Question 243016: Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the
human resource director shows the annual cost per employee per year followed the normal
probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.
a. What fraction of the employees cost more than $1,500 per year for dental expenses?
b. What fraction of the employees cost between $1,500 and $2,000 per year?
c. Estimate the percent that did not have any dental expense.
d. What was the cost for the 10 percent of employees who incurred the highest dental
expense?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a standard deviation of $420 per year.
a. What fraction of the employees cost more than $1,500 per year for dental expenses?
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a(1500) = (1500-1280)/420 = 220/420 = 0.5238
P(x > 1500) = P(z > 0.5238) = 0.30 = 30%
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b. What fraction of the employees cost between $1,500 and $2,000 per year?
z(2000) = (2000-1280)/420 = 1.7143...
P(1500 < x < 2000) = P(0.5238 < z < 1.7143) = 0.257
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c. Estimate the percent that did not have any dental expense.
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z(0) = (0-1500)/420 = -3.5714..
P(x < 0) = P(z < -3.4714) = 0.0002589..
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d. What was the cost for the 10 percent of employees who incurred the highest dental expense?
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invNorm(0.90) = z = 1.2816
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x = zs+u
x = 1.2816*420+1280
x = $1818.25
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Cheers,
Stan H.
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