STATISTICS-I
PRACTICAL PROBLEMS
BINOMIAL DISTRIBUTION
1) Fit a Binomial Distribution for the following data.
X 0 1
2
3 4 5 6
f 0 4
13 28 42 20 6
7
2
Solution: - Aim: - P(X=x) =ncx px qn-x ; x=0,1,2,…. ; p+q=1
= 0
; otherwise
Where n,p are parameters of the distribution.
Mean = 𝑿 = np => 𝒑 =
Where 𝑿=
𝑿
𝒏
𝒇𝒙
=> 𝒒=1-𝒑
𝒇
Calculations:- First we have to calculate about mean
Here n=7
𝒇𝒙
Mean =𝑿=
But 𝒑 =
𝑿
𝒏
𝒇
=
=
𝟒𝟑𝟐
𝟏𝟏𝟓
𝟑.𝟕𝟓𝟔
𝟕
= 3.756
= 0.536
 𝒒=1-𝒑 = 1-0.536=0.464
X
f
0
1
2
3
4
5
6
7
0
4
13
28
42
20
6
2
115
Probability
P(X=x) =ncx px qn-x
0.004630
0.037443
0.129759
0.249824
0.288589
0.200022
0.77020
0.012710
Expected Frequency
= N. P(X=x)
0.532≅1
4.305≈4
14.922≅15
28.729≅29
33.187≅33
23.002≅23
8.857≅9
1.461≅1
115
X
0
1
2
3
4
5
6
7
f
0
4
13
28
42
20
6
2
115
fx
0
4
26
84
168
100
36
14
432
Result: X
f0
fe
P(X=x) =7cx (0.536)x (0.464)7-x x=0,1,2,….7
0
0
1
1
4
4
2
13
15
3
28
29
4
42
33
5
20
23
6
6
9
7
2
1
2) Fit a Binomial Distribution for the following data by Recurrence Method.
X
f
0
5
1
9
2
22
3
29
4
36
5
25
6
10
7
3
8
1
Solution: - Aim: - To fit a Binomial Distribution by recurrence method.
Formulae:- P(X+1)=
𝒏−𝒙 𝒑
𝒙+𝟏 𝒒
* P(X)
P ((0) = qn 𝑿=
𝑿=
𝒇𝒙
𝒇
=
𝟒𝟗𝟖
𝟏𝟒𝟎
𝒇𝒙
𝒇
= 3.557 => 𝒑 =
𝑿
𝒏
𝒑=
𝑿
𝒏
=> 𝒒=1-𝒑
= 0.446 => 𝒒=0.554
Calculations:- First we have to calculate about mean
X
0
1
2
3
4
5
6
7
8
f
5
9
22
29
36
25
10
3
1
104
fx
0
9
44
87
144
125
60
21
8
498
X
f
probability
P(X+1)=
0
1
2
3
4
5
6
7
8
5
9
22
29
36
25
10
3
1
140
𝒏−𝒙 𝒑
𝒙+𝟏 𝒒
Expected Frequency
= N. P(X=x)
* P(X)
0.008873
0.057145
0.1610187
0.259256
0.2608915
0.1680245
0.067634
0.015568
0.001565
1.24≈1
8.00≈8
22.54≈23
36.29≈36
36.52≈37
23.52≈24
9.46≈9
2.17≈2
0.21≈0
140
Result:X
f0
fe
0
5
1
1
9
8
2
22
23
3
29
36
4
36
37
5
25
24
6
10
9
7
3
2
8
1
0
POISSON DISTRIBUTION
1) Fit a Poisson distribution to the following data with respect to the
number of red blood corpuscles (X). Find expected values also.
X
f
0
162
1
193
2
115
3
83
4
44
5
24
6
19
7
8
8
2
Solution: - Aim: - To fit a Poisson distribution by direct method.
Formulae:P(X=x)=
𝒆−𝝀
𝒙!
𝝀𝒙 ; 𝑿 =
𝟎, 𝟏, 𝟐 …
= 0
Mean = Variance =𝝀
Mean = 𝑿 = 𝝀
; otherwise
Where 𝑿=
𝒇𝒙
𝒇
(𝑵𝒐𝒕𝒆: − 𝒘𝒓𝒊𝒕𝒆 𝒕𝒉𝒊𝒔 𝝀 𝒕𝒉𝒆 𝒏𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓)
Calculations:- First we have to calculate about mean.
Here, Mean = 𝑿 = 𝝀 =
X
f
𝟏𝟏𝟓𝟒
𝟔𝟓𝟎
probability
P(X=x)=
0
1
2
3
4
5
6
7
8
162
193
115
83
44
24
19
8
2
650
0
162
129
𝒆−𝝀
𝒙!
X
0
1
2
3
4
5
6
7
8
𝒆−𝟏.𝟕𝟕𝟓∗ (𝟏.𝟕𝟕𝟓)𝒙
1
193
208
𝒙!
