Central limit theorem

**Tweety** · January 24th 2013, 05:27 AM

25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5.

I have worked out the mean [tex] E(x) = /frac{7}{2} [\tex] do I multiply this by 25?

also I tried to calculate the variance using this formula [tex] Var(X) = E(X^{2}) - E(X)^{2} [\tex] but get a negative answer.

I square all the the numbers 1-6 on the die, and than multiply my their probability and each number has 1/6 chance.

[tex] Var(X) = 1^{2} x /frac{1}{6} + 2^{2} x /frac{1}{6} + 3^{2} x/frac{1}{6} + 4^{2} x /frac{1}{6} + 5^{2} x /frac{1}{6} + 6^{2} x /frac{1}{6} - /frac{7}{2}^{2} [\tex]

= $/frac{91}{6}$

I know i have to use the formula [tex] Z = /frac{y- /mu}{/sigma} [\tex]

**abender** · January 24th 2013, 08:13 AM

Originally Posted by Tweety

25 children in a class each roll a fair die 30 times and record the number of sixes they obtain. Find an estimate of the probability that the mean number of sixes recorded for the class is less than 4.5.

"Mean number of sixes" implies a count -- it is a 6 or it is not a 6. This problem does not require any calculations with respect to the mean value of a collection of die roll values.

Think binomial.

EDIT: See my post below, please.

**abender** · January 24th 2013, 12:36 PM

Let $X$ be a RV indicating the number of sixes a child obtains from 30 (independent) single rolls of a fair die.

The probability that (one) roll of a (fair) die is 6 is given by $\hat{p}=\tfrac{1}{6}$ .

We use the sample size $n=30$ in our calculations. Why not use $n=25$ ? Because 25 is the size of the sampling distribution of sample means. The CLT's requirement on the sampling distribution (of sample means) is that it is normal, which is going to approximately be the case when the sampling distribution consists of 25 sample means. For instance, $\{\bar{X}_1, \bar{X}_2, \ldots, \bar{X}_{25}\}$ is approximately normally distributed. Same deal if there were, say, 30, or 100 sample means. All you need to glean from the number 25 is that it is a sufficient quantity of sample means from which we may infer that the sampling distribution is approximately normal.

$\mu_{X} = E[X] = n\hat{p} = (30)(\tfrac{1}{6}) = 5$

$\sigma_{X}=\sqrt{V(x)} = \sqrt{n\hat{p}(1-\hat{p})} = \sqrt{(30)(\tfrac{1}{6})(\tfrac{5}{6})} = \sqrt{4.1\overline{666}} \approx 2.04124$

$\mu_{\bar{X}}=\mu_{X}=5$

$\sigma_{\bar{X}}=\frac{\sigma_{X}}{\sqrt{n}} \approx \frac{2.04124}{\sqrt{30}} \approx 0.372678$

$P\left(\bar{X}<4.5\right) = P\left(\frac{\bar{X}-\mu_{\bar{x}}}{\sigma_{\bar{X}}}<\frac{4.5-\mu_{\bar{x}}}{\sigma_{\bar{X}}}\right) \approx P\left(Z<\frac{4.5-5}{0.372678}\right)$ $\approx P(Z<-1.34) \approx 0.0901$

**Tweety** · January 24th 2013, 12:43 PM

Thank you, but my book says the correct answer is awrt 0.11??

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