Download Exe 2. Elementary school: ¯x=648.1, Median = 655, No mode

Exe 2. Elementary school: x̄ = 648.1, Median = 655, No mode, Midrange = 666.5, Range =
1059, Sample variance = 101433, Sample standard deviation = 318.5.
Secondary school: x̄ =264.3, Median = 279.5, No mode, Midrange = 233.5, Range = 381,
Sample variance = 16968.6, Sample standard deviation = 130.3.
The numbers of elementary schools is more variable than that of secondary schools.
Exe 4.
Score
Freq. Xm fXm
2
fXm
478 − 504
4
491 1964
964324
505 − 531
6
518 3108 1609944
532 − 558
2
545 1090
594050
559 − 585
2
572 1144
654368
586 − 612
2
599 1198
717602
Sum
16
8504 4540288
Sample mean x̄ = 531.5, Variance = 1360.8, Std = 36.9, Modal class is the second class. The
shape is right-skewed.
Exe 8. Weighted mean of payoffs is 4700$.
Exe 10. These two data sets have different units so we use coefficients of variations (CV)
to compare their variation. The CV of numbers of textbooks =
5
×100% = 31.25%
16
is larger
than the CV of ages = 18.6%, so the first data set is more variable than the second one.
Exe 12. Ordered data: 3, 4, 5, 15, 16, 17, 19, 23, 24, 31, 33 (n = 11 observations). Percentile
rank of x = (Number of values below x + 0.5) / n. Percentile ranks corresponding to the
above 11 values are 5%, 14%, 23%, 32%, 41%, 50%, 59%, 68%, 77%, 86%, 95% respectively.
The value that corresponds to the 40th percentile is at the position 11×40/100 = 4.4 ≈ 5,
which is 16. Boxplot: Min = 3; Q1 = 5; Median = 17; Q3 = 24; Max = 33.
Exe 14. a). Ordered data: 400, 506, 511, 514, 517, 521 (6 observations). Q1 = 506, Q3 =
517, IQR = = Q3-Q1 = 11, (Q1-1.5IQR; Q3+1.5IQR) = (489:5; 533:5). So 400 is an outlier.
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b)-d): No outlier.
Exe 16. In order to say about the percentage of data points that fall within a particular
range, one can use the empirical rule or Chebyshev’s theorem. The empirical rule often gives
a more accurate estimate but it requires the assumption of bell-shaped distribution, while
Chebyshev’s theorem is applicable for any data set. The data set in this exercise is not
assumed to have a bell-shape, so Chebyshev’s theorem should be used.
a) The interval 47,300$-69,700$ is k=1 standard deviation from the mean. Chebyshev’s
theorem says that at least 1− k12 = 0% data points fall in this interval! This conclusion is
noninformative. Chebyshev’s theorem tells us nothing in this case.
b) 80,900$ is k = 80,900−58,500
= 2 std of the mean. According to Chebyshev’s theorem, at
11,200
leat 1− k12 = 25% data points are within k = 2 std of the mean. Therefore at most 25% data
points fall outside that interval. That is, at most 25% of workers earn more than 80,900$.
c) k = 3.705. At most 7.28%.
Exe 18. k = 2.4. At least 82.6%
Exe 20. The relative position of a data point in a data set depends on the mean and the
=
standard deviation of the data set. In the first test, a grade of 82 corresponds to z = 82−85
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−0.5. In the second test, a grade of 56 corresponds to z = −0.8. The grade in the first test
has a better relative position because it is closer to the mean than the second score.
Exe 22. “Before” data set: min = 12, Q1 = 19.5, Median = 30, Q3 = 34.25, max = 38.
“After” data set: min = 12, Q1 = 14.25, median = 18, Q3 = 23.5, max = 32
Exe 23. The distribution is bell-shaped, so the empirical rule can be used. 68% of the times
would be expected to be within 1 std of the mean, which is 23.7-35.7
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