2
115
169
f
162
193
115
83
44
24
19
8
2
650
fx
0
193
230
249
176
120
114
56
16
1154
Expected Frequency
= N. P(X=x)
𝝀𝒙 ;
0.169483
0.300833
0.266989
0.157969
0.070099
0.024885
0.007362
0.001867
0.000414
Result: - P(X=x) =
X
f0
fe
=1.775
110.16≈
195.54≈
173.54≈
102.68≈
45.56≈
16.18≈
4.79≈
1.21≈
0.27≈
650
for x=0, 1, 2…8
3
83
91
4
44
37
5
24
12
6
19
3
7
8
1
8
2
0
2) For the arrival of the patients at a doctor’s clinic has obtained the
following distribution for 445 days. Fit a Poisson distribution for the
following data by recurrence method.
No. of
0
patients(x)
No. of
153
days
1
2
3
4
5
6
169
72
31
12
6
2
Solution: - Aim: - To fit a Poisson distribution by recurrence method.
−𝝀
Formulae: - P(X=x) = 𝒆𝒙!
𝝀𝒙 ; X= 0, 1, 2….
=0
; otherwise
Recurrence formula = P(X+1) =
Calculation: - Here Mean=𝑿 = 𝝀 =
𝒇𝒙
𝒇
−𝝀
X
0
1
2
3
4
5
6
f
153
169
72
31
12
6
2
445
𝝀
𝒙+𝟏
1.114
0.557
0.3713
0.2785
0.2228
0.1857
0.1591
𝝀𝟎 = 𝒆−𝝀 = 𝒆−𝟏.𝟏𝟏𝟒 =0.3282
probability
P(X=x) = P(X+1) =
0.3282
0.3656
0.2036
0.0756
0.0211
0.0047
0.0009
𝒙+𝟏
=1.114
We have 𝝀 =1.114
Therefore P (0) = 𝒆𝟎!
𝝀
𝝀
𝒙+𝟏
p(x)
p(x)
X
0
1
2
3
4
5
6
f
153
169
72
31
12
6
2
445
fx
0
169
144
93
48
30
12
496
Expected Frequency
= N*P(X=x)
146.04≈146
162.69≈163
90.60≈91
33.64≈34
9.38≈9
2.09≈2
0.40≈0
445
Result: -
P(X=x) =
X
f0
fe
0
153
146
𝒆−𝟏.𝟏𝟏𝟒∗ (𝟏.𝟏𝟏𝟒)𝒙
𝒙!
1
169
163
for x=0, 1, 2…6
2
72
91
3
31
34
4
12
9
5
6
2
6
2
0
NEGATIVE BINOMIAL DISTRIBUTION
1) A blood bank collects B-negative blood samples only. The probability of getting Bnegative blood is ‘P’ and is treated as success. It takes only one bottle of blood from
one person and purchases ‘5’ bottles per day. The failures of 400 days before getting
5th bottle of blood of this kind were recorded as follows:
No. of failures(Not getting Bnegative blood)
No. of days
0
1
2
3
4
5
6
7
131
131
79
37
14
5
2
1
Fit a Negative binomial distribution using direct method.
Solution: - Aim: - To fit a negative binomial distribution by direct method.
𝒇𝒙
Formulae: - Mean =𝑿=
𝒇
=
𝒓𝒒
𝒑
𝑴𝒆𝒂𝒏
; Variance = 𝝈
𝒑 = 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 ; 𝒒 = 1-𝒑 P(X) =
2=
x+r-1
𝒇𝒙𝟐
𝒇
𝒓𝒒
- (𝑿)2 =
𝒑𝟐
;
Cr-1 pr qx
Calculations: - About Mean
Here r= 5 , 𝑿= 1.25 , 𝝈2 = 1.5625
𝑴𝒆𝒂𝒏
 𝒑 = 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆 =

𝒒 = 0.2
𝑿
𝝈𝟐
= 0.8
X
0
1
2
3
4
5
6
7
f
131
131
79
37
14
5
2
1
400
fx
0
131
158
111
56
25
12
7
500
fx2
0
131
316
333
224
125
72
49
1250
X
f
0
1
2
3
4
5
6
7
131
131
79
37
14
5
2
1
400
Result:-
P(X) = (
X
f0
fe
0
131
131
probability
x+r-1
P(X) =
Cr-1 pr qx
0.3276
0.3276
0.1966
0.0917
0.036
0.01321
0.004404
0.001384
x+5-1
Expected Frequency
= N*P(X)
131.04 ≈131
131.04 ≈131
78.64 ≈79
36.68 ≈37
14.4 ≈14
5.284 ≈5
1.7616 ≈2
0.5536 ≈1
400
C5-1 ) (0.8)5(0.2)x
1
131
131
2
79
79
3
37
37
4
14
14
5
5
5
6
2
2
7
1
1
NEGATIVE BINOMIAL DISTRIBUTION
1) A blood bank collects B-negative blood samples only. The probability of getting B-negative blood is ‘P’
and is treated as success. It takes only one bottle of blood from one person and purchases ‘5’ bottles
th
per day. The failures of 400 days before getting 5 bottle of blood of this kind were recorded as
follows:
No. of failures(Not getting B-negative
blood)
No. of days
0
1
2
3
4
5
6
7
131
131
79
37
14
5
2
1
X
0
1
2
3
4
5
6
7
f
131
131
79
37
14
5
2
1
400
Fit a Negative binomial distribution using direct method.
Solution: - Aim: - To fit a negative binomial distribution by direct method.
