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- Apr 14, 2013

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The daily turnover $X$ of Cafes has the expected value $ \mu_X = 600 $ Euro and the standard deviation $ \sigma_X = 30 $ Euro.

(a) How many cafes at least have to be surveyed in a random sample, so that $\overline{X}_n$ deviates from $\mu_X$ with a probability of at least $95\%$ by less than $12$ euros?

(b) After a survey of $500$ Cafes the arithmetic mean is $690$. Is this result surprising after the question (a) ?

I have done the following:

(a) From Chebyshev's inequality we have that \begin{equation*} P\left (\left |\overline{X}_n-E(\overline{X}_n)\right |< \epsilon\right )\geq 1-\frac{V(\overline{X}_n)}{\epsilon^2}\end{equation*} with $E(X_n)=\mu_X=600$ and $V(X_n)=\frac{\sigma_X^2}{n}=\frac{30^2}{n}=\frac{900}{n}$.

It must hold \begin{align*} P\left (\left |\overline{X}_n-600\right |< 12\right )\geq 95\% &\Rightarrow 1-\frac{V(\overline{X}_n)}{12^2}\geq 95\% \\ & \Rightarrow 1-\frac{\frac{900}{n}}{144}\geq 0.95 \\ & \Rightarrow 1-0.95\geq \frac{\frac{900}{n}}{144} \\ & \Rightarrow 0.05\geq \frac{25}{4n} \\ & \Rightarrow n\geq \frac{25}{4\cdot 0.05} \\ & \Rightarrow n\geq 125\end{align*}

That means that at least $125$ Cafes have to be surveyed.

Is this correct?

(b) Why does not hold when number of the surveyed Cafes is $500$ ?