A company produces one-kilogram sugar packets. The specifications on the net content are 1000 ≠ 5 grams. Assuming that the net content follows normal distribution with mean weight as 1005 grams and the process capability equal to 30 grams, find out the proportion of packets that have weight less than lower specification limit. What should be the mean if this proportion is to be reduced to 0.01?
I have no idea what is the relation between process capability and mean. How do I solve it?
The process capability will give you the $\sigma$ of the process:
Rewriting $$C_p = \frac{USL-LSL}{6\sigma}$$ we get $$ \sigma = \frac{USL-LSL}{6C_p} = \frac{1005-995}{6(30)} = \frac{1}{18} $$ So you know that the process has a normal distribution with $\mu = 1005,$ and $\sigma=1/18$. Then it's just calculating the probability that $x<=995$ for this distribution.
Part 1. You can integrate the weight $x$ using the PDF of the normal distribution over $[0, 995]$ or simply use the CDF of the normal distribution to get the answer. $$ CDF(x=995) = \frac{1}{2}\left[ 1+\text{erf}\left(\frac{x-\mu}{\sqrt{2\sigma^2}}\right)\right] $$ You should be able to find a tabulated table for the error function $\text{erf}$ or use a calculator.
Part 2. Set $CDF(x=995)= 0.01$, and solve for $\mu$.