Answer to Question #132224 in Statistics and Probability for Mami

Question

a patrol distributor records shows that sales average 3500 its per day with a stadard deviation 250 .. assume that the sales are normally distributed, find the probability of daily sales between 3300 litre 3800 litre?

Answer 1

Distribution function is $P(X\le x)=F(x)=\frac{1}{250\sqrt{2\pi}}\int\limits_{-\infty}^x e^{-\frac{(t-3500)^2}{2\cdot 250^2}}dt$

If we take $u=\frac{t-3500}{250}$, then $\frac{(t-3500)^2}{2\cdot 250^2}=\frac{u^2}{2}$ and $dt=250du$, so $F(x)=\frac{1}{250\sqrt{2\pi}}\int\limits_{-\infty}^\frac{x-3500}{250} e^{-\frac{u^2}{2}}\cdot 250du=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\frac{x-3500}{250} e^{-\frac{u^2}{2}}du=$

$=\Phi(\frac{x-3500}{250})$, where $\Phi(y)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^y e^{-\frac{u^2}{2}}du$

So the required probability is $P(3300\le X\le 3800)=F(3800)-F(3300)=$

$=\Phi(1.2)-\Phi(-0.8)$

From the table of standard normal distribution https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf we can see that $\Phi(1.2)\approx 0.88493$ and $\Phi(-0.8)\approx 0.21186$, so $P(3300\le X\le 3800)\approx0.88493-0.21186=0.67307$

Answer: $0.67307$