Lesson 6 - 1

Discrete Random Variables

Objectives• Distinguish between discrete and continuous

random variables

• Identify discrete probability distributions

• Construct probability histograms

• Compute and interpret the mean of a discrete random variable

• Interpret the mean of a discrete random variable as an expected value

• Compute the variance and standard deviation of a discrete random variable

Vocabulary• Random variable – a numerical measure of the outcome of a

probability experiment, so its value is determine by chance.

• Discrete random variable – has finite or countable number of values

• Continuous random variable – has infinitely many values

• Probability distribution of a discrete random variable – provides all possible values of the random variable and their corresponding probabilities (can be in the form of a table or graph – or a mathematical formula)

• Probability histogram – histogram with y values being probability and x axis being the random variable (similar to relative frequency histogram)

• Expected value –the mean of a random variable, E(x)

• Variance of a discrete random variable – weighted average of the squared deviations where the probabilities are the weights

Rules for a Discrete Probability Distribution

Let P(x) denote the probability that the random variable X equals x, then

• The sum of all probabilities of all outcomes must equal 1

∑ P(x) = 1 • The probability of any value x, P(x), must between 0

and 1 0≤ P(x) ≤ 1

Discrete Random Variable - Mean

The mean, or expected value [E(x)], of a discrete random variable is given by the formula

μx = ∑ [x ∙P(x)]

where x is the value of the random variable and P(x) is the probability of observing x

 

Mean of a Discrete Random Variable Interpretation:

If we run an experiment over and over again, the law of large numbers helps us conclude that the difference between x and ux gets closer to 0 as n (number of repetitions) increases

Discrete Random Variable - Variance

Variance and Standard Deviation of a Discrete Random Variable:

The variance of a discrete random variable is given by:

 

σ2x = ∑ [(x – μx)2 ∙ P(x)] = ∑[x2 ∙ P(x)] – μ2

x

and standard deviation is √σ2

 

Note: round the mean, variance and standard deviation to one more decimal place than the values of the random variable

 

Uniform PDF

An experiment is said to be a Uniform experiment provided:

1. The probability of each value of the random variable is equal (like in a six-sided die)

2. The trials are independent of each other (what happened last does not affect what happens next)

 

Uniform PDF

If X is a value of the uniform random variable, then probability formula for X is  

1P(x) = ------- x = 0, 1, 2, 3, … , n

n

where n is the total number of discrete values of the random variable x

Mean: μx = ∑ [x ∙P(x)] = (1/n)∑ x

Standard Deviation: σ2

x = ∑ [(x – μx)2 ∙ P(x)] = (1/n) ∑ [(x – μx)2

= ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx

2

 

Example 1

You have a fair 10-sided die with the number 1 to 10 on each of the faces.

 

Determine the mean and standard deviation.

Mean: ∑ [x ∙P(x)] = (1/10) (∑ x) = (1/10)(55) = 5.5

Var: ∑[x2 ∙ P(x)] – μ2x = (1/n) ∑ [x2 ] – μx

2 = (1/10) (385) - 30.25) = (38.5 – 30.25) = 8.25 St Dev = 2.8723

Example 2Below is a distribution for number of visits to a dentist

in one year.

 X = # of visits to a dentist

x 0 1 2 3 4

P(x) .1 .3 .4 .15 .05 

Determine the expected value, variance and standard deviation.

Mean: ∑ [x ∙P(x)] = (.1)(0) + (.3)(1) + (.4)(2) + (.15)(3) + (.05)(4) = 0 + .3 + .8 + .45 +.2 = 1.75

Var: ∑[x2 ∙ P(x)] – μ2x = ∑ [x2 ∙ P(x)] – μx

2 = (0 + .3 + .4(4) + .15(9) + .05(16) ) – 3.0625) = 4.05 – 3.8626 = 0.9875

St Dev = 0.9937

Example 3

What is the average size of an American family? Here is the distribution of family size according to the 1990 Census:

 

# in family 2 3 4 5 6 7

p(x) .413 .236 .211 .090 .032 .018

Mean: ∑ [x ∙P(x)] = (.413)(2) + (.236)(3) + (.211)(4) + (.09)(5) + (.032)(6) + (.018)(7)

= .826 + .708 + .844 + .45 + .192 + .126

= 3.146

Example 4

You are trying to decide whether to take out a $250 deductible which will cost you $90 per year. Records show that for this community the average cost of repair is $900. Records also show that 10% of the drivers have an accident during the year. If you have sufficient assets so that you will not be financially handicapped if you had to pay out the $900 or more for repairs, should you buy the policy?

P(accident) = 0.1 so P(no accident) = 0.9ave repair cost = $900 Yearly cost = $90

Expected Yearly with Policy: ∑ [x ∙P(x)] = (.1)(250) + (.9)(0) + 90 = 25 + 90 = $115

Expected Yearly without: ∑ [x ∙P(x)] = (.1)(900) + (.9)(0) = $90

Summary and Homework

• Summary– Discrete variables have a finite number of

values– Expected value is the mean ∑ [x ∙P(x)]

– Variance is ∑[x2 ∙ P(x)] – μ2x

• Homework– pg 323-327; 7, 10 – 15, 18, 19, 23, 31, 35