SOLUTION: According to the Internal Revenue Service, the mean tax refund for the year 2014 was $2,800. Assume the standard deviation is $450 and that the amounts refunded follow a normal pro

Question 1113523: According to the Internal Revenue Service, the mean tax refund for the year 2014 was $2,800. Assume the standard deviation is $450 and that the amounts refunded follow a normal probability distribution.

What percent of the refunds are more than $3,100? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

What percent of the refunds are more than $3,100 but less than $3,500? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)

What percent of the refunds are more than $2,250 but less than $3,500? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
According to the Internal Revenue Service, the mean tax refund for the year 2014 was $2,800. Assume the standard deviation is $450 and that the amounts refunded follow a normal probability distribution.
What percent of the refunds are more than $3,100? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)
z(3100) = (3100-2800)/450 = 2/3
P(x>3100) = P(z>2/3) = normalcdf(2/3,100) = 0.2525
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What percent of the refunds are more than $3,100 but less than $3,500? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)
z(3500) = (3500-2800)l/450 = 1.56
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P(3100< x < 3500) = P(2/3 < z < 1.56) = normalcdf(2/3,1.56) = 0.1931
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What percent of the refunds are more than $2,250 but less than $3,500? (Round the intermediate values to 2 decimal places. Round your answer to 2 decimal places.)
z(2250) = (2250-2800) = -550/450 = -1.22
P(2250 < x < 3500) = P(-1.22 < z < 1.56) = normalcdf(-1.22,1.56) = 0.8294
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Cheers,
Stan H.
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