Question

A soft-drink machine is regulated so that it discharges an average of 250 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 20 milliliters,
a) What fraction of the cups will contain more than 264 milliliters? 
b) How many cups will probably overflow if 270-milliliter cups are used for the next eight thousand drinks? 
c) Above what value do we get the largest 28% drinks?

Answer 1

mean = 250
sd = 20

As per central limit theorem,
z = (x - mu)/sigma

a)
P(X > 264)
= P(z > (264 - 250)/20)
= P(z > 0.7)
= 0.242

Hence the fraction of cupts taht will contain more than 264mm soft drink is 0.242

b)
Lets first find the probabiltiy that a cup will overflow
P(X > 270)
= P(z > (270 - 250)/20)
= P(z > 1)
= 0.1587

In case of 8000 cups,
8000 * 0.1587 = 1269.6 i.e. 1270 cups will overflow

c)
alpha = 0.28
Z(alpha) = 0.58

x = 250 + 0.58*20
x = 261.6

Hence above 261.6 mm we will get 28% of largest drink