Chapter 7 Section 7.1

Definition: a random variable is a variable whose value is a numerical outcome of a random phenomenon.

Discrete random variableA discrete random variable X has a countable number of

possible values. The probability distribution of X lists the values and their probabilities.

Value of X: x1 x2 x3 x4 ……. xk

Probability: p1 p2 p3 p4……….pk

The probabilities pi must satisfy two requirements:1. Every probability is a number between 0and 12. p1 + p2 + p3 + p4………+ .pk = 1Find the probability of any event by adding the

probabilities pi of the particular values of x1 that make up the event.

Probability distributionExample 7.1 pg 393

• Grade: 0 1 2 3 4• Probability: .10 .15 .30 .30 .15

Discrete and Continuous Variables

The height of each bar shows the probability of the outcome at its base. Probability histogram for random digits 1 to 9

The height of each bar shows the probability of the outcome at its base. Probability histogram for Benford’s law

Possible outcomes in four tosses of a coin. The random variable X is the number of heads.

• X = number of heads• P ( X= 0) TTTT = ½∙ ½∙ ½∙ ½ = 1/16 or .0625• P (X = 1) TTTH = ½∙ ½∙ ½∙ ½ = .0625

TTHT = ½∙ ½∙ ½∙ ½ = .0625 THTT = ½∙ ½∙ ½∙ ½ = .0625 HTTT = ½∙ ½∙ ½∙ ½ = .0625

P (X =1) = .0625(4) = .25

Probability histogram for the number of heads in four tosses of a coin.

• Do Problem 7.1

A spinner that generates a random number between 0 and 1.

• Do problem 7.7• (a) P(x ≤.49)• (b) P (x ≥.27)• (c) P(.27 ≤ x ≤ 1.27) =• (d) P(.1≤ x ≤ .2) or P(.8 ≤ x ≤ .9 )

=• (e) P( x < .3) or P (x >.8) =• (f) P (x = .5)

• Do problem 7.7• (a) P (x ≤.49) = .49• (b) P (x ≥.27) = .73• (c) P(.27 ≤ x ≤ 1.27) = P(.27 ≤ x ≤ 1 )

= .73• (d) P(.1≤ x ≤ .2) or P(.8 ≤ x ≤ .9 ) = .1 + .1 = .2• (e) P( x < .3) or P (x >.8) = .2 + .2 = .4• (f) P (x = .5) = 0 a continuous

distribution assigns probability of 0 to every individual outcome

Def. A continuous random variable X takes all values in an interval of numbers. The probability distribution of X is described by a density curve. The probability of any event is the area under the density curve and about the values of X that make up the event.

In the language of randomvariables, if X has the N(μ , σ)Distribution, then the Standardized

Variable Z = X – μ σIs a standard normal random variable having the distribution N(0,1)

• Do Example 7.18 pg 405

This continuous random variable takes values between 0 and 2.

The density curve for the sum of two random numbers.

End of Section 7.1

Mean of a Discrete Random Variable Suppose that X is a discrete random variable whose distribution is

Value of X: x1 x2 x3 x4 ……. xk

Probability: p1 p2 p3 p4……….pk

To find the mean of X, multiply each possible value by its probability, then add all the products:

μx = x1 p1+ x2 p2 + x3 p3 +…. + xk pk

= Σ xi pi

Locating the mean of a discrete random variable on the probability histogram for digits between 1 and 9 chosen at random.

Locating the mean of a discrete random variable on the probability histogram for digits between 1 and 9 chosen from records that obey Benford’s law

Variance of a Discrete Random VariableSuppose that X is a discrete random variable whose

distribution is Value of X: x1 x2 x3 x4 ……. xk

Probability: p1 p2 p3 p4……….pk

and that μ is the mean of X. The variance of X is σ2

x = (x1– μx)2 p1 + (x2– μx)2 p2 +…+ (xk– μx)2 pk

= Σ(xi– μx)2 pi The standard deviation σx of X is the

square root of the variance.

Do problem 7.22 and 7.26

Do Example 7.42 pg 427

Expected Value –• E (x) μ is the long term average

value of a random variable.

μ = Σ x ∙ p (x)

• Day0 Day1 Day2 prob.

1000 1300

1000 1300 1000 750 1000 750

• Day0 Day1 Day2 prob.

1000 1300 1690

1000 1300 975 1000 750 975

1000 750 562.50

• Day0 Day1 Day2 prob.

1000 1300 1690 ¼

1000 1300 975 ½1000 750 975

1000 750 562.50 ¼

• Day0 Day1 Day2 prob.

1000 1300 1690 ¼

1000 1300 975 ½1000 750 975

1000 750 562.50 ¼

P( worth >1000) = ¼ or .25Mean value (expected value) = 1690 (.25) + 975 (.5) + 562.50 (.25) = $1051

Law of Large Numbers Draw independent observations at random

from any population with finite mean μ. Decide how accurately you would like to estimate μ. As the number of observations drawn increases, the mean x bar of the observed values eventually approaches the mean μ of the population as closely as you specified and then stays that close.

The law of large numbers in action. As we increase the size of our sample, the sample mean x bar always approaches the mean μ of the population.

Rule for Means and Variances with random independent variables:

μa+bx = a + bμx μxy = μx + μy

σ2 a + bx = b2 σ2

x σ2 x+y = σ2

x + σ2 y

These patterns do not extend to standard deviation.

It is not true that σx+y = σx + σy

If x and y have a correlation ρ then

σ2x+y = σ2

x + σ2y +2 ρσx σy

σ2x-y = σ2

x + σ2y ‒2 ρσx σy

Do problem 7.32 pg 417

The normal probability calculation for Example 7.14.

Do problem 7.60 pg 433

End of Section 7.2