Statistics ⇒⇒ Exercise 13.3

Question 1

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	65 − 85	85 − 105	105 − 125	125 − 145	145 − 165	165 − 185	185 − 205
Number of consumers	4	5	13	20	14	8	4

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

$${Upper Class Limit + Lower Class Limit \over 2} $$

By using Assumed Mean method.

Let the assumed mean be a = 135

Mthl. consum	No.of consum. fi	Cumu. Freq. c.f	Class mark xi	di = (xi-a)	(fi di )
65 − 85	4	4	75	75 - 135 = - 60	-240
85 − 105	5	4 + 5 = 9	95	95 - 135 = - 40	-200
105 − 125	13 (f0)	9 + 13 = 22	115	115 - 135 = - 20	-260
125 − 145	20 (f1)	22 + 20 = 42	135	135 - 135 = 0	0
145 − 165	14(f2)	42 + 14 = 56	155	155 - 135 = 20	280
165 − 185	8	56 + 8 = 64	175	175 - 135 = 40	320
185 − 205	4	64 + 4 = 68	195	195 - 135 = 60	240
	$\Sigma f_i = $ N= 68				$\Sigma f_id_i $ = 140

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_id_i} \over {\Sigma f_i} }$$

$$ \bar{X} = 135 + { {140} \over {68} }$$

$$ \bar{X} = 135 + {( 2.058)}$$

$$ \bar{X} = 137.058 $$

Now Mode can be calculated as follows:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 20,

So, the modal class is 125-145

L (the lower limit of modal class) = 125

f₁ (frequency of the modal class) = 20

f₀ (frequency of the class preceding the modal class) = 13

f₂ (frequency of the class succeeding the modal class) = 14

h (class size) = 20

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 125 + [ { { 20 - 13} \over { 2(20) - 13 - 14} } ] ×(20) $$

$$ = 125 + [ {( 7×20) \over { 40 - 27 } } ] $$

$$ = 125 + [ { { 140} \over { 13} } ] $$

$$ = 125 + 10.76 $$

$$ Mode = 135.76 $$

Therefore, mode of this data is 135.76.

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 68

Cumulative frequency just greater than N/2 = (i.e, 68/2 = 34 ) is 42

The class corresponding to this Cumulative frequency is 125-145.

So, the Median class is 125-145

L = Lower limit of median class = 125

f = Frequency of the median class = 20

cf = Cumulative frequency of the class preceding the median class = 22

h = Class interval of the median class) = 20

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }×(h )$$

$$ = 125 + {({ { 68 \over 2} - 22 }) \over { 20} }×(20 )$$

$$ = 125 + {({ { 34} - 22}) \over { 20} }×(20 )$$

$$ = 125 + {{ { 12} × 20} \over { 20} } $$

$$ = 125 + 12 $$

$$ Median = 137 $$

Therefore, Median of this data is 137.

Question 2

If the median of the distribution is given below is 28.5, find the values of x and y.

Class interval	Frequency
0 − 10	5
10 − 20	x
20 − 30	20
30 − 40	15
40 − 50	y
50 − 60	5
Total	60

Solution :

The cumulative frequency for the given data is calculated as follows.

Class interval	Frequency	Cumu. frequency (cf)
0 − 10	5	5
10 − 20	x	5 + x
20 − 30	20	25 + x
30 − 40	15	40 + x
40 − 50	y	40 + x + y
50 − 60	5	45 + x + y
Total	N = 60

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 60

Given Median = 28.5 which lies in the class 20-30

So, the Median class is 20 - 30

L = Lower limit of median class = 20

f = Frequency of the median class = 20

cf = Cumulative frequency of the class preceding the median class = 5 + x

h = Class interval of the median class) = 10

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }× h $$

$$ 28.5 = 20 + {{ { 60 \over 2} - (5 + x) } \over { 20} }× 10 $$

$$ 28.5 = 20 + {({ { 30} - 5 - x }) \over { 20} }× 10 $$

$$ 28.5 = 20 + {({ { 25 - x}) × 10} \over { 20} } $$

$$ 28.5 - 20 = {({ { 25 - x}) } \over { 2} } $$

$$ 8.5 ×2 = ({ { 25 - x}) } $$

$$ x = { 25 - 17} $$

$$ x = 8 $$

As per question $ 45 + x + y = 60 $

$ 45 + 8 + y = 60 $

$ y= 60 - 53$

$ y = 7 $

Therefore, the values of x = 8 and y = 7.

