Info for Statisticians 100l

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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 fem
ale 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's Normal 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 0.035.

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QUESTION

The heights of adult women are normally distributed with a mean of 62.5 inches and a standard deviation of 2.5 inches. Use the Empirical Rule to determine between what two heights 95% of adult women will fall.

Solution: The interval where 95% of the heights will fall is


Ci = (x-1.96S, X+1.96S) = (62.5-1.96*2.5, 62.5.5+1.96*2.5) = (57.6, 67.4)

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