Question

A soft-drink machine is regulated so that it discharges an average of 6.8 ounces per cup. If the amount of drink is normally distributed with a standard deviation equal to 0.5 ounces,

Part a: What is the probability that a cup will contain more than 7.72 ounces?

Part b: What is the probability that a cup contains between 6.72 and 7.06 ounces?

Part c: How many cups will overflow if 7.8-ounce cups are used for the next 1000 drinks?

Part d: Below what value do we get the smallest 25% of the drinks?

Answer 1

The distribution given here is:

a) The probability here is computed as:

P( X > 7.72 )

Converting this to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.0329 is the required probability here.

b) The probability that a cup contains between 6.72 and 7.06 ounces is computed here as:

Converting this to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore 0.2621 is the required probability here.

c) First we compute here P(X > 7.8 )

Converting this to a standard normal variable, we get:

Getting it from the standard normal tables, we get:

Therefore out of 1000 drinks, number of drinks which would overflow could be modelled here as:

Therefore expected number of cups to overflow = np = 1000*0.0228 = 22.8 drinks

d) From standard normal tables, we have:

P(Z < -0.6745) = 0.25

Therefore the volume required here is:

= 6.8 -0.6745*0.5 = 6.46275 ounces.