NCERT Solutions for Class 10 Maths Chapter 14 : Statistics

Q: Question 1. State the empirical relationship between the Measures of Central Tendency.

Answer. The empirical relationship between the three measures of central tendency is as follows: 3 Median = Mode + 2 Mean

Q: Question 2. What do you understand by Mean?

Answer. Mean is a measure of the central tendency of a distribution. It refers to the most common or the average value of the given data. There are two types of mean that can be calculated. One is the Arithmetic Mean and the other one is the Geometric Mean. The three methods to calculate the Arithmetic Mean are the direct method, the step-deviation method and the assumed mean method.

Q: Question 3. What do you understand by Median?

Answer. Median refers to the middle number or value of the given set of data when the data is arranged in an ascending or descending order. When the data set contains odd numbers, the median will be the middle value or number that has half of the numbers above it and half of the number below it. However, in the case of even numbers, the median is calculated by adding the two middle numbers and dividing their sum by two. It is also a measure of central tendency.

Q: Question 4. What do you understand by Mode?

Answer. Mode is also a measure of central tendency similar to the Mean and Median. Mode refers to the number or value that is most frequent in a given set of data. In other words, Mode is that value in a given data which occurs the most number of times. Thus, a distribution of data may have only one mode, more than one mode or no mode at all.

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?

Medium

Solution

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Answer

No. of plants	No. of houses $(f_{i})$	Mid point $(X_{i})$	$f_{i} x_{i}$
$0 - 2$	$1$	$1$	$1$
$2 - 4$	$2$	$3$	$6$
$4 - 6$	$1$	$5$	$5$
$6 - 8$	$5$	$7$	$35$
$8 - 10$	$6$	$9$	$54$
$10 - 12$	$2$	$11$	$22$
$12 - 14$	$3$	$13$	$39$
Total	$20$		$162$

Calculating mean, we get

Mean,

\overset{x}{ˉ} = \frac{1}{n} \sum f_{i} x_{i}

Here,

n = 20, \sum f_{i} x_{i} = 162

Therefore, Mean,

\overset{x}{ˉ} = \frac{1 6 2}{2 0} = 8.1

plants

We have used direct method because numerical values of

f

and

x

are small.

Answer

No. of plants	No. of houses $(f_{i})$	Mid point $(X_{i})$	$f_{i} x_{i}$
$0 - 2$	$1$	$1$	$1$
$2 - 4$	$2$	$3$	$6$
$4 - 6$	$1$	$5$	$5$
$6 - 8$	$5$	$7$	$35$
$8 - 10$	$6$	$9$	$54$
$10 - 12$	$2$	$11$	$22$
$12 - 14$	$3$	$13$	$39$
Total	$20$		$162$

Calculating mean, we get

Mean,

\overset{x}{ˉ} = \frac{1}{n} \sum f_{i} x_{i}

Here,

n = 20, \sum f_{i} x_{i} = 162

Therefore, Mean,

\overset{x}{ˉ} = \frac{1 6 2}{2 0} = 8.1

plants

We have used direct method because numerical values of

f

and

x

are small.

Question 2

Consider the following distribution of daily wages of 50 workers of a factory:
Find the mean daily wages of the workers of the factory by using appropriate method.

Daily wages(in rs)	Number of workers
500-200	12
520-540	14
540-560	8
560-580	6
580-600	10

Medium

Solution

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Answer

Daily wages	$(f i)$	$(x i)$	$(f i x i)$
$600 - 20$	$12$	$510$	$6120$
$520 - 540$	$14$	$630$	$7420$
$540 - 560$	$8$	$550$	$4400$
$560 - 580$	$6$	$570$	$3420$
$580 - 600$	$10$	$590$	$$5900
	$\sum f i = 50$		$\sum f i x i \Rightarrow 27260$

mean

\frac{\sum f i x i}{\sum f i} = \frac{2 7 2 6 0}{5 0} = \frac{2 7 2 6}{5} = 545.2

∴

meam\n wage of

50

employees is

545.2 r s

Answer

Daily wages	$(f i)$	$(x i)$	$(f i x i)$
$600 - 20$	$12$	$510$	$6120$
$520 - 540$	$14$	$630$	$7420$
$540 - 560$	$8$	$550$	$4400$
$560 - 580$	$6$	$570$	$3420$
$580 - 600$	$10$	$590$	$$5900
	$\sum f i = 50$		$\sum f i x i \Rightarrow 27260$

mean

\frac{\sum f i x i}{\sum f i} = \frac{2 7 2 6 0}{5 0} = \frac{2 7 2 6}{5} = 545.2

∴

meam\n wage of

50

employees is

545.2 r s

Question 3

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is

R s . 18

. Find the missing frequency

f

.