Formulae: - Mean = =
=
=
;
; Variance =
= 1-
P(X) =
2=
x+r-1
-(
r
Cr-1 p q
2=
;
x
Calculations: - About Me
Here r= 5 ,

=

= 0.2
X
f
0
1
2
3
4
5
6
7
131
131
79
37
14
5
2
1
400
Result:X
f0
fe
= 1.25 ,
P(X) = (
0
131
131
2
= 1.5625
=
= 0.8
probability
P(X) = x+r-1 Cr-1 pr qx
0.3276
0.3276
0.1966
0.0917
0.036
0.01321
0.004404
0.001384
x+5-1
5
fx
0
131
158
111
56
25
12
7
500
Expected Frequency
= N*P(X)
131.04 ≈131
131.04 ≈131
78.64 ≈79
36.68 ≈37
14.4 ≈14
5.284 ≈5
1.7616 ≈2
0.5536 ≈1
400
x
C5-1 ) (0.8) (0.2)
1
131
131
2
79
79
3
37
37
4
14
14
5
5
5
6
2
2
7
1
1
2
fx
0
131
316
333
224
125
72
49
1250
GEOMETRIC DISTRIBUTION
1) Every day a student walks from his home to road and requests a two wheeler rider going on the road
towards his college for lift. Number of riders he requested to get lift, for 200 days, are recorded as
follows. Fit a geometric distribution.
No .of riders that
he requested(X)
1
2
3
4
5
6
No .of days (f)
140
42
12
3
2
1
Solution: - Aim: - To fit a geometric distribution.
Formulae: - Here P(X) = pq (x-1); x=0, 1, 2….
and
𝑿=
𝒇𝒙
But 𝑿=
𝒇
𝟏
𝒑
=> 𝒑 =
𝟏
X
1
2
3
4
5
6
𝑿
𝒒 = 1- 𝒑
𝑿=
Calculation: -
𝒇𝒙
𝒇
=
𝟐𝟖𝟖
𝟐𝟎𝟎
= 1.44 𝒑 =
𝟏
𝑿
=
𝟏
𝟏.𝟒𝟒
= 0.6944
𝐪 = 1- 𝒑 =1-0.6944= 0.3056
f
140
42
12
3
2
1
200
X
f
fx
1
2
3
4
5
6
140
42
12
3
2
1
200
140
84
36
12
10
6
288
probability
P(X) = pq (x-1);
0.6944
0.2122
0.0648
0.0198
0.0061
0.0027
Expected frequency
= N* P(X)
138.88 ≈139
42.44 ≈ 42
12.96 ≈13
3.96 ≈ 4
1.22 ≈1
0.54 ≈1
200
Result: -
X
f0
fe
1
140
139
2
42
42
3
12
13
4
3
4
5
2
1
6
1
1
fx
140
84
36
12
10
6
288
2) A thief steals stereo-systems from the parked cars. Every morning he starts searching parked cars until
he gets a car with stereo system. Then, he steals the stereo-system and calls it a day. The probability
that a parked car has a stereo system is P and it is treated as success. The distribution of failures before
getting a stereo-system for 100 days., is obtained as
No .of failures(X)
No .of days (f)
0
40
1
24
2
15
3
9
4
5
5
3
6
2
7
1
8
1
Fit a geometric distribution for the data by recurrence method.
Solution: - Aim: - To fit a geometric distribution by recurrence method.
Formulae: - P(X) = pq (x-1); x=0, 1, 2….
P(X+1) = q x+1 p; x= 0, 1, 2……
𝒇𝒙
𝑿=
p=

𝟏
𝟏+𝑿
𝒇
f
40
24
15
9
5
3
2
1
1
100
𝑿=
𝒒
=
𝒑
𝟏−𝒑
𝒑
Recurrence formula is P(X+1) =𝒒 P(X)
and 𝐪 = 1- 𝒑
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧𝐬:𝒇𝒙
𝑿=
=
𝒇
X
0
1
2
3
4
5
6
7
8
But
fx
0
24
30
27
20
15
12
7
8
143
𝟏𝟒𝟑
𝟏𝟎𝟎
= 1.43
p=
𝟏
𝟏+𝑿
=
𝟏
𝟏+𝟏.𝟒𝟑
probability
0.4115
0.2422
0.1425
0.0839
0.0494
0.0291
0.0171
0.0101
0.0142
= 0.4115
q= 1-0.4115 = 0.5885
Expected frequency
41.15 ≈ 41
24.22 ≈24
14.25 ≈14
8.39 ≈ 8
4.94 ≈ 5
2.91 ≈3
1.71 ≈2
1.01 ≈1
1.42 ≈1
100
Result:X
f0
fe
0
40
41
1
24
24
2
15
15
3
9
8
4
5
5
5
3
3
6
2
2
7
1
1
8
1
1
HYPERGEOMETRIC DISTRIBUTION
Out of 20 packages to be dispatched by a mail-room Clerk eight are to be sent by air mail and the rest by surface
mail. The packages got mixed up thoroughly. Five of the packages are selected randomly. The distribution of
packages marked for air mail getting into the chosen five packages, observed over period of 100 days is
X
f
0
7
1
22
2
45
3
20
4
5
5
1
Solution: - Aim: - Fit a hyper geometric distribution.
