Q: standard deviation find the probability that more than 30% A manufacturer produces a certain type of components that has a length of $2.1 \,\mathrm{cm}$. the distribution of this components is normal with mean $2.15 \,\mathrm{cm}$ and sd $0.4 \,\mathrm{cm}$. There are $50$ pieces in a bag. Find the probability that a randomly selected bag will contain more than $30\%$ of the components with length longer than $2.1 \,\mathrm{cm}$.
Thank You 

A: The probability that one piece is longer than 2.1 cm is
$$P(X> 2.1)=1-P(X\leq 2.1)=1-\Phi\left( \frac{x-\mu}{\sigma} \right)=1-\Phi\left( \frac{2.1-2.15}{0.4} \right)=1-\Phi\left( \frac{-0.05}{0.4} \right)=1-\Phi(-0.125)$$
$\Phi(Z )$ is the cdf of the standard normal distribution. Looking at the table we get for $Z=-0.12$ a value of $0.45224$ and for $Z=-0.13$ a value of $0.44828$. 
A linear interpolation gives $\Phi(-0.125)=\frac{0.45224+0.44828}{2}=0.45026\approx 45\%$
Now you can use the binomial distribution to find the probability that more than $30\%$ of the components have a length longer than 2.1cm. 
$P(Y > 15)=1-P(Y\leq 15)=1-\sum_{k=0}^{15}{50 \choose k} 0.55^k\cdot 0.45^{50-k}$
It is time consuming to calculate this. You can do an approximation by using the standard normal distribution.
$P(Y\leq 15)=\Phi\left(\frac{x+0.5-\mu}{\sigma} \right)$ $=\Phi\left(\frac{15+0.5-0.55\cdot 50}{\sqrt{50\cdot 0.55\cdot 0.45}} \right)=\Phi(-3,41)=0.00032=0.32\%$