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

Medium

Solution

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Answer

Lets take assumed mean,

A

as

16

Class interval

= 2

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 1 6}{2}

Now, we have

A = 16, \overset{x}{ˉ} = 18

and

h = 2

We know that,

Mean,

\overset{x}{ˉ} = A + h (\frac{1}{N} \sum f_{i} u_{i})

\Rightarrow 18 = 16 + 2 (\frac{2 f + 2 4}{f + 4 4})

\Rightarrow \frac{2 f + 2 4}{f + 4 4} = 1

\Rightarrow f + 44 = 2 f + 24

\Rightarrow f = 20

Hence, the value of missing frequency,

f = 20

Answer

Lets take assumed mean,

A

as

16

Class interval

= 2

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 1 6}{2}

Now, we have

A = 16, \overset{x}{ˉ} = 18

and

h = 2

We know that,

Mean,

\overset{x}{ˉ} = A + h (\frac{1}{N} \sum f_{i} u_{i})

\Rightarrow 18 = 16 + 2 (\frac{2 f + 2 4}{f + 4 4})

\Rightarrow \frac{2 f + 2 4}{f + 4 4} = 1

\Rightarrow f + 44 = 2 f + 24

\Rightarrow f = 20

Hence, the value of missing frequency,

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 summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute	$65 - 68$	$68 - 71$	$71 - 74$	$74 - 77$	$77 - 8$ 0	$80 - 83$	$83 - 86$
Number of women	$2$	$4$	$3$	$8$	$7$	$4$	$2$

Medium

Solution

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Answer

Class interval	Mid value $(x_{i})$	$(f_{i})$	$u_{i} = \frac{x _{i} - 7 5 . 5}{3}$	$f_{i} u_{i}$
$65 - 68$	$66.5$	$2$	$- 3$	$- 6$
$68 - 71$	$69.5$	$4$	$- 2$	$- 8$
$71 - 74$	$72.5$	$3$	$- 1$	$- 3$
$74 - 77$	$75.5$	$8$	$0$	$0$
$77 - 80$	$78.5$	$7$	$1$	$7$
$80 - 83$	$81.5$	$4$	$2$	$8$
$83 - 86$	$84.5$	$2$	$3$	$6$
Total		$30$		$4$

Lets take assumed mean,

A

as

75.5

Class interval

= 3

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 7 5 . 5}{3}

\overset{x}{ˉ} = A + h \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 75.5 + 3 \times \frac{4}{3 0} = 75.5 + 0.4 = 75.9

Hence, the mean heart beats per minute for these women is

75.9

Answer

Class interval	Mid value $(x_{i})$	$(f_{i})$	$u_{i} = \frac{x _{i} - 7 5 . 5}{3}$	$f_{i} u_{i}$
$65 - 68$	$66.5$	$2$	$- 3$	$- 6$
$68 - 71$	$69.5$	$4$	$- 2$	$- 8$
$71 - 74$	$72.5$	$3$	$- 1$	$- 3$
$74 - 77$	$75.5$	$8$	$0$	$0$
$77 - 80$	$78.5$	$7$	$1$	$7$
$80 - 83$	$81.5$	$4$	$2$	$8$
$83 - 86$	$84.5$	$2$	$3$	$6$
Total		$30$		$4$

Lets take assumed mean,

A

as

75.5

Class interval

= 3

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 7 5 . 5}{3}

\overset{x}{ˉ} = A + h \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 75.5 + 3 \times \frac{4}{3 0} = 75.5 + 0.4 = 75.9

Hence, the mean heart beats per minute for these women is

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	$50 - 52$	$53 - 55$	$56 - 58$	$59 - 61$	$62 - 64$
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?

