Answer to Question #265832 in Statistics and Probability for john

Question

Suppose the Production Operation Manager at the canning plant for evaporated milk is faced with a problem. Because of the designed machine, it generated a standard deviation of 7 ml in the filling of evaporated milk, but the average ml per fill can be adjusted. Under fear of revocation of license to operate and penalty charges by the Department of Trade and Industry (DTI), the management of the company has ordered the manager to set the filling machine so that there is a maximum 10% chance of a 370 ml can content less than 360 ml. What should the average be?

Step 1: Draw a figure and represent the area.

Step 2: Find the z value.

Step 3: Answer the question. 

Answer 1

If i correctly understand the question, the goal is to find such value 360 < X < 370(an average ml per can), that $P(N(X,7^2)<360)<0.1$

Step 1

We will find such X that $P(N(X,7^2)<360)=0.1$, and every value from that X to 370 will satisfy the conditions

The graphic of density function for such distribution is presented below

Step 2

$N(X,7^2)=X+7N(0,1)$

$P(X+7N(0,1)<360)=0.1\implies P(N(0,1)<{\frac {360-X} 7})=0.1\implies {\frac {360-X} 7}=-1.28\implies X=368.96ml$

Step 3

The average ml per fill must be from 368.96 to 370 ml