Homework 2 solutions
2.15
(a) x =
1
n
x
i
= 577.9/5 = 115.58. Deviations from the mean:
i
116.4 - 115.58 = .82, 115.9 - 115.58 = .32, 114.6 -115.58 = -.98,
115.2 - 115.58 = -.38, and 115.8-115.58 = .22.
(b) s2 = [(.82)2 + (.32)2 + (-.98)2 + (-.38)2 + (.22)2]/(5-1) = 1.928/4 =.482,
so s = .694.
2.16
(a) The transformed data are: 16.4 15.9 14.6 15.2 15.8.
n
Relevant quantities are:

n
xi2 1,215 .61
i 1
n
So, s 2 




2
xi  
n

i 1
i 1

xi 


n
n 1
 x  77.9
i
i 1
2

1,215 .61 
4
77 .92
5
So, s 2 .482
Notice that s 2 for the transformed data equals s 2 for the original data.
2.39
(a) ED: median = .4 (the 14th value in the sorted list of data). The lower quartile (median of the
lower half of the data, including the median, since n is odd) is
( .1+.1 )/2 = .1. The upper quartile is (2.7+2.8)/2 = 2.75. Therefore,
IQR = 2.75 - .1 = 2.65.
Non-ED: median = (1.5+1.7)/2 = 1.6. The lower quartile (median of the lower 25 observations)
is .3; the upper quartile (median of the upper half of the data) is 7.9. Therefore, IQR = 7.9 .3 = 7.6.
(b) ED: mild outliers are less than .1 - 1.5(2.65) = -3.875 or greater than 2.75 + 1.5(2.65) =
6.725. Extreme outliers are less than .1 - 3(2.65) = -7.85 or greater than 2.75 + 3(2.65) =
10.7. So, the two largest observations (11.7, 21.0) are extreme outliers and the next two
largest values (8.9, 9.2) are mild outliers. There are no outliers at the lower end of the data.
Non-ED: mild outliers are less than .3 - 1.5(7.6) = -11.1 or greater than 7.9 + 1.5(7.6) = 19.3.
Note that there are no mild outliers in the data, hence there can not be any extreme outliers
either.
(c) A comparative boxplot appears below. The outliers in the ED data are clearly visible. There
is noticeable positive skewness in both samples; the Non-Ed data has more variability then
the Ed data; the typical values of the ED data tend to be smaller than those for the Non-ED
data.
Non-ED
ED
0
10
Concentration (mg/L)
20
2.61
The transformation of the xi’s into the zi’s is analogous to ‘standardizing’ a normal distribution.
The purpose of standardizing is to reduce a distribution (or, in this exercise, a set of data) to one
that has a mean of 0 and a standard deviation of 1. To show that this transformation achieves
this goal, note that:
z
Next,
i
=
1
n 1


1
s
( xi  x ) 
(zi  z ) 2 =
1
s
 (x  x) 
i
1
s
(0)  0 , so, dividing this sum by n,
  ( x  x ) s1 s 2
1
1
n 1 s 2
2
i
=
2
= 1.
z = 0.
5.2
A Venn diagram of these three events is
(a) The event ‘at least one plant is completed by the contract date’ is represented by the shaded area
covered by all three circles:
(b) The event ‘all plants are completed by the contract date’ is the shaded area where all three circles
overlap:
(c) The event ‘none of the plants is completed by the contract date’ is the complement of the shaded area
in (a):
(d) The event ‘only the plant at site 1 is completed by the contract date’ is shown shaded:
5.9
(a)
The total of 724 +751 solder joints overstates the actual number found, since 316 solder joints were
found by both inspectors. To avoid such double-counting (i.e., the 316 is part of both the 724 joints
found by inspector A and the 751 joints found by Inspector B), 316 should be subtracted from the raw
totals, which means that 724 + 751 - 316 = 1,159 distinct joints were identified by the inspectors
together. The important point to note in this problem is that the events ‘Inspector A finds a defective
solder joint’ and ‘Inspector B finds a defective solder joint’ are not necessarily mutually exclusive, so
we can not simply add the numbers of joints (or, equivalently, the probabilities of finding defective
joints) for both inspectors.
(b) A and B contains 724 - 316 = 408 solder joints.
5.62
The probability that both pumps fail on a given day is .85