Question

(1 point) A recent poll has suggested that 66 % of Canadians will be spending money - decorations, halloween treats, etc. - to celebrate Halloween this year. 22 Canadians are randomly chosen, and the number that will be spending money to celebrate Halloween is to be counted. This count is represented by the random variable X. Part (a) Compute the probability that 16 of these Canadians indicate they will be spending money to celebrate Halloween. P(X = 16) = 0.1494138 (use four decimals in your answer) Part (b) Compute the probability that between 7 and 16 of these Canadians, inclusive, indicate they will be spending money to celebrate Halloween. P(7 < X < 16) = (use four decimals in your answer) Part (c) How many of the 22-Canadians randomly chosen would you expect to indicate they will be spending money to celebrate Halloween? Compute the standard deviation as well. E(X) = My = 14.52 . (use two decimals in your answer) SD(X) = 0x = 2.221891 !!! (use two decimals in your answer) Part (d) Compute the probability that the 15-th Canadian random chosen is the 8-th to say they will be spending money to celebrate Halloween. :: (use four decimals in your answer)

Answer 1

166 Sols 66% 46. Po 06G 9 (-0.667 .34 © By rising bidoneaf wists' Section Scienty p" * phoke P(8916) - PC *10.39 796/3 Y 1001-79638* 0.00154400 0.19991 = ® PC758516) Now we fut value - P(x= 7) + (R= 8) +(223)+(hala) -NH211) (X=12) + (4,213) (x/9)+ (A=15) Alx=/6) (0.66)*(0.375.p 27 CM (0.6) (0139, ep 22,(61.683°10:39)!* + 2,5€ (1563"premsy? a 27%9c5"260:39)" - 24,9(068}6.390 ***, *0.66) 46.39' 216,460-56 ) 10.39j* p 222, 601 663986134* 284,5*(66"0.7 (0.3996 = 0.0003726 0.06 317 6 + 06095-90 +0.024.20 +0545 0:091157 † 0:1366770053 * T 0:1366/7 A 00531458017593 +0:14.3913 15370 (X) = ln = 2*066 = 14.5% Sof(0) = N*4 PX 2 = 122*.660.3% = 1.05252 0 art = 15 here k = 8 P(N=25 = (** J* p[/= +) 14. * (0.66)® $(03437 2017 33937 * 0.0000 189 - 01649.00