Medium

Solution

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Answer

Number of mangoes	Number of boxes	$x_{i}$	$f_{i} x_{i}$
50-52	15	51	765
53-55	110	54	5940
56-58	135	57	7695
59-61	115	60	6900
62-64	25	63	1575
	$Σ f_{i} = 400$		$Σ f_{i} x_{i} = 22875$

\overset{x}{ˉ} = \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{2 2 8 7 5}{4 0 0} = 57.1875

Hence, the mean number of mangoes kept in a packing box is

57.1875

.

Answer

Number of mangoes	Number of boxes	$x_{i}$	$f_{i} x_{i}$
50-52	15	51	765
53-55	110	54	5940
56-58	135	57	7695
59-61	115	60	6900
62-64	25	63	1575
	$Σ f_{i} = 400$		$Σ f_{i} x_{i} = 22875$

\overset{x}{ˉ} = \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{2 2 8 7 5}{4 0 0} = 57.1875

Hence, the mean number of mangoes kept in a packing box is

57.1875

.

Question 6

The table below- shows the daily expenditure on food of

25

households in a locality.

Daily expenses (in Rs.)	100-150	150-200	200-250	250-300	300-350
No. of households	4	5	12	2	2

Find the mean daily expenses on food by a suitable method

Easy

Solution

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Answer

Daily Expense	Households (fi)	Mid Point (xi)	fi xi
100 - 150	4	125	500
150 - 200	5	175	875
200 - 250	12	225	2700
250 - 300	2	275	550
300 - 350	2	325	650
	25		5275

Mean

= \frac{\sum f i x i}{\sum f i} = \frac{5 2 7 5}{2 5} = 211

Answer

Daily Expense	Households (fi)	Mid Point (xi)	fi xi
100 - 150	4	125	500
150 - 200	5	175	875
200 - 250	12	225	2700
250 - 300	2	275	550
300 - 350	2	325	650
	25		5275

Mean

= \frac{\sum f i x i}{\sum f i} = \frac{5 2 7 5}{2 5} = 211

Question 7

To find out the concentration of

S O_{2}

in the air (in parts per million, i.e.,

p p m

), the data was collected for

30

localities in certain city and is presented below:

Concentration of $S O_{2}$ (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

S O_{2}

in the air.

Medium

Solution

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Answer

Answer:-

Concentration of $S O_{2}$ (in ppm)	$x_{i}$	Frequency $(f_{i})$	$f_{i} x_{i}$
0.00-0.04	0.02	4	0.08
0.04-0.08	0.06	9	0.54
0.08-0.12	0.10	9	0.90
0.12-0.16	0.14	2	0.28
0.16-0.20	0.18	4	0.72
0.20-0.24	0.22	2	0.44
		$Σ f_{i} = 30$	$Σ f_{i} x_{i} = 2.96$

mean concentration of S O_{2} in the air = \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{2 . 9 6}{3 0} = 0.0986 \approx 0.099

Answer

Answer:-

Concentration of $S O_{2}$ (in ppm)	$x_{i}$	Frequency $(f_{i})$	$f_{i} x_{i}$
0.00-0.04	0.02	4	0.08
0.04-0.08	0.06	9	0.54
0.08-0.12	0.10	9	0.90
0.12-0.16	0.14	2	0.28
0.16-0.20	0.18	4	0.72
0.20-0.24	0.22	2	0.44
		$Σ f_{i} = 30$	$Σ f_{i} x_{i} = 2.96$

mean concentration of S O_{2} in the air = \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{2 . 9 6}{3 0} = 0.0986 \approx 0.099

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:	$11$	$10$	$7$	$4$	$4$	$3$	$1$

Medium

Solution

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Answer

Answer:-

No. of days	No. of student $f_{i}$	$x_{i}$	$f_{i} x_{i}$
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
	$Σ f_{i} = 40$		$Σ f_{i} x_{i} = 499$

The mean number of days a student was absent

= \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{4 9 9}{4 0} = 12.475

Hence, the mean number of days a student was absent is

12.475 .

