I need help with this problem, please:
A company supplies pins in bulk to a customer.
The lengths of the pins made by the machine are normally distributed with a mean of $1.008$ inches and a standard deviation of $0.045$ inch.
The company supplies the pins in large batches to a customer. The customer will take a random sample of $50$ pins from the batch and compute the sample mean.
If the sample mean is within the interval $[0.99, 1.01]$ inch, then the customer will buy the whole batch.
Now suppose that the standard deviation can be reduced. what maximum value of the standard deviation would make $90$% of the parts acceptable to the customer?
Here's my approach:
$$\mathit X \sim \mathit N \,(1.008, \frac{\sigma^2}{50}) $$
$$Z_1= \frac{(1.01-1.008) \sqrt{50}}{\sigma}$$ $$Z_2= \frac{(0.99-1.008) \sqrt{50}}{\sigma}$$ $$\mathit P(-\frac{0.018 \sqrt{50}}{\sigma}\lt \mathit X \le \frac{0.002 \sqrt{50}}{\sigma}) = 0.9$$ If the probability is $0.9$, so the probability of not being in the interval (being in both tails) is $0.1$. (Each tail: ${0.1\over 2}=0.05$)
Therefore, $\mathit P(\mathit Z_1)=1-0.05=0.95$ ; the whole area under the curve excluding the right tail.
From the table, $\mathit Z_1 \le 1.645$.
And so the probability of being in the left tail is $0.05$, $\mathit P(\mathit Z_2)=0.05$
$\mathit Z_2 \le -1.645$
∴ $$\frac{(1.01-1.008) \sqrt{50}}{\sigma}\le 1.645$$ $$\frac{\vert0.99-1.008\vert \sqrt{50}}{\sigma}\ge 1.645$$ And now, it looks like a contradiction if I calculate it right; as $1.01$ and $0.99$ are not equally varied/deviated from the mean.