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A box contains 100 numbered balls - 21 with the number 1, 36 with the number 2 and 43 with the number 3. You will randomly select 10 balls from the box with replacement and you take the mean of the sum of the 10 balls

What is $E(\bar{ X})$ and $Var(\bar {X})$

I know that there are 21 different averages ranging from 1 to 3 in increments of 0.1. There are a total of 66 combinations, but there must be a trick to do this without having to write out a 2 page long equation to calculate $E(\bar X)$ and an even longer $Var(\bar X)$

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  • $\begingroup$ You're halfway there; please add the [self-study] tag & read its wiki, & we can re-open this for you. $\endgroup$
    – gung - Reinstate Monica
    Oct 24, 2014 at 18:15
  • $\begingroup$ You're quite right: but it's not a trick, it's an important technique to know. Since it is obvious you have been asked this question in a textbook or course, then you must already have been exposed to some properties of expectation and variance that allow you to infer their values for sums of random variables: use those properties to solve this problem. $\endgroup$
    – whuber ♦
    Oct 24, 2014 at 21:59
  • $\begingroup$ Are you sure that you take the mean of the sum? Or do you take just the mean? $\endgroup$
    – Sycorax ♦
    Oct 25, 2014 at 2:27
  • $\begingroup$ An easier answer is following whubers hint. What is the mean and variance of one draw? Are draws independent? What iis mean and variance of sum of independent random numbers cf en.m.wikipedia.org/wiki/Standard_error $\endgroup$
    – seanv507
    Oct 25, 2014 at 15:40

1 Answer 1

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Denote $X_1, X_2, X_3$ be the number of times that the three balls with number 1, 2, and 3 respectively, are selected in the 10 random draws. Since with replacement, $X=(X_1, X_2, X_3)$ follows a multinomial distribution with probabilities $p_1 = 0.21, p_2 = 0.36, p_3 = 0.43$. Following the distribution, we have $$\mathbf{E}(X_i) = np_i; \hspace{20pt} \mathbf{Var}(X_i) = np_i(1-p_i); \hspace{20pt} \mathbf{Cov}(X_i, X_j) = -np_ip_j ( i \neq j).$$

We then have $$\bar{X} = \frac{1*X_1 + 2*X_2 + 3*X_3} {n} = \frac{X_1 + 2X_2 + 3X_3}{10} $$ $$ \mathbf{E}(\bar{X}) = \frac{1}{10} \{ \mathbf{E}(X_1) + \mathbf{E}(2X_2) + \mathbf{E}(3X_3) \}=p_1 + 2p_2 + 3p_3$$ $$=0.21 + 2*0.36 + 3*0.43 = 2.22\\$$

Similarly, $$\mathbf{Var}(\bar{X}) = \frac{1}{100} \{\mathbf{Var}(X_1) + 4\mathbf{Var}(X_2) + 9\mathbf{Var}(X_3) +4\mathbf{Vov}(X_1, X_2) + 12\mathbf{Cov}(X_2, X_3) + 6\mathbf{Cov}(X_1, X_3)\}.$$

EDIT:

The $X_1, X_2, X_3$ are correlated, so there should be covariance parts. I just added those. Note that $\mathbf{Var}(aX + bY) = a^2\mathbf{Var}(X) + b^2\mathbf{Var}(Y) + 2ab\mathbf{Cov}(X,Y).$

END EDIT.

You can do the simple calculation by yourself.

Hope it helps.

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  • $\begingroup$ multinomial wasn't mentioned in the notes, but it was the word I was looking for. Learned it in stat mech,l but its been a few years $\endgroup$
    – Bob Unger
    Oct 25, 2014 at 7:32
  • $\begingroup$ Thanks for accepting. I just realized I missed the covariance parts, I added those. See the edited version. $\endgroup$
    – SixSigma
    Oct 25, 2014 at 13:41

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