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
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 $\sigma$