In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample?

The probability that the male sample will have at most three more days of absences than the female sample can be calculated using a binomial distribution. Since we are looking for a maximum of three days difference, the total number of possible cases is four (0 day difference, 1 day difference, 2 day difference and 3 day difference). The probability for each case can then be determined as follows:

P(0 Day Difference) = (100 choose 50) * 0.5^50 * 0.5^50 = 0.20
P(1 Day Difference) = (100 choose 51) * 0.5^51 * 0.5^49 = 0.20
P(2 Day Difference) = (100 choose 52) * 0.5^52 * 0.5^48 = 0.18
P(3 Day Difference)= (100 choose 53) *0 .5*53*0 . 5*47=0 .16

Therefore, when all these probabilities are added together, we get a final result of P(at most 3 day difference) = 20% + 20% + 18% + 16%= 74%.