Question

The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. If the benefits are normally distributed with a standard deviation of $460, find the probability that the mean benefit for a random sample of 20 patients is more than $4100. According to the U.S.Census, 67.5% of the U.S. population were born in their state of residence. In a random sample of 200 Americans, what is the probability that fewer than 125 were born in their state of residence? (Use continuity correction)

Answer 1

Solution :- (1) Given dara, Insurance ben fit = $ 4064 - Me Standard deviation = $ 460 =- random sample = 20 patients. Em "p" for more than $4100:- probability (p) more than $4100 = P(x>4100) 2 - M =P ( > 4100-2406 4 ) > 4100 - 4064 (460 20 =P(Z > 013499) So, Here = 0.5 - P(Z < 0.3499) = 0.5 -0.1368 = 0.3632 0.3199

(2) + Given, PC *<125), m = 200, p=0.0675 → using excel function Binom Dist (124,200, 0.675, true) or TI - 83/84 function binomedf (200, 0.675,124), exact answer is 0.0578132376988153 Xw Binomiat 200 , p=0.675 → Binomial can be approximated to normal with : Mamp = 200* * 0.675 5135 Grimp(1-P) - 1200 x 10.675) 61-0.675) 6.623821 6-624 * p (x <125) → Since we are approximating a descrete binomial distribution by continuous normal dostribution, values between 124.5 and 12-5 both approximate to 125. Thus, "less than 125" corresponds to continuous hormal distribution with X 2124-5 after Continuity correction → Heme, Normal Distribution, DE 194.5 M: 135 o: 6.624 this to Standard normal using we convent 1 24.5 -135 6.624 -1.5851881-59

.6.623821 (< 124-5 124.5 1.35 p(x < 124-5) = area to the left Op 124.5 P ( X 2124.5) = p (ze 1.59 = 0.0559 (from 2-table) PCs less than 125) = 5.0559) + Sucel is used instead of 7-table, then I would be 0.0564618399 Note #at f tis question is solved without contar correction, the onswer would have T p(2125): p (z < -1:51) = 0.0655 without + .......topia integer continuing cºrrea Piz + 5) = 0. oks/ - erel used instead of 7-table, it would be ...655596576