Formulae: - P(X=x) =
M
N-M
CX
N
C n-x
Cn
Calculation: - N= Total no. of packages to be dispatched = 20
M= No. of packages to be sent by air mail = 8
n= No. of packages to be selected = 5
N-M = No. of packages to be sent by surface mail = 12
X
f
probability
P(X=x) = MCX
N-M
C n-x
Expected frequency
= N* P(X)
N
Cn
0
1
2
3
4
5
7
22
45
20
5
1
100
0.0510835
0.255417
0.397316
0.238390
0.054179
0.003611
5.10 ≈5
25.54 ≈26
39.73 ≈40
23.83 ≈24
5.41 ≈5
0.36 ≈1
100
Result: X
f0
fe
0
7
5
1
22
26
2
45
40
3
20
24
4
5
5
5
1
0
NORMAL DISTRIBUTION
1) Fit a Normal distribution for the following data by areas method.
Class
60-65
65-70
70-75
75-80
80-85
85-90
Frequency 3
21
150
335
326
135
90-95
26
95-100
4
Solution: - Aim: - To fit a Normal distribution by areas method.
Formulae: -
𝒇𝒙
𝑿=
𝒇
Where = µ = Mean = 𝑿
Calculations: C.I
60-65
65-70
70-75
75-80
80-85
85-90
90-95
95-100
Mean= 𝑿 =
𝝈=
𝒇
𝟕𝟗𝟗𝟒𝟓
=
𝟏𝟎𝟎𝟎
𝑵
− √ ( ( ▒𝒇𝒙)/𝑵 )2 ( Write S.D formula)
𝝈 = S.D
f
3
21
150
335
326
135
26
4
1000
𝒇𝒙
𝒇𝒙𝟐
= 79.945
X
62.5
67.5
72.5
77.5
82.5
87.5
92.5
97.5
𝝈=
fx
187.5
1417.5
10875.0
25962.5
26895.0
11812.5
2405.0
390.0
79945
𝒇𝒙𝟐
𝑵
fx2
11718.75
95681.25
788437.5
2012093.75
2218837.5
1033593.75
222462.5
38025.0
6420850.0
− √ ( ( ▒𝒇𝒙)/𝑵 )2 = 5.445
Normal distribution takes values from -∞ to +∞ . Hence the problem is modified as
C.I
f
Lower
Z=X’ -µ
Area
∆ φ(Z) =
f(X) = N* ∆ φ(Z)
𝝈
Limit(X)
φ(Z)
φ(z+1)- φ(Z)
−∞ 𝐭𝐨 𝟔𝟎
−∞
−∞
0
0
0.0001
0.1≈0
60-65
3021
60
-3.6629
0.0001
0.003
3≈3
65-70
150
65
-2.7447
0.0031
0.0313
31.3≈31
70-75
335
70
-1.8264
0.0344
0.1497
149.7≈150
75-80
326
75
-0.9081
0.1841
0.3199
319.9≈320
80-85
135
80
0.0101
0.5040
0.3172
317.2≈317
85-90
26
85
0.9283
0.8212
0.1459
145.9≈146
90-95
4
90
1.8466
0.9671
0.03
30≈30
95-100
0
95
2.7649
0.9971
0.0028
2.8≈3
100 to ∞
100
3.6831
0.9999
1000
1000
Result: C.I.
f0
fe
−∞ 𝐭𝐨 𝟔𝟎
60-65
3
3
0
0
65-70
21
31
70-75
150
150
75-80
335
320
80-85
326
317
85-90
135
146
90-95
26
30
95-100
4
3
100 to ∞
0
2) Fit a Normal Distribution for the following data by ordinates method.
C.I.
F
150-160
9
160-170
24
170-180
51
180-190
66
190-200
72
200-210
48
210-220
21
Solution: - Aim: - To fit Normal distribution by ordinates method.
Formulae: -
𝑿=
𝒇𝒙
𝝈=
𝒇
𝒇𝒙𝟐
𝑵
Where Mean = µ = 𝑿
− √ ( ( ▒𝒇𝒙)/𝑵 )2 (Write S.D formula)
𝝈 = S.D and Z=
X- µ
𝝈
Calculations: C.I.
150-160
160-170
170-180
180-190
190-200
200-210
210-220
220-230
230-240
Mean = µ = 𝑿 =
𝝈=
𝒇𝒙𝟐
𝑵
f
9
24
51
66
72
48
21
6
3
300
𝒇𝒙
𝒇
=
𝟓𝟔𝟗𝟒𝟎
𝟑𝟎𝟎
x
155
165
175
185
195
205
215
225
235
= 189.8
− √ ( ( ▒𝒇𝒙)/𝑵 )2 = 16.15
fx
1395
3960
8925
12210
14040
9840
4515
1350
705
56940
fx2
216225
653400
156187
2258850
2737800
2017200
970725
303750
165675
10885500
220-230
6
230-240
3
C.I.
