Suppose 20 random donors come to volunteer at a local blood drive. Only about 6%

Question

Suppose 20 random donors come to volunteer at a local blood drive. Only about 6% of them are O-negative (universal donors note we are only interested in O and no other outcome matters) What are the mean and standard deviation of the number of universal donors among the 20 volunteers (mean, S.D.) a) 1.2, 1.06 b) 1.2, 1.1236 c) 6, 20 d) 0, 1 Now consider 10 donors present as volunteers and the probability of success p = 0.10. What is the probability that P(1 lessthanorequalto x lessthanorequalto 4). a) 0.3487 b) 0.7361 c) 0.6497 d) 0.9984

Explanation / Answer

Q.20 Here proportion of O donors p = 0.06

Mean = np = 20 * 0.06 = 1.2

standard deviation = sqrt [np(1-p)] = sqrt [ 0.06 * 0.94 * 20] = 1.062

Q.21 Here p = 0.1

we will calculate it by binomial distribution

Here n = 10 and p = 0.1

so P(1<=x <=4) = 10C1 (0.1)1 (0.9)9 + 10C2 (0.1)2 (0.9)8 +10C3 (0.1)3 (0.9)7 + 10C4 (0.1)4 (0.9)6

= 0.3874 + 0.1937 + 0.0574 + 0.01116 = 0.6497

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