Asoft-drink machine is regulated so that it discharges an average of 200 mililiters per cup. if the amount of drink is normally distributed with a standard deviation equal to 15 mililiters, what fraction of the cups will contain more than 224 mililiters?

0.0548

Step-by-step explanation:

Mean(μ) = 200 mm/ cup

Standard deviation (σ) = 15 mm/cup

Let X be the amount of drink distributed.

Pr(x>224) =

Since it is normally distributed

Z = (x - μ)/σ

Z = (224 - 200)/ 15

Z = 24/15

Z = 1.6

From the normal distribution table, 1.6 = 0.4452

Φ(z) = 0.4452

If Z is positive, Pr(x>a) = 0.5 - Φ(z)

Pr(x >224) = 0.5 - 0.4452

= 0.0548

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