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10% of the chocolate bars that are produced in a factory have unacceptable shape. In a sample of 1000 chocolate bars find the probability that the number of unacceptable shapes is (A) less than 80 (B) between 90 and 115 (C) 120 or more.

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  • $\begingroup$ If this is homework, please mark it as such. $\endgroup$ – Maroon Mar 8 '14 at 11:23
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    $\begingroup$ Please don't dump questions here with no source, no motivation, no sign of what you know about the terms, and worst of all no sign of the slightest effort on your part to solve the problem. $\endgroup$ – Gerry Myerson Mar 8 '14 at 11:23
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The number of unacceptable chocolate bars in a sample of $1000$ is a random variable $X$ which has the binomial distribution with parameters $n=1000$ and $p=0.10$, in symbols $$X \sim \mathrm{Bin}(n=1000,\,p=0.10)$$

To calculate the required probabilities we will use the normal approximation to the binomial distribution, that is $$X \sim \mathrm{N}\left(\mu=np, σ^2=np(1-p)\right)$$ approximately. Here $μ=np=1000\cdot0.1=100$ and $σ^2=np(1-p)=100\cdot0.9=90$. Applying also the continuity correction we have that

  1. For A) $$\begin{align*}P(X<80)&=P(X\le79) \approx P\left(\frac{X-μ}{σ}\le\frac{79+1/2-100}{\sqrt{90}}\right)=P(Z\le -2.16)=\\ \\&=Φ(-2.16)=1-Φ(2.16)=0.015\end{align*}$$ using the normal distribution tables.
  2. For B) $$\begin{align*}P(90<X<115)&=P(X\le114)-P(X\le90)\approx \\& \approx P\left(\frac{X-μ}{σ}\le\frac{114+1/2-100}{\sqrt{90}}\right)-P\left(\frac{X-μ}{σ}\le\frac{90+1/2-100}{\sqrt{90}}\right)\\&=\ldots \end{align*}$$
  3. For C) $$\begin{align*}P(X\ge120)&=1-P(X<120)=1-P(X\le119)\approx\\ \\&\approx1-P\left(\frac{X-μ}{σ}\le\frac{119+1/2-100}{\sqrt{90}}\right)=\ldots \end{align*}$$
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    $\begingroup$ Some of us are of the opinion that it's a bad idea to post complete solutions of homework problems, especially when there is no sign that the person asking the question has put the slightest amount of effort into answering it. $\endgroup$ – Gerry Myerson Mar 9 '14 at 4:22
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    $\begingroup$ @GerryMyerson Thanks for pointing it out so kindly. Personally, I do not completely agree with that opinion, because some pupils are weak in certain subjects and I believe that they need to see the complete answer. Nevertheless I accept your opinion (which you expressed kindly) as more experienced members and therefore I ensure you that in the future I will try not to give such answers and in general that I will try to act according to the guidelines of the more experienced members of the forum. $\endgroup$ – Jimmy R. Mar 9 '14 at 20:24
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    $\begingroup$ First of all , thanks to stefanos for the solution.Secondly i tried this at home but did not manage to solve it.Next time i ll try to show my effort in order not to think that i did not put the slighest amount of effort.Also this homework is optional for my lesson so a complete answer help me to become better and see my mistakes $\endgroup$ – Ioannis Mar 10 '14 at 20:45
  • $\begingroup$ @Ioannis You are welcome :). Next time, try to add one-two sentences describing some context (as for example what you said in the comment, that it is an optional homework and so on) and 1-2 snetences concerning your thoughts (for example, that this an exercise about combinations/ permutations, or normal distribution etc.). That will be enough and more helpful ;)! See you round! $\endgroup$ – Jimmy R. Mar 10 '14 at 21:18
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    $\begingroup$ Ok Stefanos , i ll do my best $\endgroup$ – Ioannis Mar 10 '14 at 21:37

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