Statistics ⇒⇒ Exercise 13.1

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?

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

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

Number of plants	Number of houses Frequency fi	Classmark xi	(fi xi )
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
	$\Sigma f_i = 20 $		$\Sigma f_ix_i = 162 $

Here , direct method has been used as the values of class marks x_i and f_i are small.

Mean can be calculated as follows:

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

$$ \bar{X} ={ {162} \over {20} }$$

Hence, Mean number of plants in each house = 8.1

Question 2

Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)	500 − 520	520 - 540	540 - 560	560 - 580	580 - 600
Number of Workers	12	14	8	6	10

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

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 = 550

Daily wages	Number of Workers fi	Class mark xi	di = (xi-a)	(fi di )
500 - 520	12	510	510 - 550 = -40	-480
520 - 540	14	530	530 - 550 = -20	-280
540 - 560	8	550	550 - 550 = 0	0
560 - 580	6	570	570 - 550 = 20	120
580 - 600	10	590	590 - 550 = 40	400
	$\Sigma f_i = 50 $			$\Sigma f_id_i = -240 $

Mean can be calculated as follows:

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

$$ \bar{X} = 550 + { {-240} \over {50} }$$

$$ \bar{X} = 550 + {( - 4.8)}$$

$$ \bar{X} = 545.20 $$

Therefore, the mean daily wage of the workers of the factory is Rs 545.20

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

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

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

Daily pocket allowance	Number of children fi	Class mark xi	pocket allowance (fi xi )
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
	$\Sigma f_i = 44 + f $		$\Sigma f_ix_i = 752 + 20f $

Mean can be calculated as follows:

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

$$ 18 = { {752 + 20f} \over {44 + f} }$$

$$ 18(44 + f) = 752 + 20f $$

$$ 792 + 18f = 752 + 20f$$

$$ 792 - 752 = 20f - 18f$$

$$ 2f = 40 $$

$$ f = 20 $$

Therefore, the missing frequency is 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 minute	65 − 68	68­ − 71	71 − 74	74 − 77	77 − 80	80 − 83	83 − 86
Number of women	2	4	3	8	7	4	2

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 = 75.5

Number of heart beats per minute	Number of women fi	Class mark xi	di = (xi-a)	(fi di )
65 − 68	2	66.5	66.5 - 75.5 = - 9	-18
68­ − 71	4	69.5	69.5 - 75.5 = - 6	-24
71 − 74	3	72.5	72.5 - 75.5 = -3	- 9
74 − 77	8	75.5	75.5 - 75.5 = 0	0
77 − 80	7	78.5	78.5 - 75.5 = 3	21
80 − 83	4	81.5	81.5 - 75.5 = 6	24
83 - 86	2	84.5	84.5 - 75.5 = 9	18
	$\Sigma f_i = 30 $			$\Sigma f_id_i = 12 $

Mean can be calculated as follows:

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

$$ \bar{X} = 75.5 + { {12} \over {30} }$$

$$ \bar{X} = 75.5 + {( 0.4)}$$

$$ \bar{X} = 75.9 $$

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

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	59 − 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?

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.

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 Step deviation method.

Let the assumed mean be a = 57 and Class size (h) of this data = 3

Nos. of mango	Nos. of box fi	Class mark xi	di = ( xi-a)	ui = $ d_i \over h $	(fi ui )
49.5 − 52.5	15	51	51 - 57 = -6	-2	-30
52.5 − 55.5	110	54	54 - 57 = -3	-1	-110
55.5 − 58.5	135	57	57 - 57 = 0	0	0
58.5 − 61.5	115	60	60 - 57 = 3	1	115
61.5 − 64.5	25	63	63 - 57 = 6	2	50
	$\Sigma f_i = 400 $				$\Sigma f_iu_i = 25 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 57 + ({ {25} \over {400} } )×3$$

$$ \bar{X} = 57 + ({ {3} \over {16}})$$

$$ \bar{X} = 57 + (0.1875)$$

$$ \bar{X} = 57.1875 $$

Therefore, the mean number of mangoes kept in a packing box is approx 57.19

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.

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 Step deviation method.

