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So I was again recently looking through Statistics Problems and Solutions Second Edition by Bassett et al., 1986 when I ran into this interesting problem:
$2$B.$4\quad$The number of matches in a box
A manufacturer claims that boxes contain, on average, $49$ matches. On checking output over a substantial period, the manager finds that $3$% of boxes contain fewer than $46$ but $\quad25$% contain $51$ or more. Making reasonable assumptions, calculate the mean and standard deviation of the number of matches per box. Do you feel like the manufacturer's claim is plausible?
What I know from this question
- Defining the numbers of matches in a box as $x$, we know that $$\operatorname{avg}(x\in\mathbb{B})=49\,\text{where }\mathbb{B}\text{ is any one box.}$$
- The probability of there being less than $46$ matches in any one box $\mathbb{B}$ can be calculated as $\operatorname{Pr}(x\lt46)=0.03$
- The probability of there being more than or equal to $51$ matches in any given box $\mathbb{B}$ can be calculated as $$\operatorname{Pr}(x\geq51)=0.25$$
My attempt to solve it
Let $\mu$ denote the mean and $\sigma$ define the mean and standard deviation. Now,$$\Phi\left(\frac{\frac{91}{2}-\mu}{\sigma}\right)=0.03\quad\text{and} \quad1-\Phi\left(\frac{\frac{101}{2}-\mu}{\sigma}\right)=0.25$$Since I know that on standard tables,$$\Phi(k_1)=0.03\text{ where }k_1\approx-1.88\text{ and }\Phi(k_2)=0.75\text{ where }k_2\approx0.67$$Solving the $2$-system equation$$\mu-1.88\sigma=45.5\text{ and }\mu+0.67\sigma=50.5$$gets us$$\mu\approx49.186$$$$\text{and}$$$$\sigma\approx1.961$$$$\text{ meaning that it would be the correct decision to believe the manufacturer's claim.}$$My question
Is my solution correct, or what steps would I need to take to attain the correct solution more easily, as well as any steps that might make it easier to solve?
To clarify
- This is different from my question here.
- I am absolutely sorry if this is a short/trivial question.
- Sorry if I accidently skipped any steps that would make it easier to solve/would get me the correct answer.
- Sorry if my tags aren't correct, they most likely are but still.