Senior Life
Problem 1
For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.
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?
(A) 0.025
(B) 0.035
(C) 0.045
(D) 0.055
(E) None of the above
Solution
The correct answer is B. The solution involves four steps. * 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'sNormal Distribution Calculator. We find that the probability of a z-score being -1.818 or less is about 0.035.
Therefore, the probability that the difference between samples will be no more than 3 days is
For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.
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?
(A) 0.025
(B) 0.035
(C) 0.045
(D) 0.055
(E) None of the above
Solution
The correct answer is B. The solution involves four steps. * 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'sNormal Distribution Calculator. We find that the probability of a z-score being -1.818 or less is about 0.035.
Therefore, the probability that the difference between samples will be no more than 3 days is