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The mean of the random of a probability distribution is     μ=X1⋅P(X1)+X2⋅P(X2)+…+Xn⋅P(Xn)~~~~~ \mu = X_1 \cdot P(X_1) + X_2 \cdot P(X_2) + \ldots + X_n \cdot P(X_n)
        =∑X⋅P(X)~~~~~~~~ = \sum X \cdot P(X) where X1,X2,…,XnX_1, X_2, \ldots, X_n are the outcomes AND
P(X1),P(X2),…,P(Xn)P(X_1), P(X_2), \ldots, P(X_n) are the corresponding probabilities.

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Mean for Discrete Variable - Example

Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below.
X123456
P(X)1/61/61/61/61/61/6

Solution:μ=∑X⋅P(X)\mu = \sum X \cdot P(X)
    =1⋅16+2⋅16+3⋅16+4⋅16+5⋅16+6⋅16~~~~ = 1 \cdot \dfrac{1} {6} + 2 \cdot \dfrac{1} {6} + 3 \cdot \dfrac{1} {6} + 4 \cdot \dfrac{1} {6} + 5 \cdot \dfrac{1} {6} + 6 \cdot \dfrac{1} {6}
    =216=3.5~~~~ = \dfrac{21} {6} = 3.5
That is, when a die is tossed many times, the theoretical mean will be 3.5.

In a family with two children, find the mean number of children who will be girls. The probability distribution is given below.
X012
P(X)1/41/21/4

Solution:μ=∑X⋅P(X)\mu = \sum X \cdot P(X)
    =0⋅14+1⋅12+2⋅14~~~~ = 0 \cdot \dfrac{1} {4} + 1 \cdot \dfrac{1} {2} + 2 \cdot \dfrac{1} {4}
    =12+12=1~~~~ = \dfrac{1} {2} + \dfrac{1} {2} = 1
That is, the average number of girls in a two-child family is 1.

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Formula for the Variance of a Probability Distribution

The variance of a probability distribution is found by multiplying the square of each outcome by its corresponding probability, summing these products, and subtracting the square of the mean.
Varians dari distribusi probabilitas ditemukan dengan mengalikan kuadrat dari setiap hasil dengan probabilitasnya yang sesuai, menjumlahkan produk-produk ini, dan mengurangkan kuadrat dari rata-rata.

The formula for the of a probability distribution is
     σ2=∑[X2⋅P(X)]−μ2~~~~~ \sigma^2 = \sum [X^2 \cdot P(X)] - \mu^2
The standard deviation of a probability distribution is
     σ=σ2~~~~~ \sigma = \sqrt{\sigma^2}

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Variance of a Probability Distribution - Example

The probability that 0, 1, 2, 3, or 4 people will be placed on hold when they call a radio talk show with four phone lines is shown in the distribution below. Find the variance and standard deviation for the data.
X01234
P(X)0.180.340.230.210.04

Solution:
XP(X)X \cdot P(X)X^2 \cdot P(X)
00.1800
10.340.340.34
20.230.460.92
30.210.631.89
40.040.160.64
μ=1.59\mu = 1.59∑2⋅P(X)=3.79\sum^2 \cdot P(X) = 3.79
σ2=3.79−1.592=1.26\sigma^2 = 3.79 - 1.59^2 = 1.26
Now, μ=(0)(0.18)+(1)(0.34)+(2)(0.23)+(3)(0.21)+(4)(0.04)=1.59\mu = (0)(0.18) + (1)(0.34) + (2)(0.23) + (3)(0.21) + (4)(0.04) = 1.59.
∑X2P(X)=(02)(0.18)+(12)(0.34)+(22)(0.23)+(32)(0.21)+(42)(0.04)=3.79\sum X^2 P(X) = (0^2)(0.18) + (1^2)(0.34) + (2^2)(0.23) + (3^2)(0.21) + (4^2)(0.04) = 3.79.
1.592=2.531.59^2 = 2.53 (rounded to two decimal places).
σ2=3.79−2.53=1.26\sigma^2 = 3.79 - 2.53 = 1.26.
σ=1.26=1.12\sigma = \sqrt{1.26} = 1.12.

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Expectation

The value of a discrete random of a probability distribution is the l average of the variable. The formula is
Nilai dari variabel acak diskrit dari distribusi probabilitas adalah rata-rata dari variabel. Rumusnya adalah

μ=E(X)=∑X⋅P(X)\mu = E(X) = \sum X \cdot P(X)
The symbol E(X)E(X) is used for the expected value

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Expectation - Example

A ski resort loses 70,000perseasonwhenitdoesnotsnowverymuchandmakes70,000 per season when it does not snow very much and makes 250,000 when it snows a lot. The probability of it snowing at least 75 inches (i.e., a good season) is 40%. Find the expected profit.
Solution:
Profit, X250,000-70,000
P(X)0.400.60
The expected profit = ($250,000)(0.40)−(−$70,000)(0.60)=$58,000(\$250,000)(0.40) - (-\$70,000)(0.60) = \$58,000.