Question

A stamping machine produces ‘can tops’ whose diameters are normally distributed with a standard deviation of 0.02 inch. At what nominal mean diameter should the machine be set, so that no more than 9 % of the ‘can tops’ produced have diameters exceeding 3.5 inches?

Answer 1

Answer on Question #45522 - Math - Statistics and Probability

Normal distribution:

$$f(x, \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$$

Where $\mu$ — mean of the distribution, $\sigma$ — standard deviation, $f(x, \mu, \sigma)$ — probability density function, $x$ — variable.

Input of the task: $\sigma = 0.02$ inch, $P(3.5 < x) \leq 0.09$ (9% corresponds to $\frac{9}{100} = 0.09$), $\mu$ — ?

$P(3.5 < x)$ — probability of $x$ to be greater than 3.5 inches; $x$ — diameter of the "can tops".

$$P(3.5 < x) = \int_{3.5}^{\infty} f(x, \mu, \sigma) \, dx = \int_{3.5}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \, dx \leq 0.09$$

$\sigma = 0.02$ (inch); $\sigma^2 = 0.02^2 = 0.0004$ (inch²);

$$\begin{aligned} & \int_{3.5}^{\infty} \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}} \, dx = \int_{3.5}^{\infty} \frac{1}{0.02\sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2 \times 0.0004}} \, dx = \frac{1}{0.02\sqrt{2\pi}} \int_{3.5}^{\infty} e^{-\frac{(x - \mu)^2}{0.0008}} \, dx \leq 0.09 \\ & \int_{3.5}^{\infty} e^{-\frac{(x - \mu)^2}{0.0008}} \, dx \leq 0.09 \times 0.02\sqrt{2\pi} \approx 0.004512 = 4.512 \times 10^{-3} \end{aligned}$$

Consider $t = \frac{x - \mu}{\sqrt{0.0008}}$, $dt = \frac{dx}{\sqrt{0.0008}}$, $dx = \sqrt{0.0008} \times dt \approx 0.0283 \, dt$.

Upper boundary: $\lim_{x \to \infty} \frac{x - \mu}{\sqrt{0.0008}} = \infty$; lower boundary: $\left. \frac{x - \mu}{\sqrt{0.0008}} \right|_{3.5} \approx \frac{3.5 - \mu}{0.0283} \approx 35.36(3.5 - \mu)$;

Thus, $\int_{3.5}^{\infty} e^{-\frac{(x - \mu)^2}{0.0008}} \, dx = 0.0283 \int_{35.36(3.5 - \mu)}^{\infty} e^{-t^2} \, dt \leq 4.512 \times 10^{-3}$;

Consider $a = 35.36(3.5 - \mu)$.

$$\begin{aligned} & 0.0283 \int_{a}^{\infty} e^{-t^2} \, dt \leq 4.512 \times 10^{-3} \\ & \int_{a}^{\infty} e^{-t^2} \, dt \leq 159.435 \times 10^{-3} \approx 0.1594 \end{aligned}$$

There is no elementary indefinite integral for $\int e^{-t^2} \, dt$, hence the only way is to use methods of numerical integration. Result: $a \geq 0.94836$.

Thus, $a = 35.36(3.5 - \mu) \geq 0.94836$; $-\mu \geq \frac{0.94836}{35.36} - 3.5$; $\mu \leq 3.5 - \frac{0.94836}{35.36}$;

$\mu \leq 3.5 - 0.02682$; $\mu \leq 3.47318$;

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