Statistics assignment 1

Business Statistics (BUS 505) Assignment I
1
Question 1:
Given Data 13 15 8 16 8 4 21 11 11 3 15
Arranging them is ascending order
3 4 8 8 11 13 15 15 16 21
1. Population mean:
 X
=
Xi

1
N
N
i
=
3  4  8  8  11  13  15  15  16  21
10
=
114
10
= 11.4
Taking sample 3 8 15 21
Sample average:
X =
Xi

1
n
n
i
=
3  4  8  8  11  13  15  15  16  21
10
=
114
10
=11.4

2
2. Median:
N = even so,
M =
th th
 n 
n

2
2
2
2











=
th th
 

2
10 2
2
10
2











=
13 11 
2
= 12
3. Mode:
Mode= 8 and 15
4. Midrange:
=
XX
L
S 2
=
3  21
2
=12
5. Geometric mean:
G = n a  a 
.....................
a1 2
n
=
10 3  4  8  8  11  13  15  15  16  21

3
1
=(3 4 8 8 11 13 15 15 16 21)10
         = 9.816
6. Harmonic mean (population):
H =
n
1
 ....................

1 1
a a a1 2
n
=
1
...............
21
1
4
1
3
10
 
=8.036
= 7
7. Range:
= X 
L X S
=21-3
= 18

4
Question 2:
21 22 27 36 22 29 22 23 22 28 36 33
Averaging them in ascending order
21 22 22 22 22 23 27 28 29 33 36 36
1. Population mean:
 X
=
Xi

1
N
N
i
=
21 22  22  22  22  23 27  28  29  33 36  36
12
=
321
12
=26.75
Sample
21 22 23 36
Sample mean:
X =
Xi
n
n
i  1

5
=
21 22  22  22  22  23 27  28  29  33 36  36
12
=
321
12
=26.75
2. Median:
=
th th
 

n n
2
2
2
2











=
th th
 

2
12 2
2
12
2











=
th th

7 6
2
=
27 23 
2
=25
3. Mode:
Mode is 22
4. Mid-range:
=
X 
L X S
2
=
36  21
2
=33.5

6
5. Harmonic mean:
H =
n
1
 ....................

1 1
a a a1 2
n
=
1
...............
36
1
22
1
21
12
 
=25.75
6. Geometric mean:
= n a  a 
.....................
a1 2
n
=12 21 22  22  22  22  23  27  28  29 33 36 36
1
=(21  22  22  22  22  23  27  28  29  33  36 
36)12
=26.23
7. Range:
= X 
L X S
= 36-21
=18

7
Question 3:
3.6, 3.1, 3.9, 3.7, 3.5, 3.7, 3.4, 3.0, 3.6
Arranging them in ascending order
3.0, 3.1, 3.4, 3.5, 3.6, 3.6, 3.7, 3.7, 3.9
1. Population mean:
 X
=
Xi

1
N
N
i
=
33.13.4 3.4 3.53.63.63.7 3.9
10
=3.13
Sample data,
3 3.4 3.6 3.9
Sample mean:
X =
Xi

n
n
i
1
=
3 3.4  3.6  3.9
4
=3.475

8
2. Median:
=
th th
 n 
n

2
2
2
2











=
th th
 

2
10 2
2
10
2











=
th th

6 5
2
=
3.5  3.6
2
=3.55
3. Mode:
Mode= 3.4, 3.6, 3.7
4. Mid range:
=
X 
L X S
2
=
3 3.9
2
=3.45

9
5. Hermonic mean:
H =
n
1
 ....................

