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The expected value of a random variable is the sum over all elements in a sample space of the product of the probability of the element and the value of the random variable at this element
Nilai harapan dari variabel acak adalah jumlah dari semua elemen dalam ruang sampel dari hasil kali probabilitas elemen dan nilai variabel acak pada elemen ini
→ The expected value is a weighted average of the values of a random variable
Nilai harapan adalah rata-rata tertimbang dari nilai variabel acak
The expected value, also called the expectation or mean, of the random variable XX on the sample space SS is equal to
Nilai harapan, juga disebut ekspektasi atau rata-rata, dari variabel acak XX pada ruang sampel SS sama dengan
     E(X)=∑s∈Sp(s)⋅X(s)~~~~~ E(X) = \sum_{s \in S} p(s) \cdot X(s)
The deviation of XX at s∈Ss \in S is X(s)−E(X)X(s) - E(X), the difference between the value of XX and the mean of XX.
Deviasi dari XX pada s∈Ss \in S adalah X(s)−E(X)X(s) - E(X), perbedaan antara nilai XX dan rata-rata XX.

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Example

Let X be the number that comes up when a fair die is rolled. What is the expected value of X?
Solution:The random variable X takes the values 1, 2, 3, 4, 5, or 6, each with probability 16\dfrac{1}{6}.
It follows that     E(X)=∑s∈Sp(s)⋅X(s)~~~~~ E(X) = \sum_{s \in S} p(s) \cdot X(s)
                =16⋅1+16⋅2+16⋅3+16⋅4+16⋅5+16⋅6=216=72~~~~~~~~~~~~~~~~ = \dfrac{1} {6} \cdot 1 + \dfrac{1} {6} \cdot 2 + \dfrac{1} {6} \cdot 3 + \dfrac{1} {6} \cdot 4 + \dfrac{1} {6} \cdot 5 + \dfrac{1} {6} \cdot 6 = \dfrac{21} {6} = \dfrac{7} {2}.

A fair coin is flipped three times. Let S be the sample space of the eight possible outcomes,and let X be the random variable that assigns to an outcome the number of heads in this outcome. What is the expected value of X?
Solution:Because the coin is fair and the flips are independent, the probability of each outcome is 18\dfrac{1}{8}.
It follows that     E(X)=18[X(HHH)+X(HHT)+X(HTH)+X(THH)+X(TTH)~~~~~ E(X) = \dfrac{1}{8} [X(HHH) + X(HHT) + X(HTH) + X(THH) + X(TTH)                           +X(THT)+X(HTT)+X(TTT)]~~~~~~~~~~~~~~~~~~~~~~~~~~ + X(THT) + X(HTT) + X(TTT)]
                =18[3+2+2+2+1+1+1+0]=128=32~~~~~~~~~~~~~~~~ = \dfrac{1} {8} [3 + 2 + 2 + 2 + 1 + 1 + 1 + 0] = \dfrac{12} {8} = \dfrac{3} {2}.

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Theorem 2

If XX is a random variable and p(X=r)p(X = r) is the probability that X=rX = r, so that p(X=r)=∑s∈S:X(s)=rp(s)p(X = r) = \sum_{s \in S: X(s) = r} p(s), then
Jika XX adalah variabel acak dan p(X=r)p(X = r) adalah probabilitas bahwa X=rX = r, sehingga p(X=r)=∑s∈S:X(s)=rp(s)p(X = r) = \sum_{s \in S: X(s) = r} p(s), maka
     E(X)=∑r∈X(S)p(X=r)⋅r~~~~~ E(X) = \sum_{r \in X(S)} p(X = r) \cdot r

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Theorem 3

The expected number of successes when nn mutually independent Bernoulli trials are performed, where pp is the probability of success on each trial, is npnp.

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Variance

Let XX be a random variable on a sample space SS. The variance of XX, denoted by V(X)V(X), is
Biarkan XX menjadi variabel acak pada ruang sampel SS. Varians dari XX, dilambangkan dengan V(X)V(X), adalah
     V(X)=∑s∈S[X(s)−E(X)]2⋅p(s)~~~~~ V(X) = \sum_{s \in S} [X(s) - E(X)]^2 \cdot p(s)
That is, V(X)V(X) is the weighted average of the square of the deviation of XX.
Artinya, V(X)V(X) adalah rata-rata tertimbang dari kuadrat deviasi XX.
The standard deviation of XX, denoted σ(X)\sigma(X), is defined to be V(X)\sqrt{V(X)}.

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Theorem 4

If XX is a random variable on a sample space SS, then
Jika XX adalah variabel acak pada ruang sampel SS, maka
     V(X)=E(X2)−E(X)2~~~~~ V(X) = E(X^2) - E(X)^2