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AveGali [126]
3 years ago
6

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?
Business
1 answer:
Fiesta28 [93]3 years ago
5 0

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

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