1 1
a a a1 2
n
=
1
...............
3.9
1
3.1
1
3
10
 
=3.515
6. Geomatric mean:
= n a  a 
.....................
a1 2
n
=10 3*3.1*3.4*3.4*3.5*3.6*3.6*3.7*3.7*3.9
= 3.47
7. Range:
= X 
L X S
=3.9
= 0.9

10
Question 4
Population: 2, 4, 2, 3, 5, 4, 3, 2
1. Population Mean:
N=10 observations, so the Mean is

x
i 1
N
μx =
2 + 4 + 2 + 3 + 5 + 4 + 3 + 2
25
= 3.125
2. Median: 2, 2, 2, 3, 4, 4, 5
Median =
=
= 3


 
 
 
 
8 2
8



4 5 
3 3
3. Mode: 2, 2, 2, 3, 4, 4, 5
Mode = 2
4. Midrange:
Midrange =
XS+ XL
2
=
2+5
2
=
7
2
= 3.5
N
i
8
2
2
2
2
th th
N N








2
2
2
th th







2
2

 th th
8

11
5. Harmonic Mean:
HM =
N
1
푎1
+
1
푎2
1
푎푁
+−−−−−−−−−−−−−−−−−−−+
= 8
1
2+1
4+1
2+1
3+1
5+1
4+1
3+1
2
=
8
.5+.25+.5+.33+.2+.25+.33+.5
= 2.7
6. Geometric Mean:
N
G. M. = √푎1 × 푎2 × 푎3 × − − − − − − − × 푎푁
= √2 × 4 × 2 × 3 × 5 × 4 × 3 × 2 8
= 2.95
7. Range:
Range = XL – XS = 5 – 2 = 3

12
Question No. 5
Population: 42, 29, 21, 37, 40, 33, 38, 26, 39, 47
1. Population Mean:
The population contains N=10 observations, so the Mean is
μx =
=
42 + 29 + 21 + 37 + 40 + 33 + 38 + 26 + 39 + 47
=

x
i 1
N
352
= 35.2
2. Median:
Ascending order: 21, 26, 29, 33, 37, 38, 39, 40, 42, 47
Median =
=
=


 
 
10


 
 
5 6 
= 37.5
3. Mode:

10 2

37 38
The Mode of a set of observations is the value that occurs most frequently. So, there is no Mode.
4. Midrange:
Midrange =
XS+ XL
2
=
21+47
2
=
68
2
= 37.5
N
i
10
2
2
2
2
th th
N N







2
2
2
th th







2
2

 th th
10

13
5. Harmonic Mean:
H. M. =
푁
1
푎1
+
1
푎2
+−−−−−−−−−−−−−−−+
1
푎푁
= 10
1
42+ 1
29+ 1
21+ 1
37+ 1
40+ 1
33+ 1
38+ 1
26+ 1
39+ 1
47
=
10
.02+.03+.04+.02+.02+.03+.02+.03+.02+.02
= 40
6. Geometric Mean:
퐍
G. M. = √푎1 × 푎2 × 푎3 × − − − − − − − × 푎푁
= √42 × 29 × 21 × 37 × 40 × 33 × 38 × 26 × 39 × 47 ퟏퟎ
= 34.31
7. Range:
Range = XL – XS = 47 – 21 = 26

14
Question No. 6
Population: 10.2, 3.1, 5.9, 7.0, 3.7, 2.9, 6.8, 7.3, 8.2, 4.3
1. Population Mean (Average):
μx =
=
10.2 + 3.1 + 5.9 + 7.0 + 3.7 + 2.9 + 6.8 + 7.3 + 8.2 + 4.3
=

x
i 1
N
4.59
= 5.94
2. Median: Arranging N=10 observations in ascending order, we have
Ascending Order: 2.9, 3.1, 3.7, 4.3, 5.9, 6.8, 7.0, 7.3, 8.2, 10.2
Median =
=
=


 
 
10


 
 