X
I Z = X- µ I
Ordinate
Value φ(Z)
0.0396
0.1238
0.2637
0.3825
0.3790
0.2565
0.1182
0.0371
0.0079
𝝈
150-160
160-170
170-180
180-190
190-200
200-210
210-220
220-230
230-240
155
2.15
1.537
0.917
0.297
0.322
0.942
1.562
2.181
2.80
165
175
185
195
205
215
225
235
f(x)= N* φ(Z) *h
𝝈
7.36 ≈ 7
23.02 ≈23
49.03 ≈49
71.13 ≈71
70.5 ≈71
47.70 ≈ 48
21.98 ≈22
6.89 ≈7
1.5 ≈2
300
Result: C.I.
f0
fe
150-160
160-170
170-180
180-190
190-200
200-210
210-220
220-230
230-240
9
7
24
23
51
49
66
71
72
71
48
48
21
22
6
7
3
2
EXPONENTIAL DISTRIBUTION
The study of divorced cases in the western countries, the following distribution is obtained for the time interval
(in years) between the day of their marriage and day of their divorce.
No .of years
0-3
3-6
No. of persons
190
70
Fit an exponential distribution.
6-9
25
9-12
10
12-15
4
15 & above
1
Solution: - Aim: - To fit an exponential distribution.
Formulae: - Mean= 𝑿=
𝒇𝒙
𝒇
𝟏
; 𝜽 =𝑿
P (a<X<b) = e –θ.a – e-θ.b Where ‘a’ is the lower limit and ‘b’ is the upper limit.
Calculation: No .of years
0-3
3-6
6-9
9-12
12-15
No .of persons
190
70
25
10
4
Midvalue(X)
1.5
4.5
7.5
10.5
13.5
fX
285
315
187.5
105
54
15 & above
1
16.5
16.5
300
𝑿=
𝒇𝒙
=
𝒇
𝟗𝟔𝟑
= 3.21
𝟑𝟎𝟎
No .of years
0-3
3-6
6-9
9-12
12-15
15 & above
𝜽 =
𝟏
𝑿
=
𝟏
𝟑.𝟐𝟏
= 0.3115
No .of persons
190
70
25
10
4
1
300
P (a<X<b)
3-6
70
72
6-9
25
28
Expected frequency
182.16 ≈182
71.55 ≈72
28.11 ≈28
11.04 ≈11
4.35 ≈4
2.79 ≈3
300
0.6072
0.2385
0.0937
0.0368
0.0145
0.0093
Result: No .of years
f0
fe
0-3
190
182
9-12
10
11
12-15
4
4
15& above
1
3
CAUCHY DISTRIBUTION
1) In air force operations, suppose a pilot-less helicopter is flying at 1 KM height from the origin.
It has a sophisticated machine gun which identifies the enemy crossing the border and fires at
him. It can uniformly turn in between (
−𝝅
𝟐
,
𝝅
𝟐
).
It was reported that 200 terrorists were
killed at different places along the border as given below:
Distance
No .of terrorists killed
-∞ to -25
2
-25 to -19
1
-19 to -13
2
-13 to-7
4
-7 to -1
41
-1 to 5
137
5 to 11
7
11 to 17
2
17 to 23
1
23 to +∞
3
Fit Cauchy distribution.
Solution: - Aim: - To fit a Cauchy distribution.
𝟏
𝟏
Formulae: - F(X) = 𝟐 + 𝝅 𝐭𝐚𝐧−𝟏 (𝑿) Where 𝝅 =180 0 P (a<X<b) = F(b) – F(a)
Calculation: Distance
-∞ to -25
-25 to -19
-19 to -13
-13 to-7
-7 to -1
-1 to 5
5 to 11
11 to 17
17 to 23
23 to +∞
Frequency
2
1
2
4
41
137
7
2
1
3
P(a<X<b)
0.0127
0.0040
0.007
0.0208
0.2048
0.6872
0.0339
0.0102
0.0049
0.0138
200
Expected frequency
2.54 ≈3
0.8 ≈1
1.4 ≈1
4.16 ≈4
40.96 ≈41
137.44 ≈137
6.78 ≈7
2.04 ≈2
0.98 ≈1
2.76 ≈3
200
Result:Distance
-∞ to -25
-25 to -19
-19 to -13
-13 to-7
-7 to -1
-1 to 5
5 to 11
11 to 17
17 to 23
23 to +∞
Frequency
2
1
2
4
41
137
7
2
1
3
Expected frequency
3
1
1
4
41
137
7
2
1
3
2) In air force operations, suppose a pilot-less helicopter is flying at 5KM height and 2.5 KM away from the
origin to the right. It has a sophisticated machine gun which identifies the enemy crossing the border
−𝝅 𝝅
and fires at him. It can uniformly turn in between ( 𝟐 , 𝟐 ). It was reported that 1000 terrorists were
killed at different places along the border as given below:
Distance
-∞ to -25
-25 to -20
-20 to -15
-15 to -10
-10 to -5
-5 to 0
0 to 5
5 to 10
10 to 15
15 to 20
20 to 25
25 to 30
30 to ∞
Fit a Cauchy distribution.
No. of terrorists killed
56
14
20
13
66
165
300
160
60
38
20
12
58
Solution: - Aim: - To fit Cauchy distribution.
