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

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 $z=\dfrac{x-\mu}{\sigma}$

The z-value corresponds to x=224

$z=\dfrac{224-200}{15}=1.6$

Required probability :

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

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

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