Stat 211  Tutorial 1 Solutions
1. A closing price (in dollars) at Dec 31, 2011 of 30 stocks in a portfolio are given below:
52.78
48.27
49.82
55.85
54.60
52.87
52.85
54.48
50.73
51.46
55.88
40.72
51.69
47.59
57.19
54.83
50.21
49.54
52.29
53.14
51.12
52.39
54.59
52.48
49.04
53.34
50.49
49.03
54.97
52.16
You are also given that the sum of the above values is 1556.40 and the sum of their
squares is 81041.7816.
(a)
i. Calculate the mean/average closing price and the standard deviation of
the prices.
Solution:
Denote the data by
x
1
, x
2
, . . . , x
3
0
.
Mean
= ¯
x
=
∑
30
i
=1
x
i
30
=
1556
.
40
30
= $51
.
88
Variance
=
s
2
=
∑
30
i
=1
x
2
i

30¯
x
2
30

1
=
81041
.
7816

30(51
.
88)
2
29
= 10
.
1983
Standard Deviation
=
s
=
√
10
.
1983 = $3
.
1935
ii. Determine the range, median, and the 1st and 3rd quartiles. Hence, deter
mine the interquartile range.
Solution:
Range
=
Largest Value

Smallest Value
= 57
.
19

40
.
72 = $16
.
47
Sort the 30 data points: (The data will already be sorted for you in an exam)
52.78
55.88
51.12
48.27
40.72
52.39
49.82
51.69
54.59
55.85
47.59
52.48
54.6
57.19
49.04
52.87
54.83
53.34
52.85
50.21
50.49
54.48
49.54
49.03
50.73
52.29
54.97
51.46
53.14
52.16
Median (or 50th percentile) :
i
= (50
/
100)
·
30 = 15
=
⇒
Median
=
Average of the 15th and 16th sorted values
=
52
.
29 + 52
.
39
2
= 52
.
34
1
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1st Quartile (or 25th percentile) :
i
= (25
/
100)
·
30 = 7
.
5
=
⇒
Q
1
=
8th sorted value
= 50
.
21
3rd Quartile (or 75th percentile) :
i
= (75
/
100)
·
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 Winter '12
 Javid
 Standard Deviation

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