𝟏
𝟏
𝟐
𝝅
Formulae: - - F(X) = +
𝑿−𝒂
𝐭𝐚𝐧−𝟏 (
𝒃
) Where 𝝅 =180 0 P (a<X<b) = F (b) – F (a)
Calculation: -
Distance
No. of terrorists killed
56
14
20
13
66
165
300
160
60
38
20
12
58
-∞ to -25
-25 to -20
-20 to -15
-15 to -10
-10 to -5
-5 to 0
0 to 5
5 to 10
10 to 15
15 to 20
20 to 25
25 to 30
30 to ∞
P (a<X<b) = F (b) – F (a)
0.0572
0.0124
0.019
0.0325
0.0661
0.1652
0.2952
0.1652
0.0661
0.0325
0.0189
0.0125
0.0572
1000
Expected frequency
57.2 ≈57
12.4 ≈12
19 ≈19
32.5 ≈33
66.1 ≈66
165.2 ≈165
295.2 ≈ 295
165.2 ≈165
66.1 ≈66
32.5 ≈33
18.9 ≈19
12.5 ≈13
57.2 ≈57
1000
Result: -
Distance
-∞ to -25
-25 to -20
-20 to -15
-15 to -10
-10 to -5
-5 to 0
0 to 5
5 to 10
10 to 15
15 to 20
20 to 25
25 to 30
30 to ∞
f0
56
14
20
13
66
165
300
160
60
38
20
12
58
fe
57
12
19
33
66
165
295
165
66
33
19
13
57
PROBLEMS ON MOMENTS
For the frequency distribution of scores in maths of 50 candidates selected at random from among those appearing at a
certain examination. Compute the first four moments about mean of distribution. Find also the corrected values of the
oments after Sheppard’s correction is applied.
Scores
50-60
60-70
70-80
80-90
90-100
100-110
Frequency
Scores
Frequency
Scores
Frequency
1
110-120
1
170-180
10
0
120-130
0
180-190
11
0
130-140
4
190-200
4
1
140-150
4
200-210
1
1
150-160
2
210-220
1
2
160-170
5
220-230
2
Solution: - Aim: - To find moments about mean and applying Sheppard’s correction.
Formulae: - Central moments in terms of raw moments given by
µ2 = µ 2 1 - µ1 1
2
µ3 = µ3 1 - 3 µ2 1 µ1 1 + 2 µ 1 1
3
2
µ4 = µ4 1 - 4 µ 3 1 µ1 1 + 6 µ 2 1 µ1 1 - 3 µ 1 1
4
Raw moments are given by
µ1 = h *
1
Σ fd
N
µ2 = h 2 *
1
Σ fd2
N
µ3 = h 3 *
1
Σ fd3
N
µ4 = h 4 *
1
Σ fd4
N
Sheppard’s corrections are given by
µ2( corrected) = µ2 - h2 / 12
µ4(corrected) = µ4 - h2 / 2 µ2 + 7 / 240 h4
Calculations: -
C.I
50-60
60-70
70-80
80-90
90-100
100-110
110-120
120-130
130-140
140-150
150-160
160-170
170-180
180-190
190-200
200-210
210-220
220-230
f
1
0
0
1
1
2
1
0
4
4
2
5
10
11
4
1
1
2
50
X
55
65
75
85
95
105
115
125
135
145
155
165
175
185
195
205
215
225
di =Xi –A/h
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
fd
-8
0
0
-5
-4
-6
-2
0
0
4
4
15
40
55
24
7
8
18
fd2
64
0
0
25
16
18
4
0
0
4
8
45
160
275
144
49
64
162
fd3
-512
0
0
-125
-64
-54
-8
0
0
4
16
135
640
1375
864
343
512
1458
fd4
4096
0
0
625
256
162
16
0
0
4
32
405
2560
6875
5184
2401
4096
13122
150
1038
4584
39834
Raw moments:-
µ1 1 = h *
µ2 1 = h 2 *
1
1
Σ fd = 10 *
* 150 = 30
N
50
1
1
Σ fd2 = 100 *
* 1038 = 2076
N
50
1
1
Σ fd3 = 1000 * * 4584 = 91,680
N
50
µ3 1 = h 3 *
µ4 1 = h 4 *
1
1
Σ fd4 = 10000 * * 39838 = 7,966,800
N
50
Central moments:-
µ2 = µ 2 1 - µ1 1
2
= 2076 - (30)2 = 2076 - 900 = 1176
µ3 = µ3 1 - 3 µ 2 1 µ1 1 + 2 µ 1 1
3
= 91860 -3(2076) (30) + 2(30)3 = -41,160
2
µ4 = µ4 1 - 4 µ31 µ11 + 6 µ21 µ11 - 3 µ11
4
= 7966,800 – 4(30) (91680) + 6 (900) (2076) – 3(30)4 = 5,745,600
Sheppard’s correction:µ2( corrected) = µ2 - h2 / 12 = 1167.67
µ4(corrected) = µ4 - h2 / 2 µ2 + 7 / 240 h4 = 5745,891.618
2) Calculate first four moments of the following distribution about mean and hence find β 1 and β2
X
f
0
1
1
8
2
28
3
56
4
70
5
56
6
28
7
8
8
1
olution:- Aim:- To calculate moments about mean.
