Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
    
If this is homework, you should also tag it as such. –  bgins Mar 12 '12 at 9:05
add comment

1 Answer

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|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
add comment

Your Answer

 
discard

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