $y - y_1 = m (x- x_1) \mathrm{\ We\ get\ } y = -0.25x + 0.75$

\[P(x) = \begin{cases} \text{0.5,} &\quad 0 \leq x \leq 1\\ \text{-0.25x + 0.75} &\quad 1 \leq x \leq 3 \\ \end{cases} \] \begin{align*} E(X) &= \int_0^3 x P(x) dx\\ &= \int_0^1 0.5x dx + \int_1^3 x (-0.25x + 0.75) dx\\ &= 0.5 \left[\dfrac{x^2}{2}\right]_0^1 + (-0.25)\left[\dfrac{x^3}{3}\right]_1^3 + 0.75\left[\dfrac{x^2}{2}\right]_1^3\\ &= 0.25 - (0.25)(8.67) + 0.75(4)\\ &= 0.25 - 2.167 + 3 = 1.083 \end{align*}

Method II: Using concept of Centroid.

We will calculate the x coordinate of the centroid of the combined shape.

\begin{align*} \mathrm{Rectangle}: \\A_1 &= 0.5,\\ x_1 &= 0.5\\ \mathrm{Triangle}: \\A_2&= 0.5 \times 2 \times 0.5 = 0.5,\\ x_2 &= 1 + \dfrac{1}{3} (2) = 1.67 \end{align*} \begin{align*} \bar{x} &= \dfrac{A_1 x_1 + A_2 x_2}{A_1 + A_2}\\ &= \dfrac{(0.5)(0.5)+(0.5)(1.667)}{(0.5)+(0.5)}\\ &= 1.083 \end{align*}

Correct Answer: B

Solved Example: 9-3-02

If x is a random variable with the expected value of 5 and the variance of 1, then the expected value of x$^2$ is:

A. 24

B. 25

C. 26

D. 36

\begin{align*} Var(X) &= E(X^2) - [E(X)]^2\\ 1 &= E(X^2) - 5^2\\ E(X^2) &= 1 + 25\\ &= 26 \end{align*}

Correct Answer: C