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Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If the data conforms to a normal distribution and a household is selected at random, the probability of its generating between 28 and 30 pounds per month is about __?

User Mathiasfc
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1 Answer

2 votes

Answer: 0.3413

Explanation:

Given: an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds.

Let
\mu=\text{ 28 pounds and }\sigma=\text{ 2 pounds} .

Let x be the random variable that denotes the amount of newspaper for garbage generated by each household each month.

If the data conforms to a normal distribution and a household is selected at random, the probability of its generating between 28 and 30 pounds per month is about


P(28<x<30)=P((28-28)/(2)<(x-\mu)/(\sigma)<(30-28)/(2))\\\\=P(0<z<1)\ \ \ [\because\ z=(x-\mu)/(\sigma)]\\\\ =P(z<1)-P(z<0)\\\\=0.8413-0.5=0.3413

Hence, the required probability is 0.3413.

User Jonatan Dragon
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