P(withdraw) = 0.2
n = 20
P(X = x) = 20Cx * 0.2x * (1 - 0.2)20-x
a) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 20C0 * 0.20 * 0.820 + 20C1 * 0.21 * 0.819 + 20C2 * 0.22 * 0.818
= 0.2061
b) P(X = 4) = 20C4 * 0.24 * 0.816 = 0.2182
c) P(X > 3) = 1 - P(X < 3)
= 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
= 1 - (20C0 * 0.20 * 0.820 + 20C1 * 0.21 * 0.819 + 20C2 * 0.22 * 0.818 + 20C3 * 0.23 * 0.817)
= 1 - 0.4114
= 0.5886
d) Expected number = 20 * 0.2 = 4
Problem 1 A university found that in an introductory statistics course, about 40% of the enrolled...
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