Nov. 1, 2005
ECON 240A
1
L. Phillips
Midterm
1
Answer all five questions
1.
(15 points) The following Box plots describe the midterm scores for Econ 240A
for the past three years. The total potential number of points was 75 each year.
Note the numbers of students taking the midterm were 45 in 2002, 30 in 2003,
and 35 in 2004, so note that the scale differs from box plot to box plot.
2002
Smallest = 34
Q1 = 55
Median = 61
Q3 = 65.5
Largest = 74
IQR = 10.5
Outliers: 34,
2003
Smallest = 49
Q1 = 59.75
Median = 64
Q3 = 67.25
Largest = 73
IQR = 7.5
Outliers:
2004
Smallest = 18.75
Q1 = 34.5
Median = 40.5
Q3 = 52.5
Largest = 70.5
IQR = 18
Outliers:
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Nov. 1, 2005
ECON 240A
2
L. Phillips
Midterm
1
a.
On average, which year’s class appears
to do the best? Explain the criterion
(ia) that you used.
The class of 2003 has the highest median. The question is
does this mean they did the best, or are differences among years obscured by
the grading of different TAs, etc.?
b.
Which year’s class was most closely bunched, i.e. had the smallest dispersion?
The class of 2003 has the smallest intequartile range.
c.
Which year’s class(es) did not have any outliers? Would it have been possible
,
given these distributions (i.e. numbers), to have an outlier at the upper end of
the distribution in any of the three years?
The classes of 2003 and 2004 had no
outliers. Since Q
3
+ 1.5* IQR is the potential borderline for outliers, this
borderline in the various years was:
2002: 65.5 + 1.5* 10.5 = 81.25
2003: 67.25 + 1.5* 7.5 = 78.5
2004: 52.5+ 1.5* 18 = 79.5
Every year this boundary is above the maximum score possible so no outliers
at the upper end.
d.
How is an outlier calculated in 2002?
Q
3
+ 1.5*IQR, Q
1
–1.5*IQR
e.
Do you think it would be fair to grade each year’s class on an absolute scale,
i.e. based on your score as a percent of 75 points, versus grading on a curve?
Justify your answer. What can vary from year to year in addition to average
student performance?
Comparing the median or 2004 with the other years , a
curve seems more appropriate. See part a for a possible reason.
2.
(15 points) You conduct an experiment by throwing a fair die three times. Each
throw is independent.
a.
What is the probability of observing one or more sixes?
One or more sixes is
the complement of no sixes. The probability of the latter is (5/6)(5/6)(5/6) = 125/216.
So the answer is 1125/216 = 91/216.
b.
What is the probability of observing exactly one six?
Using the binomial, 3!/
(1!2!) (1/6)(5/6)
2
= 75/216
c.
What is the probability of observing three sixes?
(1/6)(1/6)(1/6) = 1/216.
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 Spring '08
 Staff
 Normal Distribution, personal income, L. Phillips, UC Budget

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