Chapter 4 – Discrete Random Variables

Random variables describe the outcomes of statistical experiments.

Random Variable Notation

• Uppercase letters ($X$ or $Y$) denote random variables
• Lowercase letters ($x$ or $y$) denote the values of random variables
• Example:
  • $X=$ the number of heads when three fair coins are tossed (words)
  • $x=0,1,2,3$ (numbers)

Probability Distribution Function (PDF) for Discrete Random Variables

For discrete PDFs:

• Each probability is between 0 and 1 (inclusive)
• The sum of probabilities is 1

Example

Jeremiah has basketball practice 2x/week. 90% of the time, he attends both practices. 8% of the time, he attends one practice. 2% of the time, he does not attend either practice.

Identify $X$ and the values it takes on.

$X=$ the number of practices that Jeremiah will attend

| $x$ | $P(X=x)$ | | --- | ----------- | | 0 | $P(0)=0.02$ | | 1 | $P(1)=0.08$ | | 2 | $P(2)=0.9$ |

$x=0,1,2$

This is a discrete PDF because:

• Each probability is between 0 and 1 (inclusive)
• The sum of the probabilities is 1 (0.02 + 0.08 + 0.9 = 1)

Mean, Expected Value, Standard Deviation

• Expected value is referred to as the long-term mean/average (${\mu }_X$)
  • After conducting the experiment many times, this value is expected

$$E(X)={\mu }_X=\Sigma (x_i\cdot P(x_i))$$

The standard deviation (${\sigma }_X$) of the PDF has the formula

$$SD(X)={\sigma }_X=\sqrt{\Sigma [(x_i-{\mu }_X{)}^2\cdot P(x_i)]}$$

Earthquake

On May 11, 2013 at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about 1.08%. You bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let $X$ = the amount of profit from a bet.

Find the mean and standard deviation of $X$.

| $x$ | $P(X=x)$ | $x\cdot P(x)$ | $(x-\mu {)}^2\cdot P(x)$ | | ------ | -------- | ------------- | --------------------- | | $100 | 0.0108 | 1.08 | 127.873 | | ($10) | 0.9892 | (9.892) | 1.396 |

$\mu =1.08-9.892=-8.812$ $\sigma =\sqrt{127.873+1.396}=11.370$

Transforming Random Variables

Multiplication (division) by some constant $c$

• Mean is multiplied by $c$
• Stdev is multiplied by $c$

Addition (subtraction) of some constant $c$

• $c$ is added to the mean
• Stdev not affected

Combining Random Variables

• Can only be combined if independent
• Mean is combined (addition or subtraction)
• Stdev cannot be combined
• Variance can be added
  • Take sqrt of combined variance to get stdev

Example

Let $S=$ the number of shirts Albert sells on a given day and $J=$ the number of jackets albert sells on a given day.

Day	$S$	$J$
1	1	29
2	4	12
3	3	14
4	9	25
5	3	16
6	8	19
7	3	21

• What are the means of shirts and pants sold on any given day?

${\mu }_S=\frac{1+4+3+9+3+8+3}{7}=4.286$ ${\mu }_J=\frac{29+12+14+25+16+19+21}{7}=19.429$

• What are the standard deviations?

${\sigma }_S=2.718$ ${\sigma }_J=5.628$

• Shirts sell for $20 and jackets sell for $50. What is the mean total revenue?

$R(S)=20\cdot S$ ${\mu }_{R_S}=20\cdot {\mu }_S=96.52$

$R(J)=50\cdot J$ ${\mu }_{R_J}=50\cdot {\mu }_J=971.45$

$R(T)=\Sigma R$ ${\mu }_R={\mu }_{R_S}+{\mu }_{R_J}=1067.97$

• What is the standard deviation of the total revenue?

${\sigma }_{R_S}=20\cdot {\sigma }_S=54.36$ ${\sigma }_{R_J}=50\cdot {\sigma }_J=281.4$

${\sigma }_R=\sqrt{({\sigma }_{R_S}{)}^2+({\sigma }_{J_S}{)}^2}=286.602$

Binomial Distribution

Characteristics of a binomial experiment:

• The number of trials is fixed

• There are only two possible outcomes

  • $p=$ success-
  • $(1-p)=q=$ failure

• The $n$ trials are independent and repeated under identical conditions

  • $p$ and $q$ remain the same between trials

• Mean ($\mu$) = $np$

• Variance (${\sigma }^2$) = $npq=np(1-p)$

• Standard deviation ($\sigma$) = $\sqrt{npq}=\sqrt{np(1-p)}$

An experiment with $n=1$ trial is called a Bernoulli trial.

Notation

$X\sim B(n,p)$ is read as ”$X$ is a random variable with a binomial distribution”.

• $n=$ number of trials
• $p=$ probability of success

Example

It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?

Let $X$ = the number of workers who have a high school diploma but do not pursue any further education.

$X$ takes on the values 0, 1, 2, …, 20 where $n$ = 20, $p$ = 0.41, and $q$ = 1 – 0.41 = 0.59. $X\sim B(20,0.41)$

Find $P(x\le 12)$. (calculator or computer)

>>> BinomialCD(12, 20, 0.41)
0.9738

$\mu =np=(20)(0.41)=8.2$ $\sigma =\sqrt{npq}=\sqrt{(20)(0.41)(0.59)}=2.20$

Geometric Distribution

Characteristics of a geometric experiment:

• Repeated until success occurs

• Repeated trials are independent

• $p$ and $q$ remain the same between trials

• $X=$ number of trials until the first success

• Mean ($\mu$) = $\frac{1}{p}$

• Stdev ($\sigma$) = $\sqrt{\frac{1-p}{p^2}}$ or $\frac{\sqrt{1-p}}{p}$

$X\sim G(p)$ where $p$ is the probability of a success for each trial.

Example

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?

Let $X$ = the number of computer components tested until the first defect is found.

$X$ takes on the values 1, 2, 3, … where $p$ = 0.02. Find $P(x=7)$.

>>> GeoPD(7, 0.02)
0.0177