Special discrete prob. distrib

20(a) Expected mark of each question when wild guesing

= (1/4)(2) + (3/4)(-1)

= -0.25

So, the total expected mark is -2.5

(b) Total expected mark = 10[(0.7)(2) + (0.3)(-1)] = 11

21(a) C(4,2)(0.55)^2(0.45)^2 = 0.3675

(b) Expected value = 50 + 4(2 * 0.55 - 2 * 0.45) = 50.8

Standard deviation = 4 * 2 * √(0.55 * 0.45) = 3.98

22 (0.98)^n < 0.15

n > 93.9

So, the min. no. is 94