Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let x1 x2 x3 x4 be random sample from the population that satisfies an binomial distribution with n = 3 and p = 1/4

a) Find the mean and variance for Sum Y = x1 + x2 + x3 + x4.

b) Find the mean and variance for W = 2x1 - 3x2 - 2x4.

I am trying to solve it and I was just wondering if I was going the right way I found the C.D.F for binomial distribution for n = 3 p = 1/4 with lower bound being 1 and upper being 4. Which is .57 and I got the variance to be (.57)^2. Am I correct ? And for part B I am completely lost can someone please either confirm my answer or explain to me the steps to solving this problem. Thank you for your help :)

share|cite|improve this question

I think you want to write these as random variables: $$ X_i \sim \text{Binom}\left(n=3,~p=\frac14\right) \qquad \text{for} \qquad 1 \le i \le 4, $$ so that the mean is $E[X_i]=np=\frac34$ and the variance is $\text{Var}(X_i)=np(1-p)=\frac9{16}$.

So for $W = 2X_1 - 3X_2 - 2X_4$, we have mean $$ E[W] = 2E[X_1] - 3E[X_2] - 2E[X_4] = \left(2-3-2\right) \frac34 = -\frac94 $$ and variance $$ \text{Var}(W)=(2^2+3^3+2^2)\text{Var}(X_i)=17\cdot\frac9{16}=\frac{153}{16}, $$ which follow from the facts that (for any $X,Y$ for which the quantities below are defined) $$ \eqalign{ &E[aX+bY] = a \, E[X] + b \, E[Y]\\ &\text{Var}(aX+bY) = a^2\text{Var}(X)+b^2\text{Var}(Y)+2ab \, \text{Cov}(X,Y)\\ } $$ and that the $X_i$ are all pairwise independent, and so all the covariances $\text{Cov}(X_i,Y_j)=0$.

In the case of $Y=\sum_{i=1}^4 X_i$, you should therefore get $E[Y]=\sum_{i=1}^4 E[X_i]=4\cdot\frac34$ and $\text{Var}(Y)=\sum_{i=1}^4 1^2\cdot\text{Var}(X_i)=4\cdot\frac9{16}$.

The cumulative distribution function is not really useful here; it doesn't directly apply to the sum of four different (independent) instances of a random variable.

share|cite|improve this answer
Yes, thanks, I saw that. – bgins Mar 12 '12 at 8:55
Thank you for your help :) – user26738 Mar 12 '12 at 8:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.