NCERT Class 10 Maths Solutions Chapter - 14 Statistics, Ex - 14.3

Ex - 14.3

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)	Number of consumers
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

Solution

We may find class marks by using the relation 

 
Taking 135 as assumed mean (a) we may find d_i, u_i, f_iu_i, according to step deviation method as following 

Monthly consumption  (in units)	Number of consumers (f i)	xi class mark	di = xi - 135		fiui
65 - 85	4	75	- 60	- 3	- 12
85 - 105	5	95	- 40	- 2	- 10
105 - 125	13	115	- 20	- 1	- 13
125 - 145	20	135	0	0	0
145 - 165	14	155	20	1	14
165 - 185	8	175	40	2	16
185 - 205	4	195	60	3	12
Total	68				7

From the table we may observe that

Now from table it is clear that maximum class frequency is 20 belonging to class interval 125 - 145.
Modal class = 125 - 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14 

 
We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.

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

We may find cumulative frequency for the given data as following 

Class interval	Frequency	Cumulative frequency
0 - 10	5	5
10 - 20	x	5 + x
20 - 30	20	25 + x
30 - 40	15	40 + x
40 - 5	y	40 + x + y
50 - 60	5	45 + x + y
Total (n)	60

It is clear that n = 60 

45 + x + y = 60 

x + y = 15        (1)
Median of data is given as 28.5 which lies in interval 20 - 30.
So, median class = 20 - 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10

From equation (1)
    8 + y = 15
    y = 7
Hence values of x and y are 8 and 7 respectively.

Question 3.  

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

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

Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below 

Age (in years)	Number of policy holders (fi)	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
Total (n)

Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than  is 78 belonging to interval 35 - 40
So, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45

So, median age is 35.76 years.

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 fi
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.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5 - 171.5 - 180.5)

Solution

The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract 

  to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below 

Length (in mm)	Number or leaves fi	Cumulative frequency
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

From the table we may observe that cumulative frequency just greater then 
  is 29, belonging to class interval 144.5 - 153.5.

Median class = 144.5 - 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17


 
So, median length of leaves is 146.75 mm.

Question 5.  

Find the following tables 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

We can find cumulative frequencies with their respective class intervals as below - 

Life time	Number of lamps (fi)	Cumulative frequency
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

Now we may observe that cumulative frequency just greater than  is 216 belonging to class interval 3000 - 3500.
Median class = 3000 - 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500 

 
So, median life time of lamps is 3406.98 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	1 - 4	4 - 7	7 - 10	10 - 13	13 - 16	16 - 19
Number of surnames	6	30	40	6	4	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

We can find cumulative frequencies with their respective class intervals as below 

Number of letters	Frequency (fi)	Cumulative frequency
1 - 4	6	6
4 - 7	30	30 + 6 = 36
7 - 10	40	36 + 40 = 76
10 - 13	16	76 + 16 = 92
13 - 16	4	92 + 4 = 96
16 - 19	4	96 + 4 = 100
Total (n)	100

Now we may observe that cumulative frequency just greater than  is 76 belonging to class interval 7 - 10.
Median class = 7 - 10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3

Now we can find class marks of given class intervals by using relation 
 

Taking 11.5 as assumed mean (a) we can find d_i, u_i and f_iu_i according to step deviation method as below. 

Number of letters	Number of surnames	xi	xi - a		fiui
1 - 4	6	2.5	-9	-3	-18
4 - 7	30	5.5	-6	-2	-60
7 - 10	40	8.5	-3	-1	-40
10 - 13	16	11.5	0	0	0
13 - 16	4	14.5	3	1	4
16 - 19	4	17.5	6	2	8
Total	100				-106

f_iu_i = - 106
f_i = 100 
 

We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.

Question 7.  

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

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

Solution

We may find cumulative frequencies with their respective class intervals as below 


Cumulative frequency just greater than   is 19, belonging to class interval 55 - 60.
Median class = 55 - 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5 

 

So, median weight is 56.67 kg.

Ex-14.2

Ex-14.4