In a store, when the quantity of any product falls below $20$ units, the order for new products is made. It is assumed that the product order has a normal distribution with an average of $15$ and a standard deviation of $6$.
$a)$ Determine the probability that there will be a shortage of the product, ie determine the probability $P (X> 20)$.
$b)$ Determine the number of products, so that the probability of having a shortage is less than 3%.
My attemp is: we make the transition of variable $X(15,6)$ to the variable $Y$ $$Y=\frac{X-15}{6}$$
for par $a)$ $$P (X> 20)=P(\frac{X-15}{6}>\frac{20-15}{6})=P(Y>0.83)=1-P(Y<0.83)=1-F_Y(0.83)=\cdots$$
I don't know if this part is good, and I don't know how to calculate it $F_Y(0.83)=???$
For part $b)$ $$P(X<x)=0.03$$ Through the variable Y we do the transition $$Y=\frac{X-15}{6}$$ $$P(X<x)=P(\frac{x-\mu}{\sigma})=F_Y\frac{x-\mu}{\sigma}=0.03$$ $$F_Y(\frac{x-\mu}{\sigma})=0.03/^{-1}$$ $$(\frac{x-\mu}{\sigma})=F_Y^{-1}(0.03)$$ $$x=\mu+\sigma F_Y^{-1}(0.03) $$ $$x=15+6\cdot ?$$ I don't know if this part is good, and I don't know how to calculate it $F_Y^{-1}(0.03)=???$
Please help me. Thanks for your attetnion and your help.