Question

A nationwide survey of 100 boys and 50 girls in the first grade found that the daily average number of boys who are absent from school is 15, with a standard deviation of 7. For girls, the average number of absentees is 10, with a standard deviation of 6. The standard deviation of the difference is

Answer 1

Te difference of 2 standard deviation of a population n1 & n2 is given by the formula:
sigma (difference)=√(sigma1/n1 + sigma2/n2), Plug:

sigma(d)= √(49/100 + 36/50)

Sigma(d=difference) =1.1

Answer 2

Find the mean difference (male absences minus female absences) in the population.

μd = μ1 - μ2 = 15 - 10 = 5

Find the standard deviation of the difference.

σd = sqrt( σ12 / n1 + σ22 / n2 )

σd = sqrt(72/100 + 62/50) = sqrt(49/100 + 36/50) = sqrt(0.49 + .72) = sqrt(1.21) = 1.1

Find the z-score that produced when boys have three more days of absences than girls. When boys have three more days of absences, the number of male absences minus female absences is three. And the associated z-score is

z = (x - μ)/σ = (3 - 5)/1.1 = -2/1.1 = -1.818

Find the probability. This problem requires us to find the probability that the average number of absences in the boy sample minus the average number of absences in the girl sample is less than 3. To find this probability, we enter the z-score (-1.818) into Stat Trek's Normal Distribution Calculator. We find that the probability of a z-score being -1.818 or less is about 0.035