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.
<span>The Probability Distribution for X = Amount Won or Lost<span><span>X+$2+$1-$1</span><span>Probability1/61/64/6</span></span></span><span><span>E(X)=$2(<span>16</span>)+$1(<span>16</span>)+(−$1)(<span>46</span>)=$<span><span>−1</span>6</span>=−$0.17</span><span>E(X)=$2(<span>16</span>)+$1(<span>16</span>)+(−$1)(<span>46</span>)=$<span><span>−1</span>6</span>=−$0.17</span></span>
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.
<span>Probabilty Distribution for Number of Tattoos Each Student Has in a Population of Students<span><span>Tattoos01234</span><span>Probability.850.120.015.010.005</span></span></span><span><span>E(X)=0(.85)+1(.12)+2(.015)+3(.010)+4(.005)=.20</span><span>E(X)=0(.85)+1(.12)+2(.015)+3(.010)+4(.005)=.20</span></span>
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 <span><span>x<span>¯¯¯</span></span><span>x¯</span></span> and the standard deviation by <span>ss</span>. 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 <span>μμ</span> (lower case "mu") and the population standard deviation by <span>σσ</span>(lower case "sigma").
<span><span> Sample StatisticPopulation Parameter</span><span>Mean<span><span>x<span>¯¯¯</span></span><span>x¯</span></span><span>μμ</span></span><span>Variance<span><span>s2</span><span>s2</span></span><span><span>σ2</span><span>σ2</span></span></span><span>Standard Deviation<span>ss</span><span>σσ</span></span></span>Also recall that the standard deviation is equal to the square root of the variance. Thus, <span><span>σ=<span><span>(<span>σ2</span>)</span><span>−−−−</span>√</span></span><span>σ=<span>(<span>σ2</span>)</span></span></span>
Standard Deviation of a Discrete Random VariableKnowing 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
<span><span><span>σ2</span>=∑[(<span>xi</span>−μ<span>)2</span><span>pi</span>]</span><span><span>σ2</span>=∑[(<span>xi</span>−μ<span>)2</span><span>pi</span>]</span></span>
Standard Deviation for a Discrete Random Variable
<span><span>σ=<span><span>∑[(<span>xi</span>−μ<span>)2</span><span>pi</span></span><span>−−−−−−−−−−−</span>√</span>]</span><span>σ=<span>∑[(<span>xi</span>−μ<span>)2</span><span>pi</span></span>]</span></span>
<span><span>xi</span><span>xi</span></span>= value of the i<span>th </span>outcome
<span><span>μ=E(X)=∑<span>xi</span><span>pi</span></span><span>μ=E(X)=∑<span>xi</span><span>pi</span></span></span>
<span><span>pi</span><span>pi</span></span> = probability of the ith outcome
In these expressions we substitute our result for E(X) into <span>μμ</span> because <span>μμ</span> 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
<span><span><span>σ2</span>=[∑(<span>x2i</span><span>pi</span>)]−<span>μ2</span></span><span><span>σ2</span>=[∑(<span>xi2</span><span>pi</span>)]−<span>μ2</span></span></span>
Standard Deviation for a Discrete Random Variable
<span><span>σ=<span><span>[∑(<span>x2i</span><span>pi</span>)]−<span>μ2</span></span><span>−−−−−−−−−−−−</span>√</span></span><span>σ=<span>[∑(<span>xi2</span><span>pi</span>)]−<span>μ2</span></span></span></span><span>
</span>
<span><span>xi</span><span>xi</span></span>= value of the i<span>th </span>outcome
<span><span>μ=E(X)=∑<span>xi</span><span>pi</span></span><span>μ=E(X)=∑<span>xi</span><span>pi</span></span></span>
<span><span>pi</span><span>pi</span></span> = 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 <span>μμ</span> is outside of the summation.
ExampleGoing 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:
<span>The Probability Distribution for X = Amount Won or Lost<span><span>X+$2+$1-$1</span><span>Probability1/61/64/6</span></span></span><span><span><span>σ2</span>=[∑<span>x2i</span><span>pi</span>]−<span>μ2</span>=[<span>22</span>(<span>16</span>)+<span>12</span>(<span>16</span>)+(−1<span>)2</span>(<span>46</span>)]−(−<span>16</span><span>)2</span></span><span><span>σ2</span>=[∑<span>xi2</span><span>pi</span>]−<span>μ2</span>=[<span>22</span>(<span>16</span>)+<span>12</span>(<span>16</span>)+(−1<span>)2</span>(<span>46</span>)]−(−<span>16</span><span>)2</span></span></span>
<span><span>=[<span>46</span>+<span>16</span>+<span>46</span>]−<span>136</span>=<span>5336</span>=1.472</span><span>=[<span>46</span>+<span>16</span>+<span>46</span>]−<span>136</span>=<span>5336</span>=1.472</span></span>
The variance of this discrete random variable is 1.472.
<span><span>σ=<span><span>(<span>σ2</span>)</span><span>−−−−</span>√</span></span><span>σ=<span>(<span>σ2</span>)</span></span></span>
<span><span>σ=<span>1.472<span>−−−−</span>√</span>=1.213</span><span>σ=1.472=1.213</span></span>
The standard deviation of this discrete random vairable is 1.213. hope this helps