Let the assumed mean be a = 225 and Class size (h) of this data = 50

Daily exp.	Number of house holds fi	Class mark xi	di = (xi-a)	ui = $ d_i \over h $	(fi ui )
100 − 150	4	125	125 - 225 = -100	-2	-8
150 − 200	5	175	175 - 225 = -50	-1	-5
200 − 250	12	225	225 - 225 = 0	0	0
250 − 300	2	275	275 - 225 = 25	1	2
300 − 350	2	325	325- 225 = 100	2	4
	$\Sigma f_i = 25 $				$\Sigma f_iu_i = -7 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 225 + ({ {-7} \over {25} } )×50$$

$$ \bar{X} = 225 + ({ {-7} } )×(2)$$

$$ \bar{X} = 225 - 14 $$

$$ \bar{X} = 211 $$

Therefore, mean daily expenditure on food is Rs 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)	0.00 − 0.04	0.04 - 0.08	0.08 - 0.12	0.12 - 0.16	0.16 - 0.20	0.20 - 0.24
Frequency	4	9	9	2	4	2

Find the mean concentration of SO₂ in the air.

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 Step deviation method.

Let the assumed mean be a = 0.10 and Class size (h) of this data = 0.04

Conc. of SO2	Frequency fi	Class mark xi	di = (xi-a)	ui = $ d_i \over h $	(fi ui )
0.00 − 0.04	4	0.02	0.02 - 0.10 = - 0.08	-2	-8
0.04 - 0.08	9	0.06	0.06 - 0.10 = - 0.04	-1	-9
0.08 - 0.12	9	0.10	0.10 - 0.10 = 0	0	0
0.12 - 0.16	2	0.14	0.14 - 0.10 = 0.04	1	2
0.16 - 0.20	4	0.18	0.18 - 0.10 = 0.08	2	8
0.20 - 0.24	2	0.22	0.22 - 0.10 = 0.12	3	6
	$\Sigma f_i = 30 $				$\Sigma f_iu_i = -1 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 0.10 + ({ {-1} \over {30} } )×0.04$$

$$ \bar{X} = 0.10 + ( - 0.3333333 )×0.04$$

$$ \bar{X} = 0.10 - 0.0013333 $$

$$ \bar{X} = 0.09866 $$

$$ \bar{X} = 0.099 $$

Therefore, the mean concentration of SO₂ in the air is 0.099 ppm.

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

Solution :

The following relation is used to find the classmarks x_i

$$ Class marks (x_i) = $$

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

Nos. of days	Number of students fi	Class mark xi	(fi xi )
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
	$\Sigma f_i = 40 $		$\Sigma f_ix_i = 499 $

Mean can be calculated as follows:

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

$$ \bar{X} = { {499} \over {40} }$$

$$ \bar{X} = 12.475 $$

$$ \bar{X} = 12.48 $$

Therefore, the mean number of days a student was absent is 12.48 days.

Question 9

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

Literacy rate	45 − 55	55 − 65	65 − 75	75 − 85	85 − 95
Number of cities	3	10	11	8	3

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 Step deviation method.

Let the assumed mean be a = 70 and Class size (h) of this data = 10

Literacy rate	Number of cities fi	Class mark xi	di = (xi-a)	ui = $ d_i \over h $	(fi ui )
45 − 55	3	50	50 - 70 = -20	-2	-6
55 − 65	10	60	60 - 70 = -10	-1	-10
65 − 75	11	70	70 - 70 = 0	0	0
75 − 85	8	80	80 - 70 = 10	1	8
85 − 95	3	90	90- 70 = 20	2	6
	$\Sigma f_i = 35 $				$\Sigma f_iu_i = -2 $

Mean can be calculated as follows:

Mean $$ \bar{X} = a + { {\Sigma f_iu_i} \over {\Sigma f_i} } ×h $$

$$ \bar{X} = 70 + [{ {-2} \over {35} } ]×(10) $$

$$ \bar{X} = 70 + [{ {-20} \over {35} } ] $$

$$ \bar{X} = 70 + [{ {-4} \over {7} } ] $$

$$ \bar{X} = 70 -0.57 $$

$$ \bar{X} = 69.43 $$

Therefore, the mean literacy rate is 69.43% .