Question

A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. if the amount of drink is normally distributed with a standard deviation equal to 15 milliliters, what is the probability that a cup contains between 191 and 209 milliliters?

Answer 1

The probability that a cup contains between 191 and 209 milliliters will be 0.4514 or 45.14%

Explanation

A soft-drink machine is regulated so that it discharges an average of 200 milliliters per cup. So, the mean$(\mu)$ = 200 milliliters/cup

Given that, the standard deviation$(\sigma)$ = 15 milliliters

For finding the probability that a cup contains between 191 and 209 milliliters, first we need to find the z-scores at $X=191$ and $X= 209$

Formula for z-score:  $z= \frac{X-\mu}{\sigma}$

So,  $z(X>191)= \frac{191-200}{15}=-0.6$

and $z(X<209)=\frac{209-200}{15}=0.6$

According to the normal distribution table, probability for z-score of -0.6 is 0.2743 and probability for z-score of 0.6 is 0.7257

So......

$P(191<X<209)\\ \\ = P(-0.6<z<0.6)\\ \\ = 0.7257-0.2743\\ \\ =0.4514$

Thus, the probability that a cup contains between 191 and 209 milliliters will be 0.4514 or 45.14%