Calculations:-
X
0
1
2
3
4
5
6
7
8
f
1
8
28
56
70
56
28
8
1
256
di= Xi-A
-4
-3
-2
-1
0
1
2
3
4
Raw moments:-
µ1 1 =
µ2 1 =
µ3 1 =
µ4 1 =
1
Σ fd = 1/250 * 0 = 0
N
1
Σ fd2 = 1/256 *512 = 2
N
1
Σ fd3 = 1/256 * 0 = 0
N
1
Σ fd4 = 1/256 * 2816 = 11
N
Central moments:-
fd
-4
-24
-56
-56
0
56
56
24
4
0
fd2
16
72
112
56
0
56
112
72
16
512
fd3
-64
-216
-224
-56
0
56
224
216
64
0
fd4
256
648
448
56
0
56
448
648
256
2816
µ2 = µ21 - µ11
2
= 2-0 = 2
µ3 = µ3 1 - 3 µ2 1 µ1 1 + 2 µ 1 1
3
= 0 – 3(0) (2) + 2(0) = 0
2
µ4 = µ4 1 - 4 µ 3 1 µ1 1 + 6 µ 2 1 µ1 1 - 3 µ1 1
β1 = µ3
2
4
= 11 – 4(0) (0) + 6 (2) (0) – 3(0) =11
/ µ23 = 0
β2 = µ4 / µ22 = 11/4 = 2.75
3) Obtain Karl Pearson’s coefficient of skew ness for the following data.
Value
f
5-10
6
10-15
8
15-20
17
20-25
21
25-30
15
30-35
11
35-40
12
Solution:- Aim:- To find Karl Pearson’s coefficient of skew ness for the following data.
Formulae:- Karl Pearson’s coefficient of skew ness SK is given by
Skew ness =
σ= h*
√
3(M-Md) / σ
Where M = mean Md = median σ = standard deviation
 fidi  fidi 2
-(
)
N
N
Mean = A +
σ= h*
Calculations:-
√
h
Σ fidi
N
Md = l 1 +
 fidi  fidi 2
-(
)
N
N
h N
( - c)
f 2
C.I
5-10
10-15
15-20
20-25
25-30
30-35
35-40
Mean = A +
Md = l 1 +
f
6
8
17
21
15
11
12
80
X
7.5
12.5
17.5
22.5
27.5
32.5
37.5
C.F
6
14
31
52
67
78
80
180
di =Xi –A/h
-3
-2
-1
0
1
2
3
fidi
-18
-16
-17
0
15
22
6
-8
h
Σ fidi = 22.5 + 5/80 * (-8) = 22
N
h N
( - c) = l1 = 20 , h = 5 , f=21 , c= 31
f 2
= 22.142
σ= h*
√
 fidi  fidi 2
-(
) = 5 * √ 2.24 = 5 * 1.496 = 7.48
N
N
Skew ness =
3(M-Md) / σ
= 3( 22 – 22.14 ) / 7.48 = -0.056
Result:- Karl Pearson’s coefficient of skew ness SK is given by SK = -0.056
Hence the given distribution is negatively skewed.
4) Calculate Bow ley’s coefficient of skew ness for the following data.
Solution:- Aim:- To calculate Bow ley’s coefficient of skew ness.
Formulae:-
Skew ness =Q3 + Q1 - 2 Md / Q3 – Q1 where Q1 = l 1 +
h N
(
-c)
f 4
Md = l 1 +
h N
( - c)
f 2
Q3 = l 1 +
h 3N
(
- c)
f
4
fidi2
54
32
17
0
44
88
18
103
Calculations:- Md =
N
= 130/2 = 65 ;
2
l1 = 10 , f = 26 , c = 45
Md = l 1 +
h N
( - c) = 13. 84
f 2
Q1 = l 1 +
h N
(
- c ) ; l1 = 5, f = 45,c = 0 => 8.3578
f 4
Q3 = l 1 +
h 3N
(
- c) = 23.26
f
4
Skew ness =Q3 + Q1 - 2 Md / Q3 – Q1 = 3.9378 / 14.903 = 0. 264228
1) Draw a histogram for the following data.
Variable
Frequency
100-110
11
110-120
28
120-130
36
130-140
49
140-150
33
150-160
20
160-170
8
170-180
0
frequency
histogram
60
50
40
30
20
10
0
frequency
100 110
to
to
110 120
120
to
130
130 140
to
to
140 150
150
to
160
160 170
to
to
170 180
variables
2) Represent the data by Histogram
C.I
10-15
15-20
20-25
FREQ
7
19
27
ADJ.FRE 7
19
27
25-30
15
15
Weekly wages
HISTOGRAM
30
25
20
15
10
5
0
Adj.frequency
10
to
15
15
to
20
20
to
25
25
to
30
30
to
40
No .of w orkers
40
to
60
60
to
80
30-40
12
6
40-60
12
3
60-80
8
2
3) Draw a histogram and frequency polygon to the following data.
Length of
leaves
No .of
leaves
6.5-7.5
7.5-8.5
8.5-9.5
9.5-10.5
10.5-11.5
11.5-12.5
12.5-13.5
5
12
25
48
32
6
1
FREQUENCY POLYGON
NO. OF LEAVES
60
50
40
30
NO .OF LEAVES
20
10
0
0
2
4
6
8
LENGTH OF LEAVES
4)
Draw less than ogive for the following data.
Marks
No. of students
10-20
4
20-30
6
30-40
10
40-50
20
50-60
18
60-70
2
NO.OF STUDENTS
L.C.F
70
60
50
40
30
20
10
0
L.C.F
0
20
40
60
80
MARKS
5) Draw more than ogive for the following data.
