The *mean *of a discrete random variable *X* can be explained as a weighted average of the possible values that the random variable contains. Contrary to the sample mean, which gives each observation equal weight, random variable mean weights each outcome *x _{i}* as per its probability,

*p*. The commonly used symbol for the mean or

_{i}*expected value*of

*X*) is μ, and it is formally defined by

*μ* = *E*(*x*)= ∑*x _{i}p*(

*x*)

_{i}where x_{i}= the value of the random variable and P(x_{i}) = the probability corresponding to a particular x value.

Example: Find the mean for the following probability distribution shown in the table below:

Value of X |
0 1 2 3 |

Probability |
0.064 0.288 0.432 0.216 |

The mean *µ* of *X* is

*µ* = (0*0.064) + (1*0.288) + (2*0.432) + (3*0.216)

= 1.8

The variance of a discrete random variable X measures the spread, or variability, of the distribution, and is defined by A weighted average of squared deviation about the mean

σ^{2} = *E*[(*x _{i}* –

*μ*)

^{2}] = ∑ (

*x*–

_{i}*μ*)

^{2}

*p*(

*x*)

_{i}Standard deviation: S.D =

**Example:** Find the variance for the table above:

The variance *σ*^{2} of *X* is

*σ*^{2} = 0.064*(0−1.8)^{2} + 0.288*(1−1.8)^{2} + 0.432*(2−1.8)^{2} + 0.216*(3−1.8)^{2}

= 0.20736 + 0.18432 + 0.01728 + 0.31104

= 0.72

the square root of Var X provides the SD: SD = 0.85

You can use the following table for calculating the variables:

Variance and mean are related to another relationship:

σ^{2}=E(X_{i} – μ)^{2} = E(X_{i}^{2} – 2X_{i}μ + μ^{2})

= E(X_{i}^{2} – 2[E(X_{i})]^{2} + [E(μ)]^{2}

= E(X_{i}^{2}) – [E(μ)]^{2}

E(X_{i}^{2}) =

**Example:**

Two cards are drawn one after another without replacement from a well-shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of queens.

**Solution:** Let Q be the number of queens in a draw of two cards. Therefore, Q is the random variable that assumes either 0, 1 or 2 values.

P(Q=0) = $^{48}$C$_{2}$/$^{52}$C$_{2}$ = $\frac{\frac{48!}{2!(48-2)!}}{\frac{52!}{2!(52-2)!}}=\frac{48\times 47}{52\times 51}=\frac{188}{221}$

P(Q = 1) = $^{4}$C$_{1}$? $^{48}$C$_{1}$ / $^{52}$C$_{2}$$\frac{4\times 48\times 2}{52\times 51}=\frac{32}{221}$

And P(Q = 2) = $^{4}$C$_{2}$/$^{52}$C$_{2}$ = $\frac{4\times 3}{52\times 51}=\frac{1}{221}$

Hence we can summarize the results in a probability distribution table for Q such that

Q | 0 | 1 | 2 |

P(Q) | 188/221 | 32/221 | 1/221 |

Mean of Q = E(Q) = $0\times \frac{188}{221}+1\times \frac{32}{221}+2\times \frac{1}{221}=\frac{34}{221}$

E(Q^{2}) = =

Var(Q) = E(Q^{2}) – [E(Q)]^{2}= $0^{2}\times \frac{188}{221}+1^{2}\times \frac{32}{221}+2^{2}\times \frac{1}{221}=\frac{36}{221}$

S.D = $\sqrt{\frac{6800}{221^{2}}}=0.37$