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IgorC [24]
3 years ago
14

Bob and Roberta are married and have two children. They are covered by a comprehensive medical and major medical policy with a $

200 deductible per person, an 80/20 coinsurance provision, and an annual stop loss of $3,000. Bob dislocated his shoulder in November and needed surgery to fix the injury, which cost $35,000. Bob, Roberta, and each of the children had an annual physical exam earlier in January, which cost $250 each. Aside from these expenses, there were no other health care costs incurred by the family this year. How much will Bob have to pay out of pocket to cover the costs of his surgery?
Mathematics
1 answer:
Novay_Z [31]3 years ago
4 0

Answer:

Bob will have to pay $2,160 to cover the cost of his surgery

Step-by-step explanation:

Coinsurance on the surgery = 20% * $35,000

Coinsurance on the Surgery = 20/100 * $35,000

Coinsurance on the Surgery = 0.2 * $35,000

Coinsurance on the Surgery = $7,000

Annual Stop Loss = $3,000

Annual Physical Exam costs = $250 * 4

Annual physical Exam = $1000

Coinsurance on physical exam = 80% * $1,000

Coinsurance= 80/100 * $1,000

Coinsurance = 0.8 * $1,000

Coinsurance = $800

Deductible paid for Physical Exam = 20% *$200

Deductible = 20/100 * $200

Deductible = 0.2 * $200

Deductible = $40

Total payables = Annual Loss - Coinsurance on Physical Exam - Deductible for Physical Exam

Total Payables = $3,000 - $800 - $40

Total Payables = $2,160

The insurance company will pay the rest of the medical expenses.

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he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w
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Answer:

90% confidence interval is ( -149.114, -62.666   )

Step-by-step explanation:

Given the data in the question;

Sample 1                                Sample 2

x"₁ = 259.23                            x"₂ = 365.12

s₁  = 34.713                              s₂ = 48.297

n₁ = 5                                       n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

S_p^2 = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

S_p^2 = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

S_p^2 = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

S_p^2 = [ 4819.9692 + 20993.4 ] / [  13 ]

S_p^2 = 25813.3692 / 13

S_p^2 = 1985.64378

Now the Standard Error will be;

S_{x1-x2 = √[ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

S_{x1-x2 = √[ 397.128756 + 198.564378 ]

S_{x1-x2 = √595.693134

S_{x1-x2 = 24.4068

Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

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So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

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<h2><u>______________________</u></h2>

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