Find sigma, given Probability of normal distribution and mean.
In a factory sugar is packed in bags of normal weight $1000\text{ g}$. The actual weight $X$ (in $\text{g}$) can be modelled as a $N(\mu, \sigma_2)$-distributed random variable. What is the largest value of the standard deviation $\sigma$, assuming $\mu = 1000$, to guarantee that$$\Pr(950 \le X \le 1050) \ge 0.98\text?$$
Can someone please help me with this? I don't know what to do after:$$0.98<\Pr\left(X<\frac{50}σ\right)-\left(1-\Pr\left(X<-\frac{50}σ\right)\right)$$
$\endgroup$2 Answers
$\begingroup$You need to move from normal distribution to standard normal distribution as so:
$$ Z = \frac{X - \mu}{\sigma}$$
We need to assure that:
$$ \mathbb{P} ( 950 \leq X \leq 1050) = \mathbb{P}( \frac{950-1000}{\sigma} \leq Z \leq \frac{1050-1000}{\sigma}) = \phi(\frac{50}{\sigma}) - \phi(-\frac{50}{\sigma}) \\ = \phi( \frac{50}{\sigma}) \geq 0.99$$
Looking at the $Z$-table we get that the corresponding value is $2.33$ and so:
$$ \frac{50}{\sigma} \geq 2.33 \\ \sigma \leq \frac{50}{2.33} \approx 21.46$$
$\endgroup$ $\begingroup$Hint:
$$P ( 950 \leq X \leq 1050)=\Phi\left(\frac{1050-1000}{\sigma} \right)-\Phi\left(\frac{950-1000}{\sigma} \right)$$ $$=\Phi\left(\frac{50}{\sigma} \right)-\Phi\left(\frac{-50}{\sigma} \right)=2\Phi\left(\frac{50}{\sigma} \right)-1\geq 0.98$$
The latter equality sign is a consequence of the symmetry of the standard normal distribution. I leave it to you to solve the inequality for $\sigma$
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