Answer to Question #267346 in Statistics and Probability for Dawnnn

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 356 ml. What should the average be?

Answer 1

Let the random variable $X$represent the amount of evaporated milk filled in a can. The distribution of this random variable is, $X\sim N(\mu, \sigma^2)$. The value of the variance is given as, $\sigma^2=7^2$. We need to determine the value of the mean $\mu$.

Now, the maximum probability that a 370 ml can is filled with evaporated milk of less than 356 ml is 0.1.

Therefore, we can express this probability as,

$p(X\lt 356)\leqslant 0.1$ and set this probability equal to 0.1 as it would give the the least desired mean that is; $p(X\lt356)=0.1$. Since the random variable $X$is normally distributed, we standardize this variable to a standard normal variable and find the required value for the mean as follows.

$p(X\lt356)=p((X-\mu)/\sigma\lt (356-\mu)/\sigma)=0.1=p(Z\lt(356-\mu)/\sigma)=0.1,$ We then substitute for $\sigma$.

$p(Z\lt (356-\mu)/7)=0.1$.

Let $Z_0=(356-\mu)/7.......(i)$, such that $p(Z\lt Z_0)=0.1$. To find the value of $Z_0$ we enter the following command in $R.$

 y=qnorm(0.1) and the output is ,

> y

[1] -1.281552

The output is the value of $Z_0$.

Therefore, $Z_0=-1.281552$.

Putting this value in $equation\space (i)$,

$Z_0=-1.281552=(356-\mu)/7$, which we can solve for $\mu$.

Making $\mu$ the subject,

$\mu=356+(1.281552*7)= 364.9709$

Therefore, the average should be $\mu=364.9709$.