Question

A journal polled its readers to determine the proportion of wives who, if given a second chance, would marry their husband. According to the responses from the survey, 80% of wives would marry their husbands again. Suppose a random and independent sample of n = 16 wives was collected. Define the random variable X as the number of the 16 wives who would marry their husbands again.

(a)   How many of the 16 wives sampled do we expect to say they would marry their husbands again?   

(b) A Statistics student was asked to calculate the probability that less than 14 wives would marry their husbands again. The student’s attempt at the question is given below.

X is the number of wives who would marry their husbands again To Bin' C16, 0.80) P(X = 4) = ''( • go"*):) = (120) 0.04398) (0.04) = 0.2111004

Comment on the student’s attempt at the question by commenting on:

(i) the suitability of the probability distribution used and accuracy of the parameters stated

(ii) the accuracy of the solutions. If incorrect, show the correct solution.   

Answer 1

Solution Given that, 0:16, P = 0.80 This is a binomial distribution. Poob a since, in Binomial dist ECX) = np - 16X0.80 [E(x) = 12.8~13 Then we can say that approximately 13 wives are expected to mootory their husbands again 0=16, P=0.80, X = 14 9, Distribution and parameters selected by the student is suitable for the ansun ut the answer is incorrect, becomes be were less than 14 wives not exactly 14 wing their husbands again, them P(XL14) = ? P(x = x) = nC, PR (1-P) 0-x b) But than 14, less are they marry PLX214) = 1-P(x314) -1- (PLX=14)+ PLX 21S)+PCX=16)]

-1-($ ¢, 60-80)91–0.8014}+{ "8,210-80) "11-0-80"} + 60.80)" (1-0-80°} I lf120 x 10.80) 460.203%} + { 1660-80)'s (0.2013 + { 1. 10.80) 6.13 PLXZ14) - 1-[0.2011 +0.1126+0.0281] = 1- (0.3518) PLXL 14) = 0.6402) e the probability that less than 14 wives husbands again = 0.6402 husbande their maroy