5 6 
= 6.35

10 2

5.9 6.8
3. Mode:
4. Midrange:
Midrange =
XS+ XL
2
N
i
10
2
2
2
2
th th
N N







2
2
2
th th







2
2

 th th
10

15
=
2.9+10.2
2
=
13.1
2
= 6.55
5. Harmonic Mean:
H. M. =
N
1
푎1
+
1
푎2
+−−−−−−−−−−−−−−−+
1
푎푁
=
10
1
10.2
+
1
3.1
+
1
5.9
1
7.0
+
+
1
3.7
+
1
2.9
+
1
6.8
1
7.3
+
+
1
8.2
+
1
4.3
=
10
.09+.32+.16+.14+.27+.34+.14+.13+.12+.23
= 4.85
6. Geometric Mean:
퐍
G. M. = √푎1 × 푎2 × 푎3 × − − − − − − − × 푎푁
= √10.2 × 3.1 × 5.9 × 7.0 × 3.7 × 2.9 × 6.8 × 7.3 × 8.2 × 4.3 ퟏퟎ
= 5.48
7. Population Range:
Range = XL – XS = 10.2 – 2.9 = 7.3

16
Question No. 7
Population: 7.3, 10.2, 13.1, 15.0, 15.8, 16.9, 18.2, 24.7, 25.3, 28.4, 29.3, 34.7
1. Population Mean:
μx =
=
=

x
i 1
N
15.8 + 7.3 + 28.4 + 18.2 + 15.0 + 24.7 + 13.1 + 10.2 + 29.3 + 34.7 + 16.9 + 25.3
9. 238
= 19.90
Ascending Order: 7.3, 10.2, 13.1, 15.0, 15.8, 16.9, 18.2, 24.7, 25.3, 28.4, 29.3, 34.7
Median =
=
=


 
 
12


 
 
6 7 
= 17.55

12 2

16.9 18.2
3. Mode:
4. Population Midrange:
Midrange =
XS+ XL
2
N
i
12
2
2
2
2
th th
N N







2
2
2
th th







2
2

 th th
Here,
XS = Smallest observations
XL = Largest observations
12

17
=
7.3+34.7
2
=
42
2
= 21
5. Harmonic Mean:
H. M. =
푁
1
푎1
+
1
푎2
+−−−−−−−−−−−−−−−+
1
푎푁
=
12
1
15.8
+
1
7.3
+
1
28.4
+
1
18.2
+
1
15
+
1
24.7
+
1
13.1
1
10.2
+
1
29.3
+
1
34.7
+
+
1
16.9
+
1
25.3
=
12
.06+.13+.03+.05+.06+.04+.07+.09+.03+.02+.05+.03
= 18.18
6. Geometric Mean:
퐍
G. M. = √푎1 × 푎2 × 푎3 × − − − − − − − × 푎푁
= √15.8 × 7.3 × 28.4 × 18.2 × 15.0 × 24.7 × 13.1 × 10.2 × 29.3 × 34.7 × 16.9 × 25.3 ퟏퟐ
= 18.15
7. Range:
Range = XL – XS = 34.7 – 7.3 = 27.4

18
Question No. 16
Population: 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 16, 21
1. Population Mean:
μx =
=
12 + 7 + 4 + 16 + 21 + 5 + 9 + 3 + 11 + 14 + 10 + 6
=

x
i 1
N
118
= 9.83
Ascending Order: 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 16, 21
Median =
=
=

12

6 7 
= 9.5



 
  
 
12 2


9 10
3. Mode:
4. Midrange:
Midrange =
XS+ XL
2
=
3+21
2
=
24
2
= 12
N
i
12
2
2
2
2
th th
N N







2
2
2
th th







2
2

 th th
12

19
5. Harmonic Mean:
H. M. =
푁
1
푎1
+
1
푎2
1
푎푁
+−−−−−−−−−−−−−−−+
=
12
1
12
+
1
7
+
1
4
1
16
+
1
21
+
+
1
5
1
9
+
+
1
3
+
1
11
+
1
14
+
1
10
+
1
6
=
12
.08+.14+.25+.06+.04+.2+.11+.13+.09+.07+.1+.16
= 7.40
6. Geometric Mean:
퐍
G. M. = √푎1 × 푎2 × 푎3 × − − − − − − − × 푎푁
= √12 × 7 × 4 × 16 × 21 × 5 × 9 × 3 × 11 × 14 × 10 × 6 ퟏퟐ
= 6.95
7. Range:
Range = XL – XS = 21 – 3 = 18

Statistics assignment 1

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Statistics assignment 1