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?

Respuesta :

Answer:0.0548

Explanation:

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

As per given , we have

[tex]\mu=200[/tex]   [tex]\sigma=15[/tex]

Since [tex]z=\dfrac{x-\mu}{\sigma}[/tex]

The z-value corresponds to x=224

[tex]z=\dfrac{224-200}{15}=1.6[/tex]

Required probability :

P-value : [tex]P(x>224)=P(z>1.6)=1-P(z<1.6)[/tex]

[tex]=1- 0.9452007=0.0547993\approx0.0548[/tex]

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

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