Question 3

A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Solution :

The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

Class interval	Frequency	Cumulative frequency (cf)
18 − 20	2	2
20 − 25	6 - 2 = 4	6
25 − 30	24 - 6 = 18	24
30 − 35	45 − 24 = 21	45
35 − 40	78 − 45 = 33	78
40 − 45	89 − 78 = 11	89
45 − 50	92 − 89 = 3	92
50 − 55	98 − 92 = 6	98
55 − 60	100 − 98 = 2	100

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 100

Cumulative frequency just greater than N/2 = (i.e, 100/2 = 50 ) is 78

The class corresponding to this Cumulative frequency is 35 − 40.

So, the Median class is 35 − 40

L = Lower limit of median class = 35

f = Frequency of the median class = 33

cf = Cumulative frequency of the class preceding the median class = 45

h = Class interval of the median class) = 5

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }× h $$

$$ = 35 + {({ { 100 \over 2} - 45 }) \over { 33} }×(5 )$$

$$ = 35 + {({ { 50} - 45}) \over { 33} }×(5 )$$

$$ = 35 + {({ { 5} ×5}) \over {33} } $$

$$ = 35 + {25 \over {33} } $$

$$ = 35 + 0.76 $$

$$ Median = 35.76 $$

Therefore, Median of this data is 35.76

Question 4

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :

Length (in mm)	Number or leaves
118 − 126	3
127 − 135	5
136 − 144	9
145 − 153	12
154 − 162	5
163 − 171	4
172 − 180	2

Find the median length of the leaves.

Solution :

It can be observed that given series is inclusive and class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, to converting series in exclusive $ 1 \over 2 $ has to be added to the upper class limit and $ 1 \over 2 $ has to be subtracted from the lower class limit of each interval.

Length (in mm)	Number of leaves Freq.	Cumulative frequency (cf)
117.5 − 126.5	3	3
126.5 − 135.5	5	3 + 5 = 8
135.5 − 144.5	9	8 + 9 = 17
144.5 − 153.5	12	17 + 12 = 29
153.5 − 162.5	5	29 + 5 = 34
162.5 − 171.5	4	34 + 4 = 38
171.5 − 180.5	2	38 + 2 = 40
	N = 40

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 40

Cumulative frequency just greater than N/2 = (i.e, 40/2 = 20 ) is 29

The class corresponding to this Cumulative frequency is 144.5 − 153.5.

So, the Median class is 144.5 − 153.5

L = Lower limit of median class = 144.5

f = Frequency of the median class = 12

cf = Cumulative frequency of the class preceding the median class = 17

h = Class interval of the median class) = 9

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }× h $$

$$ = 144.5 + {({ { 40 \over 2} - 17 }) \over { 12} }× 9 $$

$$ = 144.5 + {({ { 20} - 17}) \over { 12} }× 9 $$

$$ = 144.5 + {({ { 3} ×9}) \over {12} } $$

$$ = 144.5 + {27 \over {12} } $$

$$ = 144.5 + 2.25 $$

$$ Median = 146.75 $$

Therefore, Median of this data is 146.75

Question 5

Find the following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)	Number of lamps
1500 − 2000	14
2000 − 2500	56
2500 − 3000	60
3000 − 3500	86
3500 − 4000	74
4000 − 4500	62
4500 − 5000	48

Find the median life time of a lamp .

Solution :

The cumulative frequency for the given data is calculated as follows.

Life time	Number of lamps	Cumulative frequency (cf)
1500 − 2000	14	14
2000 − 2500	56	14 + 56 = 70
2500 − 3000	60	70 + 60 = 130
3000 − 3500	86	130 + 86 = 216
3500 − 4000	74	216 + 74 = 290
4000 − 4500	62	290 + 62 = 352
4500 − 5000	48	352 + 48 = 400
Total	N = 400

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 400

Cumulative frequency just greater than N/2 = (i.e, 400/2 = 200 ) is 216

The class corresponding to this Cumulative frequency is 3000 − 3500.

So, the Median class is 3000 − 3500

L = Lower limit of median class = 3000

f = Frequency of the median class = 86

cf = Cumulative frequency of the class preceding the median class = 130

h = Class interval of the median class = 500

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }× h $$

$$ = 3000 + {({ { 400 \over 2} - 130 }) \over { 86} } × 500 $$

$$ = 3000 + {({ { 200} - 130}) \over { 86} }× 500 $$

$$ = 3000 + {({ { 70} ×500}) \over {86} } $$

$$ = 3000 + 406.97 $$

$$ Median = 3406.97 $$

Therefore, the median lifetime of a lamp is 3406.97 hours

Question 6

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	Number of surnames
1 − 4	6
4 − 7	30
7 − 10	40
10 − 13	16
13 − 16	4
16 − 19	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames

Solution :

The cumulative frequency for the given data is calculated as follows.

