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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.
μ_{d} = μ_{1}  μ_{2} =
15  10 = 5

Find the standard deviation of the difference.
σ_{d} =
sqrt( σ_{1}^{2} / n_{1} +
σ_{2}^{2} / n_{2} )
σ_{d} =
sqrt(7^{2}/100 + 6^{2}/50) =
sqrt(49/100 + 36/50) = sqrt(0.49 + .72) = sqrt(1.21) = 1.1

Find the
zscore
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 zscore
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 zscore (1.818) into
Stat Trek's
Normal Distribution Calculator.
We find that the probability of a zscore 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.