Answer to Question #275483 in Statistics and Probability for un3344087

Question

The percentage X of a particular compound contained in a rocket

fuel follows the normal distribution with mean 33 and SD of 3,

though the specification for X is that it should lie between 30

and 35. The manufacturer will get a net profit ( per unit of the

fuel) of Rs. 100, if 30<X<35 ; Rs.50, if 25<X≤30 or 35≤X<40

and incur a loss of Rs. 60 per unit of the fuel otherwise.

Calculate the expected profit of the manufacturer.

Answer 1

To determine the expected profit of the manufacturer, we proceed as follows,

We first determine the following probabilities,

$p(30\lt X\lt 35)=p((30-33)/3\lt Z\lt(35-33)/3)$

$=p(-1\lt Z\lt 0.67)=\phi(0.67)-\phi(-1)=0.7486-0.1587=0.5899$

$p(25\lt X\leqslant 30)=p((25-33)/3\lt Z\lt(30-33)/3)$

$=p(-2.67\lt Z\lt -1)=\phi(-1)-\phi(-2.67)=0.1587-0.0038=0.1549$

$p( 35\leqslant X\lt40)=p((35-33)/3\lt Z\lt (40-33)/3)$

$=p(0.67\lt Z\lt2.33)=\phi(2.33)-\phi(0.67)=0.9901-0.7486=0.2415$

$p(25\lt X\leqslant30\space or \space (35\leqslant X\lt 40))$

$=p(25\lt X\leqslant30)+p (35\leqslant X\lt 40)=0.1549+0.2415=0.3964,$ since the two ranges are mutually exclusive.

$p(X\lt25 \space or X\gt 40)=1-(0.5899+0.3964)=0.0137$

With these probabilities, let us form the table below.

Profit per unit probability

Rs 100 0.5899

Rs 50 0.3964

Rs -60 0.0137

Thus, the expeceted profit per unit is given as,

$E(profit \space per \space unit)=Rs(100*0.5899+50*0.3964-60*0.0137)=Rs\space 77.988\approx Rs\space 78$

Therefore, the expected profit of the manufacturer is Rs 78.