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Firstly, I am studying the basic concepts of statistics and so any explanations, advice and suggestions are more than appreciated. Onto the problem- I am given the central limit theorem and understand its intuition (that the distribution of the means of any distribution converge to the normal distribution with increasing number of samples), but I do not know how to apply it to this scenario:

The Central Limit Theorem

Let $X_1$, $X_2$, . . . , $X_n$ be independent, identically distributed random variables with mean μ and variance $σ^2$. Then:

􏰁 a. $\sum_{i=1}^n$$X_i$ ∼ N(nμ, n$σ^2$), approximately.

b. $\bar{X}$ ∼ N(μ,$\frac{σ^2}{n}$), approximately.

The approximation improves as n → ∞.

Question 1) The number of typing errors made on a page follows a Poisson distribution with mean 2. Use the central limit theorem to calculate (approximately) the probability that there are more than 950 typing errors in a 450 page book.

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  • $\begingroup$ We want (roughly) the probability that a normal with mean $900$ and variance $900$ is greater than $950$. $\endgroup$ – André Nicolas May 29 '16 at 21:17
  • $\begingroup$ The error rate per page is 2; the error rate for a 450 pg book is 900. Because of the CLT, POIS(rate=900) is approximately normal with mean 900 and sd 30. According to R the exact Poisson probability is given by 1 - ppois(950,900) which returns 0.0471, and the approximating normal probability is 1 - pnorm(950.5, 900, 30) which returns 0.0462 (both rounded to four places). $\endgroup$ – BruceET May 29 '16 at 22:16
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Let $X_i$ be the number of errors on a single page in a sample of $n=450$ pages. We want to calculate the probability that $\sum X_i>950$

According to CLT, with $n=450$, $\sigma=\sqrt{2}$ and $\mu =2$, $\sum X_i$~$N(900,900)$. And $\bar{X}$ ~ $(2,\frac{2}{450})$

So, $Pr(\bar{X}>\frac{950}{450})=Pr(\frac{\bar{X}-2}{\sqrt{2}/\sqrt{450}}>\frac{\frac{950}{450}-2}{\sqrt{2}/\sqrt{450}})=Pr(Z>\frac{\frac{950}{450}-2}{\sqrt{2}/\sqrt{450}})=0.04779035$

So the probability would be $0.04779035$.

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  • $\begingroup$ I know this is a rather old answer, but I was just wondering, shouldn't that be $Pr(\bar{X} > \frac{950}{450} + \frac{1}{900})$? You are using a continuous distribution to approximate a discreet one, so I would have thought there would be a continuity correction $\endgroup$ – DividedByZero Jun 12 '18 at 21:54

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