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The manager of XYZ shop orders 110 cartridges per week but runs out cartridges every 1 out of 4 weeks. He knows that on average the shop uses 95 cartridges per week.

What is the standard deviation of this distribution

What i did:

$np = 95$

$110(p) = 95$

$q=3/22$

$sd = \sqrt{95(\frac{3}{22})}$

the answer sheet says 22.26

please help

bryan
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  • Presumably, if he has cartridges left over at the end of the week, he throws them away? – Arthur Aug 28 '19 at 13:48
  • What are $n$ and $p$? Do we assume our random variable to follow a distribution among a particular family? – David Aug 28 '19 at 13:48

1 Answers1

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Let's assume $X$ (number of cartridges sold) is a random variable following a $N(\mu, \sigma)$ normal distribution. We know:

$$\mu=95$$ $$ P(X>110)=\frac{1}{4}$$

We need an expression for $P(X>100)$ Let $Z$ follow a $N(0,1)$ standard normal distribution. Then:

$$P(X>110) = P(Z>\frac{110-95}{\sigma}) = \frac{1}{4}$$

Of course, $Z$ is just "standardized $X$". We subtract $X$'s mean and divide by its standard deviation Now you need to find $\sigma$ such that $$P(Z>\frac{15}{\sigma}) = \frac{1}{4}$$

Using a chart (or a piece of software) you can see that $P(Z>t) = \frac{1}{4} \iff t=0.6784$. You can now solve for $\sigma$

David
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