Exercise 13.1 Statistics Math's Class 10

Q1. 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

Answer:

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

Since the values of $x_{i}$ and $f_{i}$ are small, the direct method is convenient here.

Class marks: $x_{i} = 1, 3, 5, 7, 9, 11, 13$

$x_{i}$	$f_{i}$	$f_{i} x_{i}$
1	1	1
3	2	6
5	1	5
7	5	35
9	6	54
11	2	22
13	3	39
Total	20	162

$\overset{ˉ}{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}} = \frac{162}{20} = 8.1$

Mean number of plants per house = 8.1. The direct method was used because the numerical values are small and easy to multiply.

Q2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹)	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.

Answer:

Mean daily wages of 50 workers

Daily wages (in ₹)	500–520	520–540	540–560	560–580	580–600
Number of workers	12	14	8	6	10

Class sizes are equal and values are large, so the step-deviation method is suitable. Take $a = 550$ and $h = 20$ .

Class	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 550$	$u_{i} = \frac{x_{i} - 550}{20}$	$f_{i} u_{i}$
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

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 550 + (\frac{- 12}{50}) \times 20$

$\overset{ˉ}{x} = 550 - 4.8 = 545.2$

Mean daily wage = ₹ 545.20.

Q3. 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 ₹)	11–13	13–15	15–17	17–19	19–21	21–23	23–25
Number of children	7	6	9	13	$f$ f	5	4

Answer:

Finding the missing frequency f

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

Given: mean = ₹ 18.

Class marks: $12, 14, 16, 18, 20, 22, 24$

$\sum f_{i} = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$

$\sum f_{i} x_{i} = 7 (12) + 6 (14) + 9 (16) + 13 (18) + f (20) + 5 (22) + 4 (24)$

$\sum f_{i} x_{i} = 84 + 84 + 144 + 234 + 20 f + 110 + 96 = 752 + 20 f$

Using the mean formula:

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

$752 + 20 f = 18 (44 + f) = 792 + 18 f$

$20 f - 18 f = 792 - 752$

$2 f = 40$

$f = 20$

The missing frequency is f = 20.

Q4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Answer: Mean heartbeats per minute of 30 women

Number of heartbeats 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

The class size is $h = 3$ and the values are moderate, so the step-deviation method is suitable. Take the assumed mean $a = 75.5$ (class mark of 74–77).

Class	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 75.5$	$u_{i} = \frac{x_{i} - 75.5}{3}$	$f_{i} u_{i}$
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

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 75.5 + (\frac{4}{30}) \times 3$

$\overset{ˉ}{x} = 75.5 + 0.4 = 75.9$

Mean heartbeats per minute = 75.9.

Q5. 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

Answer:

Mean number of mangoes per packing box

Number of mangoes	50–52	53–55	56–58	59–61	62–64
Number of boxes	15	110	135	115	25

The classes have a gap of 1 between the upper limit of one class and the lower limit of the next, so we first convert them into continuous classes by adjusting the limits by 0.5: 49.5–52.5, 52.5–55.5, 55.5–58.5, 58.5–61.5, 61.5–64.5.

The class size is $h = 3$ . Take the assumed mean $a = 57$ .

Class (adjusted)	$f_{i}$	$x_{i}$	$u_{i} = \frac{x_{i} - 57}{3}$	$f_{i} u_{i}$
49.5–52.5	15	51	−2	−30
52.5–55.5	110	54	−1	−110
55.5–58.5	135	57	0	0
58.5–61.5	115	60	1	115
61.5–64.5	25	63	2	50
Total	400			25

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 57 + (\frac{25}{400}) \times 3$

$\overset{ˉ}{x} = 57 + 0.1875 = 57.1875$

Mean number of mangoes per box ≈ 57.19. The step-deviation method was chosen because the frequencies are large and the class size is common, which keeps the arithmetic simple.

Q6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in ₹)	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: Mean daily expenditure on food of 25 households

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

The class size is $h = 50$ . Take the assumed mean $a = 225$ .

Class	$f_{i}$	$x_{i}$	$u_{i} = \frac{x_{i} - 225}{50}$	$f_{i} u_{i}$
100–150	4	125	−2	−8
150–200	5	175	−1	−5
200–250	12	225	0	0
250–300	2	275	1	2
300–350	2	325	2	4
Total	25			−7

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 225 + (\frac{- 7}{25}) \times 50$

$\overset{ˉ}{x} = 225 - 14 = 211$

Mean daily expenditure on food = ₹ 211.

Q7. 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 (number of localities)	4	9	9	2	4	2

Answer:

Mean Concentration of SO₂ in the Air

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 (number of localities)	4	9	9	2	4	2

The class size is uniform ( $h = 0.04$ ), so the step-deviation method is convenient. Take the assumed mean $a = 0.10$ (class mark of 0.08–0.12).

Class	$f_{i}$	$x_{i}$	$u_{i} = \frac{x_{i} - 0.10}{0.04}$	$f_{i} u_{i}$
0.00–0.04	4	0.02	−2	−8
0.04–0.08	9	0.06	−1	−9
0.08–0.12	9	0.10	0	0
0.12–0.16	2	0.14	1	2
0.16–0.20	4	0.18	2	8
0.20–0.24	2	0.22	3	6
Total	30			−1

Applying the step-deviation formula:

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 0.10 + (\frac{- 1}{30}) \times 0.04$

$\overset{ˉ}{x} = 0.10 - \frac{0.04}{30}$

$\overset{ˉ}{x} = 0.10 - 0.00133 \dots \approx 0.0987$

Mean concentration of SO₂ in the air ≈ 0.0987 ppm (approximately 0.099 ppm).

Q8. 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:

Mean number of days students were 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

The class sizes are unequal, so the step-deviation method with a common $h$ is not convenient. The assumed mean method is suitable. Take $a = 17$ (class mark of 14–20).

Class	$f_{i}$	$x_{i}$	$d_{i} = x_{i} - 17$	$f_{i} d_{i}$
0–6	11	3	−14	−154
6–10	10	8	−9	−90
10–14	7	12	−5	−35
14–20	4	17	0	0
20–28	4	24	7	28
28–38	3	33	16	48
38–40	1	39	22	22
Total	40			−181

$\overset{ˉ}{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$

$\overset{ˉ}{x} = 17 + \frac{- 181}{40}$

$\overset{ˉ}{x} = 17 - 4.525 = 12.475$

Mean number of days absent ≈ 12.48 days.

Q9. 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	85–95
Number of cities	3	10	11	8	3

Answer:

Mean literacy rate of 35 cities

Literacy rate (in %)	45–55	55–65	65–75	75–85	85–95
Number of cities	3	10	11	8	3

The class size is uniform ( $h = 10$ h=10), so the step-deviation method is suitable. Take $a = 70$ a=70.

Class	$f_{i}$	$x_{i}$	$u_{i} = \frac{x_{i} - 70}{10}$	$f_{i} u_{i}$
45–55	3	50	−2	−6
55–65	10	60	−1	−10
65–75	11	70	0	0
75–85	8	80	1	8
85–95	3	90	2	6
Total	35			−2

$\overset{ˉ}{x} = a + (\frac{\sum f_{i} u_{i}}{\sum f_{i}}) \times h$

$\overset{ˉ}{x} = 70 + (\frac{- 2}{35}) \times 10$

$\overset{ˉ}{x} = 70 - \frac{20}{35} = 70 - \frac{4}{7}$

$\overset{ˉ}{x} \approx 69.43$

Mean literacy rate ≈ 69.43%.

Exercise 13.1 Statistics Math’s Class 10

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