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The amount of soda in a 16-ounce can is normally distributed with a mean of 16 ounces and a standard deviation of 0.20 ounce. What percentage of these cans will have less than 15.6 ounces

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Answer: The percentage of these cans will have less than 15.6 ounces = 2.28%

Explanation:

Let x = amount of soda in a 16-ounce can which is normally distributed.


\mu=16\text{ ounces},\ \ \ \ \sigma=0.20\text{ ounces}

The probability that cans will have less than 15.6 ounces:


P(X<15.6)=P((X-\mu)/(\sigma)<(15.6-16)/(0.20))\\\\=P(z<-2)=1-P(Z<2)\\\\=1-0.9772\ \ \ [\text{By p-value table}]\\\\=0.0228=2.28\%

Hence, the percentage of these cans will have less than 15.6 ounces = 2.28%

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