Class 10 NCERT Solutions Maths Chapter 13 - Statistics

Let us find class marks (x_i) for each interval by using the relation.

Now we may compute x_i and f_ix_ias following

From the table, we may observe that
 

So, the mean number of plants per house is 8.1.
We have used here direct method as values of class marks (x_i) and f_i are small.

Let us find class mark for each interval by using the relation.


Class size (h) of this data = 20

Now taking 550 as assured mean (a) we may calculate di, u_i and f_iu_ias following.

Daily wages (in Rs)	Number of workers (fi)	xi	di = xi - 550		fiui
500 - 520	12	510	- 40	-2	- 24
520 - 540	14	530	- 20	-1	- 14
540 - 560	8	550	0	0	0
560 - 580	6	570	20	1	6
580 - 600	10	590	40	2	20
Total	50				-12

From the table we may observe that

So mean daily wages of the workers of the factory is Rs. 545.2.

We may find class mark (x_i) for each interval by using the relation.


Given that mean pocket allowance = Rs.18
Now taking 18 as assured mean (a) we may calculate d_i and f_id_ias following.

Daily pocket allowance (in Rs.)	Number of children fi	Class mark xi	di = xi - 18	fidi
11 - 13	7	12	- 6	- 42
13 - 15	6	14	- 4	- 24
15 - 17	9	16	- 2	- 18
17 - 19	13	18	0	0
19 - 21	f	20	2	2 f
21 - 23	5	22	4	20
23 - 25	4	24	6	24
Total				2f - 40

From the table we may obtain

Hence the missing frequency f is 20.

We may find class mark of each interval (x_i) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following.

Number of heart beats per minute	Number of women fi	xi	di = xi -75.5		fiui
65 - 68	2	66.5	- 9	- 3	- 6
68 - 71	4	69.5	- 6	- 2	- 8
71 - 74	3	72.5	- 3	- 1	- 3
74 - 77	8	75.5	0	0	0
77 - 80	7	78.5	3	1	7
80 - 83	4	81.5	6	2	8
83 - 86	2	84.5	9	3	6
Total	30				4

Now we may observe from table that

So mean heart beats per minute for these women are 75.9 beats per minute.

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add to upper class limit and subtract from lower class limit of each interval.
And class mark (x_i) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following -

Class interval	fi	xi	di = xi - 57		fiui
49.5 - 52.5	15	51	-6	-2	-30
52.5 - 55.5	110	54	-3	-1	-110
55.5 - 58.5	135	57	0	0	0
58.5 - 61.5	115	60	3	1	115
61.5 - 64.5	25	63	6	2	50
Total	400				25

Now we may observe that

 
Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of f_i, d_i are big and also there is a common multiple between all d_i.

We may calculate class mark (x_i) for each interval by using the relation 

Class size = 50

Now taking 225 as assumed mean (a) we may calculate d_i, u_i, f_iu_i as following

Daily expenditure (in Rs)	fi	xi	di = xi - 225		fiui
100 - 150	4	125	-100	-2	-8
150 - 200	5	175	-50	-1	-5
200 - 250	12	225	0	0	0
250 - 300	2	275	50	1	2
300 - 350	2	325	100	2	4
Total	25				-7

Now we may observe that -

We may find class mark of each interval by using the relation

 

Now taking 16 as assumed mean (a) we may calculate di and fidi as following 

Now we may observe that

 
So, mean number of days is 12.38 days, for which a student was absent.

We may find class marks by using the relation

 

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate d_i, u_i, and f_iu_i as following

Literacy rate (in %)	Number of cities fi	xi	di= xi - 70		fiui
45 - 55	3	50	-20	-2	-6
55 - 65	10	60	-10	-1	-10
65 - 75	11	70	0	0	0
75 - 85	8	80	10	1	8
85 - 95	3	90	20	2	6
Total	35				-2

Now we may observe that

So, the mean literacy rate is 69.43%.

We may compute class marks (x_i) as per the relation

 
Now taking 30 as assumed mean (a) we may calculate d_i and f_id_i as following.

From the table we may observe that


Clearly, mean of this data is 35.38. It represents that on an average the age of a patient admitted to hospital was 35.38 years.
As we may observe that maximum class frequency is 23 belonging to class interval 35 - 45.
So, modal class = 35 -  45
Lower limit (l) of modal class = 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class = 14


Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.
So, modal class = 60 - 80
Lower class limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class = 38
Class size (h) = 20


 
So, modal lifetime of electrical components is 65.625 hours.

We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 intervals.
So, modal class = 1500 - 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding modal class = 24
Frequency (f₂) of class succeeding modal class = 33
Class size (h) = 500

 

Now we may find classmark as


Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate d_i, u_i and f_iu_i as following

Expenditure (in Rs)	Number of families fi	xi	di = xi - 2750		fiui
1000 - 1500	24	1250	-1500	-3	-72
1500 - 2000	40	1750	-1000	-2	-80
2000 - 2500	33	2250	-500	-1	-33
2500 - 3000	28	2750	0	0	0
3000 - 3500	30	3250	500	1	30
3500 - 4000	22	3750	1000	2	44
4000 - 4500	16	4250	1500	3	48
4500 - 5000	7	4750	2000	4	28
Total	200				-35

Now from the table, we may observe that


So, mean monthly expenditure was Rs.2662.50.

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 - 35.
So, modal class = 30 - 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding modal class = 9
Frequency (f₂) of class succeeding modal class = 3


It represents that most of states/U.T have a teacher  student ratio as 30.6

Now we may find class marks by using the relation


Now taking 32.5 as assumed mean (a) we may calculate d_i, u_i and f_iu_i as following.

So mean of data is 29.2
It represents that an average teacher-student ratio was 29.2.

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 - 5000.
So, modal class = 4000 - 5000
Lower limit (l) of modal class = 4000
Frequency (f₁) of modal class = 18
Frequency (f₀) of class preceding modal class = 4
Frequency (f₂) of class succeeding modal class = 9
Class size (h) = 1000

So mode of given data is 4608.7 runs.

From the given data we may observe that maximum class frequency is 20 belonging to 40 - 50 class intervals.
So, modal class = 40 - 50
Lower limit (l) of modal class = 40
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 12
Frequency (f₂) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.

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.

We may find cumulative frequency for the given data as following

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.

Here the class width is not the 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

Now from the 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, the median age is 35.76 years.

The given data is not having continuous class intervals. We can observe that the difference between the 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 

From the table, we may observe that cumulative frequency just greater than
  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, the median length of leaves is 146.75 mm.

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

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, the median lifetime of lamps is 3406.98 hours.

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

Now we may observe that cumulative frequency just greater than is 76 belonging to the 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.

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

                         Total = 30

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 class preceeding the median class = 13
Class size (h) = 5



So, the median weight is 56.67 kg.