Number of letters	Number of surnames	Cumulative frequency (cf)
1 − 4	6	6
4 − 7	30	6 + 30 = 36
7 − 10	40	36 + 40 = 76
10 − 13	16	76 + 16 = 92
13 − 16	4	92 + 4 = 96
16 − 19	4	96 + 6 = 100
Total	N = 100

MEDIAN :Now Median can be calculated as follows:

From the above table, we get

N (Sum of frequencies) = 100

Cumulative frequency just greater than N/2 = (i.e, 100/2 = 50 ) is 76

The class corresponding to this Cumulative frequency is 7 − 10.

So, the Median class is 7 − 10

L = Lower limit of median class = 7

f = Frequency of the median class = 40

cf = Cumulative frequency of the class preceding the median class = 36

h = Class interval of the median class = 3

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }×(h )$$

$$ = 7 + {({ { 100 \over 2} - 36 }) \over { 40} }×(3 )$$

$$ = 7 + {({ { 50} - 36}) \over { 40} }×(3)$$

$$ = 7 + {({ { 14} ×3}) \over {40} } $$

$$ = 7 + 1.05 $$

$$ Median = 8.05 $$

Therefore, the median of the data is 8.05

Now Mean can be calculated as follows:

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

$${Upper Class Limit + Lower Class Limit \over 2} $$

By using Direct Mean method.

Number of letters	Number of surnames (fi )	Class mark xi	(fi xi )
1 − 4	6	2.5	15
4 − 7	30 (f0 )	5.5	165
7 − 10	40 (f1 )	8.5	340
10 − 13	16 (f2 )	11.5	184
13 − 16	4	14.5	58
16 − 19	4	17.5	70
	$\Sigma f_i = 100 $		$\Sigma f_ix_i $ = 832

Mean can be calculated as follows:

Mean $$ \bar{X} = { {\Sigma f_ix_i} \over {\Sigma f_i} }$$

$$ \bar{X} = { {832} \over {100} }$$

$$ \bar{X} = {( 8.32)}$$

Therefore, the mean of the data is 8.32

Now Mode can be calculated as follows:

The class having maximum frequency is the modal class.

It can be observed that the maximum class frequency is 40

The class corresponding to this frequency is 7 − 10.

So, the modal class is 7 − 10

L (the lower limit of modal class) = 7

f₁ (frequency of the modal class) = 40

f₀ (frequency of the class preceding the modal class) = 30

f₂ (frequency of the class succeeding the modal class) = 16

h (class size) = 3

$$ Mode = L + [ { { f_1 - f_0} \over { 2f_1 - f_0- f_2} } ] ×h $$

$$ = 7 + [ { { 40 - 30} \over { 2(40) - 30 - 16} } ] ×(3) $$

$$ = 7 + [ {( 10×3) \over { 80 - 46 } } ] $$

$$ = 7 + [ { { 30} \over { 34} } ] $$

$$ = 7 + 0.88 $$

$$ Mode = 7.88 $$

Therefore, mode of this data is 7.88

Question 7

The distribution below gives the weights of 30 students of a class. Find the median weight of the students

Weight (in kg)	Number of students
40 − 45	2
45 − 50	3
50 − 55	8
55 − 60	6
60 − 65	6
65 − 70	3
70 − 75	2

Solution :

The cumulative frequency for the given data is calculated as follows.

Weight (in kg)	Number of students	Cumulative frequency (cf)
40 − 45	2	2
45 − 50	3	2 + 3 = 5
50 − 55	8	5 + 8 = 13
55 − 60	6	13 + 6 = 19
60 − 65	6	19 + 6 = 25
65 − 70	3	25 + 3 = 28
70 − 75	2	28 + 2 = 30
Total	N = 30

MEDIAN : Now Median can be calculated as follows

From the above table, we get

N (Sum of frequencies) = 30

Cumulative frequency just greater than N/2 = (i.e, 30/2 = 15 ) is 19

The class corresponding to this Cumulative frequency is 55 − 60.

So, the Median class is 55 − 60

L = Lower limit of median class = 55

f = Frequency of the median class = 6

cf = Cumulative frequency of the class preceding the median class = 13

h = Class interval of the median class = 5

$$ Median = L + {({ { N \over 2} - cf }) \over { f} }× h $$

$$ = 55 + {({ { 30 \over 2} - 13 }) \over { 6} }×(5 )$$

$$ = 55 + {({ { 15} - 13}) \over { 6} }×(5 )$$

$$ = 55 + {({ { 2} ×5}) \over {6} } $$

$$ = 55 + 1.67 $$

$$ Median = 56.697 $$

Therefore, the median weight is 56.67 kg.