# Statistics & Probability -Distribution problem

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.

• If this is homework, please mark it as such. – Maroon Mar 8 '14 at 11:23
• 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. – Gerry Myerson Mar 8 '14 at 11:23

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*}