Statistics

NCERT Exercise 14.1

Question 1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Answer:

Class Interval	fi	xi	fixi
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	Σ f_i = 20		Σ f_ix_i = 162

Mean can be calculated as follows:

`x=(Σf_ix_1)/(Σf_1)=(162)/(20)=8.1`

In this case, the values of fi and xi are small hence direct method has been used.

Question 2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer: In this case, value of xi is quite large and hence we should select the assumed mean method.

Let us take assumed mean a = 150

Class Interval	fi	xi	di = xi - a	fidi
100-120	12	110	-40	-480
120-140	14	130	-20	-280
140-160	8	150	0	0
160-180	6	170	20	120
180-200	10	190	40	400
	Σ f_i = 50			Σ f_id_i = -240

Now, mean of deviations can be calculated as follows:

`d=(Σf_i\d_i)/(Σf_i)=(-240)/(50)=-4.8`

Mean can be calculated as follows:

`x = d + a = -4.8 + 150 = 145.20`

Question 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Daily pocket allowance (in Rs)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

Answer:

Class Interval	fi	xi	fixi
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
	Σ f_i = 44 + f		Σ f_ix_i = 752 + 20f

We have;

`x=(Σf_i\x_i)/(Σf_i)=18`

Or, `18=(752+20f)/(44+f)`

Or, `18(44+f)=752+20f`
Or, `2f=40`
Or, `f=20`

Question 4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per min	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of women	2	4	3	8	7	4	2

Answer:

Class Interval	fi	xi	di = xi - a	fidi
65-68	2	66.5	-9	-18
68-71	4	69.5	-6	-24
71-74	3	72.5	-3	-9
74-77	8	75.5	0	0
77-80	7	78.5	3	21
80-83	4	81.5	6	24
83-86	2	84.5	9	18
	Σ f_i = 30			Σ f_id_i = 12

Now, mean can be calculated as follows:

`d=(Σf_i\d_i)/(Σf_1)=12/30=0.4`

`x=d+a=0.4+75.5=75.9`

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50-52	53-55	56-58	89-61	62-64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer:

Class Interval	fi	xi	di = x - a	fidi
50-52	15	51	-6	90
53-55	110	54	-3	-330
56-58	135	57	0	0
59-61	115	60	3	345
62-64	25	63	6	150
	Σ f_i = 400			Σ f_id_i = 75

Mean can be calculated as follows:

`d=(Σf_i\d_i)/(Σf_1)=75/400=0.1875`

`x=d+a=0.1875+57=57.1875`

In this case, there are wide variations in fi and hence assumed mean method is used.

Question 6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)	100-150	150-200	200-250	250-300	300-350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

Answer:

Class Interval	fi	xi	di = xi - a	ui = di/h	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
	Σ f_i = 25				Σ f_iu_i = -7

Mean can be calculated as follows:

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=225+(-7)/(25)xx50=211`

Question 7. To find out the concentration of SO₂ in the air (in parts per million, i.e. ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO₂ (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

Find the mean concentration of SO₂ in the air.

Answer:

Class Interval	fi	xi	fixi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
	Σ f_i = 30		Σ f_ix_i = 2.96

Mean can be calculated as follows:

`x=(Σf_i\x_i)/(Σf_i)=(2.96)/(30)=0.099 pp\m`

Question 8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of students	11	10	7	4	4	3	1

Answer:

Class Interval	fi	xi	fixi
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
	Σ f_i = 40		Σ f_ix_i = 499

Mean can be calculated as follows:

`x=(Σf_i\x_i)/(Σf_i)=(499)/(40)=12.4` (approx)

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-98
Number of cities	3	10	11	8	3

Answer:

Class Interval	fi	xi	di = xi - a	ui = di/h	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
	Σ f_i = 35				Σ f_iu_i = -2

Mean can be calculated as follows:

`x=a+(Σf_i\u_i)/(Σf_i)xxh`

`=70+(-2)/(35)xx10=69.42`