Marks
No. of students
10-20
4
20-30
6
30-40
10
40-50
20
NO.OF STUDENTS
M.C.F
70
60
50
40
30
20
10
0
M.C.F
0
20
40
MARKS
60
80
50-60
18
60-70
2
6) Determine the median wage graphically from the following data.
Wages(in 700-800
800-900
900-1000 1000-1100 1100-1200
Rs)
Workers
4
6
10
16
12
1200-1300
1300-1400
7
3
Ogive Curves
70
60
Workers
50
40
L.C.F
30
M.C.F
20
10
0
0
500
1000
1500
Wages
Solution: - Two cumulative frequency curves one by the less than method and another by more
than method. From the point where both these curves meet draw a perpendicular on the X-axis
and the point where it meets the X-axis is median.
7) The frequency distribution of weekly wages in a certain family is as follows.
Weekly 23-27
wages
in Rs
No .of
2
workers
28-32
32-37
38-42
43-47
48-52
53-57
58-62
63-67
68-72
6
9
14
32
16
12
6
2
1
Draw the ogive and find from the ogive i) 1st Quartile ii) Median iii) 3rd Quartile
Solution:- Since it is upper limit exclusive method the real limits of the classes shall be 22.5 , 27.5 ,
32.5 etc.
Q1 = size of
N
th item = 100 / 4 = 25th item
4
Q3 = size of 3N / 4 item = 75th item
Median (N/2) = 100/2 = 50th item.
Wages less than
22.5
27.5
32.5
37.5
42.55
47.5
52.5
57.5
62.5
67.5
72.5
No .of workers
2
8
17
31
63
79
91
97
99
100
0
Ogive
No .of workers
120
100
80
No .of
w orkers
60
40
20
0
0
50
100
Wages
8) The Monthly production of maruthi Udhyog limited for the 1 st six months of the year 1985 are
given below.
Month
Jan
Feb
Mar
Apr
May
June
Production 250
300
340
320
270
240
Production
Bar diagram
400
350
300
250
200
150
100
50
0
Production
Jan
Feb
Mar
Apr
Months
May
June
9) The number of students in various courses like B.Com, M.Com, B.A, and M.A in various colleges may be represented by a sub divid
bar diagram. While constructing such that the various components in each bar should be kept in same order. During 1992-93 to 1994
the no .of students in
Year
1991
1992
1993
1994
university are as
Sales in
120
135
140
150
follows. Represent th
thousands
data by a suitable
Gross profit in
40
45
55
60
diagram.
thousands
Net profit in
20
30
35
40
thousands
Year
1992-93
1993-94
1994-95
Arts
20,000
26,000
31,000
Science
10,000
9,000
9,500
Law
5,000
7,000
7,500
Total
35,000
42,000
48,000
Multiple bar diagram
No .of students
120,000
100,000
Total
80,000
Law
60,000
Science
40,000
Arts
20,000
0
1992-93
1993-94
1994-95
Years
10) Draw a suitable diagram from the following data.
Solution: - Same as above.
11) Represent the following by sub-divided bar diagram drawn on the percentage bars.
Particulars
1986
1987
1988
1) Cost per chair
a) Wages
b) Other cost
c) Polishing
Total
2) Proceeds per chair
3) profit(+)
Loss(-)
9
6
3
18
20
2
15
10
5
30
30
-
21
14
7
42
40
-2
Solution: - Same as above. Take the sale price per chair as 100 and express the other figures in percentages. The
percentage is obtained as given below.
Particulars
1986
1987
1988
1) Cost per chair
a) Wages
b) Other cost
c) Polishing
Total
2) Proceeds per chair
3) profit(+)
Loss(-)
45
30
15
100
100
+10.0
50
33.3
16.7
100
100
-
52.5
35
17.5
105.5
100
-5.0
12) Present the following data by means of pie diagram.
Material
Bricks
Cement
Steel
Timber
Cost of constructing a house(in %)
15%
20%
15%
10%
Wages
Super result
25%
15%
Solution: - We are given here the percentage expenditure on various components. The appropriate diagram is pie
diagram. Before doing that it is necessary to convert the percentage into angles of different degrees. Since the total
is 100 1 percentage = 3.6. Therefore we have to multiply each of the above percentage by 3.6.
Material
Bricks
Cement
Steel
Timber
Wages
Super result
Cost of constructing a house (in degrees)
54
72
54
36
90
54
Pie chart
54, 15%
54, 15%
Bricks
Cement
90, 25%
72, 20%
Steel
Timber
Wages
36, 10%
54, 15%
Super result
13) The following data relate to the expenditure of three families per month
Item of expenditure
Food
Clothing
Education
Lighting
Miscellaneous
Rent
Family A
400
200
100
50
50
200
Family B
600
300
400
100
200
400
Family C
1600
1000
800
300
800
1500
Item of expenditure
Family A
Rs
Degrees
400
144
200
72
200
72
100
36
50
18
50
18
Family B
Rs
Degrees
600
108
400
72
300
54
400
72
100
36
200
72
Family C
Rs
Degrees
1600
96
1500
60
1000
60
800
48
300
18
800
48
Solution: -
Food
Rent
Clothing
Education
Lighting
Miscellaneous
Same as above