Question

A soft-drink machine is regulated so that it dischargesan average of 200 milliliters per cup. If theamount of drink is normally distributed with a standarddeviation equal to 15 milliliters,(a) what fraction of the cups will contain more than224 milliliters?

Answer 1

Answer:0.0548

Step-by-step explanation:

Let x denotes the random variable that represents the amount of drink in cups.

As per given , we have

`\mu=200`
`\sigma=15`

Since
`z=(x-\mu)/(\sigma)`

The z-value corresponds to x=224

`z=(224-200)/(15)=1.6`

Required probability :

P-value :
`P(x>224)=P(z>1.6)=1-P(z<1.6)`

`=1- 0.9452007=0.0547993\approx0.0548`

Hence, the fraction of the cups will contain more than 224 milliliters = 0.0548