hw6_sol - Statistics 528 – Homework 6 Solutions 4.10 (a)...

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Statistics 528 – Homework 6 Solutions 4.10 (a) The sample space would be {female, male}. (b) The sample space would be {6, 7, …, 20}. (c) The sample space would be the interval [2.5, 6]/min. (d) For living human beings, heart rate is probably in the range of 40-200 and would be a whole number. The theoretical range would be positive whole numbers. 4.14 The probability they both have type O is the product of the probabilities for each country, because two persons are independently chosen. This is 0.45*0.35=0.1575. To find the probability they have the same blood type requires doing the above calculation for each type. Then, because the types are disjoint, we can add the results. The probability they both have type A is 0.40*0.27=0.108. Both type B :0.11*0.26=0.0286. Both type AB: 0.04*0.12=0.0048. Therefore, P(both have the same blood type) = P(both have type O)+P(both have type A)+P(both have type B) +P(both have type AB) = 0.1575+0.108+0.0286+0.0048=0.2928. 4.20
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This note was uploaded on 07/26/2011 for the course STA 528 taught by Professor Calder during the Winter '09 term at Ohio State.

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hw6_sol - Statistics 528 – Homework 6 Solutions 4.10 (a)...

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