2014 SCORING COMMENTARY
AP
®
STATISTICS
Question 3
Overview
The primary goals of this question were to assess a student’s ability to (1) perform a probability calculation
f
rom a normal distribution; (2) explain an implication of examining the distribution of a sample mean rather
than the distribution of a single measurement; and (3) perform a probability calculation involving
independent events using the multiplication rule.
Sam
ple: 3A
Score: 4
In p
art (a) the response indicates the use of the normal distribution with a sketch as well as with a correct
z-score equation and z-score inequality. The values of the parameters are identified with correct substitution
into the z-score equation and z-score inequality. Additionally, the mean is again identified in the sketch.
With the normal distribution indicated and the parameter values identified the first component is satisfied.
The boundary value is also identified in the z-score equation and z-score inequality and the correct normal
probability is computed. Thus, the second and third components are satisfied. With all three components
satisfied, part (a) was scored as essentially correct. In part (b) the response begins by stating the correct
answer, “High School A would be less likely to lose funding using this plan than it would in part (a).” The next
statement refers to the distribution of the sample mean, indicates that this distribution remains centered at
120, and indicates that its standard deviation is smaller, so each of the three components is satisfied. With
the correct answer and each of the three components satisfied, part (b) was scored as essentially correct. In
part (c) the response correctly uses the multiplication rule with the correct probability and reports the correct
answer. Since the multiplication rule is used correctly with work shown, part (c) was scored as essentially
correct. Because all three parts were scored as essentially correct, the response earned a score of 4.
Sam
ple: 3B
Score: 3
In p
art (a) the response indicates the use of the normal distribution and identifies the parameter values with a
well-labeled sketch. Thus, the first component is satisfied. The inequality indicates that the probability that
X exceeds 140 is being computed, satisfying the second component. However, an incorrect probability is
reported, so the third component is not satisfied. With two of the three components satisfied, part (a) was
scored as partially correct. In part (b) the response correctly calculates the mean and standard deviation for
the distribution of the total number of absences for 3 randomly selected days and correctly computes the
probability that the total number of absences for 3 randomly selected days exceeds
=420 3(140).
This is
mathematically equivalent to computing the probability that the mean number of absences for 3 randomly
selected days exceeds 140. The response gives the correct answer, “less likely” and gives a correct
comparison to the probability computed in part (a). Since the correct answer, the correct probability, and a
correct comparison to the probability in part (a) are given, part (b) was scored as essentially correct. In part (c)
the response correctly uses the binomial probability formula with
= 3n
and the correct value of
= =
2
0.4
5
p
to compute the probability that only Mondays or Fridays are selected. Since the correct
probability is computed with work shown, part (c) was scored as essentially correct. Because two parts were
scored as essentially correct and one part was scored as partially correct, the response earned a score of 3.
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