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According to this formula, we take each observed X value and multiply it by its respective probability. We then add these products to reach our expected value. You may have seen this before referred to as a weighted average. It is known as a weighted average because it takes into account the probability of each outcome and weighs it accordingly. This is in contrast to an unweighted average which would not take into account the probability of each outcome and weigh each possibility equally.

Let's look at a few examples of expected values for a discrete random variable:

Example

A fair six-sided die is tossed. You win $2 if the result is a “1,” you win $1 if the result is a “6,” but otherwise you lose $1.

The Probability Distribution for X = Amount Won or LostX+$2+$1-$1Probability1/61/64/6

E(X)=$2(16)+$1(16)+(−$1)(46)=$−16=−$0.17E(X)=$2(16)+$1(16)+(−$1)(46)=$−16=−$0.17

The interpretation is that if you play many times, the average outcome is losing 17 cents per play. Thus, over time you should expect to lose money.

Example

Using the probability distribution for number of tattoos, let's find the mean number of tattoos per student.

Probabilty Distribution for Number of Tattoos Each Student Has in a Population of StudentsTattoos01234Probability.850.120.015.010.005

E(X)=0(.85)+1(.12)+2(.015)+3(.010)+4(.005)=.20E(X)=0(.85)+1(.12)+2(.015)+3(.010)+4(.005)=.20

The mean number of tattoos per student is .20.

Symbols for Population Parameters

Recall from Lesson 3, in a sample, the mean is symbolized by x¯¯¯x¯ and the standard deviation by ss. Because the probabilities that we are working with here are computed using the population, they are symbolized using lower case Greek letters. The population mean is symbolized by μμ (lower case "mu") and the population standard deviation by σσ(lower case "sigma").

Sample StatisticPopulation ParameterMeanx¯¯¯x¯μμVariances2s2σ2σ2Standard Deviationssσσ

Also recall that the standard deviation is equal to the square root of the variance. Thus, σ=(σ2)−−−−√σ=(σ2)

Standard Deviation of a Discrete Random Variable

Knowing the expected value is not the only important characteristic one may want to know about a set of discrete numbers: one may also need to know the spread, or variability, of these data. For instance, you may "expect" to win $20 when playing a particular game (which appears good!), but the spread for this might be from losing $20 to winning $60. Knowing such information can influence you decision on whether to play.

To calculate the standard deviation we first must calculate the variance. From the variance, we take the square root and this provides us the standard deviation. Conceptually, the variance of a discrete random variable is the sum of the difference between each value and the mean times the probility of obtaining that value, as seen in the conceptual formulas below:

Conceptual Formulas

Variance for a Discrete Random Variable

σ2=∑[(xi−μ)2pi]σ2=∑[(xi−μ)2pi]

Standard Deviation for a Discrete Random Variable

σ=∑[(xi−μ)2pi−−−−−−−−−−−√]σ=∑[(xi−μ)2pi]

xixi= value of the ith outcome
μ=E(X)=∑xipiμ=E(X)=∑xipi
pipi = probability of the ith outcome

In these expressions we substitute our result for E(X) into μμ because μμ is the symbol used to represent the mean of a population .

However, there is an easier computational formula. The compuational formula will give you the same result as the conceptual formula above, but the calculations are simplier.

Computational Formulas

Variance for a Discrete Random Variable

σ2=[∑(x2ipi)]−μ2σ2=[∑(xi2pi)]−μ2

Standard Deviation for a Discrete Random Variable

σ=[∑(x2ipi)]−μ2−−−−−−−−−−−−√σ=[∑(xi2pi)]−μ2

xixi= value of the ith outcome
μ=E(X)=∑xipiμ=E(X)=∑xipi
pipi = probability of the ith outcome

Notice in the summation part of this equation that we only square each observed X value and not the respective probability. Also note that the μμ is outside of the summation.

Example

Going back to the first example used above for expectation involving the dice game, we would calculate the standard deviation for this discrete distribution by first calculating the variance:

The Probability Distribution for X = Amount Won or LostX+$2+$1-$1Probability1/61/64/6

σ2=[∑x2ipi]−μ2=[22(16)+12(16)+(−1)2(46)]−(−16)2σ2=[∑xi2pi]−μ2=[22(16)+12(16)+(−1)2(46)]−(−16)2

=[46+16+46]−136=5336=1.472=[46+16+46]−136=5336=1.472

The variance of this discrete random variable is 1.472.

σ=(σ2)−−−−√σ=(σ2)

σ=1.472−−−−√=1.213σ=1.472=1.213

The standard deviation of this discrete random vairable is 1.213. hope this helps

7 0

4 years ago

Read 2 more answers

HURRY NEED HELP NOW 42 POINTS NEED 1 PARAGRAPH FOR EACH QUESTION there are 2 photos I put up and that where the questions are.

Shtirlitz [24]

1. All you have to do to find the area of the figure. IS first find the area of the rectangle. This is easy because all you have to do is base times height.

22 x 17 = 374

Now to find the area of the triangle all you have to do is do base times height like a rectangle, but then divide it by 2 afterwords.
19 x 12 = 228 ÷ 2 = 114.

Now add the two area's together:

374 + 114 = 488.

Just do the same with the one below, I've altered the pic a bit to make it easier.

4 0

3 years ago