Answer

Answer:-

No. of days	No. of student $f_{i}$	$x_{i}$	$f_{i} x_{i}$
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
	$Σ f_{i} = 40$		$Σ f_{i} x_{i} = 499$

The mean number of days a student was absent

= \frac{Σ f _{i} x _{i}}{Σ f _{i}} = \frac{4 9 9}{4 0} = 12.475

Hence, the mean number of days a student was absent is

12.475 .

Question 9

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

Medium

Solution

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Answer

To find the class mark

(x_{i})

for each interval, using the following relation:

(x_{i}) = \frac{U p p e r c l a s s l i m i t + L o w e r c l a s s l i m i t}{2}

Class size

(h)

for this data

= 10

Taking 70 as assumed mean

(a)

, calculating

d_{i}, u_{i}, f_{i} u_{i}

:
From the table:

\sum f_{i} = 35

\sum f_{i} u_{i} = - 2

Mean,

\overset{x}{ˉ} = a + (\frac{\sum f _{i} u _{i}}{\sum f _{i}}) \times h

= 70 + (\frac{- 2}{3 5}) \times 10

= 70 - \frac{2 0}{3 5}

= 70 - \frac{4}{7}

= 70 - 0.57

= 69.43

Therefore, mean literacy rate is

69.43

%

Answer

To find the class mark

(x_{i})

for each interval, using the following relation:

(x_{i}) = \frac{U p p e r c l a s s l i m i t + L o w e r c l a s s l i m i t}{2}

Class size

(h)

for this data

= 10

Taking 70 as assumed mean

(a)

, calculating

d_{i}, u_{i}, f_{i} u_{i}

:
From the table:

\sum f_{i} = 35

\sum f_{i} u_{i} = - 2

Mean,

\overset{x}{ˉ} = a + (\frac{\sum f _{i} u _{i}}{\sum f _{i}}) \times h

= 70 + (\frac{- 2}{3 5}) \times 10

= 70 - \frac{2 0}{3 5}

= 70 - \frac{4}{7}

= 70 - 0.57

= 69.43

Therefore, mean literacy rate is

69.43

%

Question 10

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	$5 - 15$	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$
Number of patients	$6$	$11$	$21$	$23$	$14$	$5$

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Medium

Solution

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Answer

Mode: The class which have highest frequency.

In this case, class interval

35 - 45

is the modal class.

Now,

Lower limit of modal class,

l = 35, h = 10, f_{1}, = 23, f_{0} = 21, f_{2} = 14

We know that,

Mode

= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h

= 35 + (\frac{2 3 - 2 1}{2 ( 2 3 ) - 2 1 - 1 4}) \times 10

= 35 + \frac{2}{1 1} \times 10

= 35 + 1.818

Mode

= 36.8

Lets take assumed mean,

A

as

30

Class interval

= 10

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 3 0}{1 0}

\overset{x}{ˉ} = A + h \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 30 + 10 \times \frac{4 3}{8 0}

= 30 + 5.375

= 35.375

\approx 35.37

Hence, Mode

= 36.8

years, Mean

= 35.37

years.

Maximum number of patients admitted in the hospital are of age group

36.8

years, while on an average the age of a patient admitted to the hospital is

35.37

years.

Answer

Mode: The class which have highest frequency.

In this case, class interval

35 - 45

is the modal class.

Now,

Lower limit of modal class,

l = 35, h = 10, f_{1}, = 23, f_{0} = 21, f_{2} = 14

We know that,

Mode

= l + (\frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}}) \times h

= 35 + (\frac{2 3 - 2 1}{2 ( 2 3 ) - 2 1 - 1 4}) \times 10

= 35 + \frac{2}{1 1} \times 10

= 35 + 1.818

Mode

= 36.8

Lets take assumed mean,

A

as

30

Class interval

= 10

∴ u_{i} = \frac{x _{i} - A}{h} = \frac{x _{i} - 3 0}{1 0}

\overset{x}{ˉ} = A + h \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 30 + 10 \times \frac{4 3}{8 0}

= 30 + 5.375

= 35.375

\approx 35.37

Hence, Mode

= 36.8

years, Mean

= 35.37

years.

Maximum number of patients admitted in the hospital are of age group

36.8

years, while on an average the age of a patient admitted to the hospital is

35.37

years.

NCERT Solutions for Class 10 Maths Chapter 14 : Statistics

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