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) Whatfractionof the cups will contain more than 264 milliliters? b) How many cups will probably overflow if 270-milliliter cups are used for the nexteight thousand drinks? c)Abovewhat 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

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