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I have a problem to solve.

A factory makes bolds (following the normal distribution) with an average weight= $\mu$ gr and s.d =$\sigma$ gr. A bold is considered faulty if it weights less than $z$ gr. A store buys $100$ of these bolds, what is the probability that $2$ are faulty?

What troubles me here, how do I combine the information of $2$ and $100$? The new $\mu$ and $\sigma$ that I will use to get the $Z$ should be calculated for the $2$ or the $100$ objects ?

I am a person
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Just_Newbie
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1 Answers1

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This will involve combining the notions of normal and binomial distributions. Let $X$ be the weight of a bolt; the probability that it is faulty is, with $Z$ being standard normal, $$\begin{split}\Pr(X<z)&=\Pr\left(Z<\frac{z-\mu}{\sigma}\right)\end{split}$$

Call the above probability $p$. The probability that two bolts in one hundred are faulty is

$${100\choose 2}p^2(1-p)^{98}$$

Vons
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