Data Handling

The first five prime numbers are 2, 3, 5, 7 and 11.

Sum of all observations = 2 + 3 + 5 + 7 + 11 = 28

Number of observations = 5

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= \frac{28}{5} \\[1em] = 5.6\end{aligned}$$

Hence, the arithmetic mean of first five prime numbers is 5.6.

Sum of all observations = 18 + 35 + 2 + 27 + 40 + 0 + 21 + 33 + 27 + 8 + 36 + 23 = 270

Number of observations = 12

Mean marks = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= \frac{270}{12} \\[1em] = 22.5\end{aligned}$$

Hence, the mean marks is 22.5.

Sum of all observations = 6 + 10 + 16 + 20 + 8 + 19 + 30 + 57 + 2 + 0 + 8 = 176

Number of observations = 11

Mean score = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= \frac{176}{11} \\[1em] = 16\end{aligned}$$

Hence, the mean score is 16.

Arranging the given data in ascending order, we get :

35, 36, 38, 39, 40, 42, 44, 45, 46, 49, 53, 55

(i) The last number in our sorted list is the maximum value i.e., 55.

Hence, the age of the eldest teacher is 55 years.

(ii) The first number in our sorted list is the minimum value i.e., 35.

Hence, the age of the youngest teacher is 35 years.

(iii) Range = Highest observation - Lowest observation

= 55 years - 35 years

= 20 years.

Hence, the range of the ages of the teachers is 20 years.

(iv) Sum of all observations = 35 + 36 + 38 + 39 + 40 + 42 + 44 + 45 + 46 + 49 + 53 + 55 = 522

Number of observations = 12

Mean age = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= \frac{522}{12} \\[1em] = 43.5\end{aligned}$$

Hence, the mean age of the teachers is 43.5 years.

(i) Sum of all observations = 195 + 185 + 145 + 155 + 135 + 180 + 175 + 200 + 150 + 125 + 190 + 180 + 170 + 175 + 190 = 2550

Number of observations = 15

Mean daily wage = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= ₹ \left(\frac{2550}{15}\right) \\[1em] = ₹ 170\end{aligned}$$

Hence, the mean daily wage is ₹ 170.

(ii) Range = Highest observation - Lowest observation

= ₹ 200 - ₹ 125

= ₹ 75

Hence, the range of the data is ₹ 75.

(i) Sum of all observations = 28.9 + 32.6 + 24.6 + 26.1 + 29.2 + 30 + 27.4 = 198.8

Number of observations = 7

Mean temperature = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}= \left(\frac{198.8}{7}\right) ^{\circ} C \\[1em] = 28.4 ^{\circ} C\end{aligned}$$

Hence, the mean temperature is 28.4 °C.

(ii) Range = Highest observation - Lowest observation

= 32.6 °C - 24.6 °C

= 8 °C

Hence, the range of the data is 8 °C.

Given:

Observations are 4, 6, x, 9, 10, 5

Mean = 7

Sum of all observations = 4 + 6 + x + 9 + 10 + 5 = 34 + x

Number of observations = 6

Mean = $\frac{\text{Sum of all observations}}{\text{Number of observations}}$

$$\begin{aligned}⇒ 7 = \frac{34 + x}{6} \\[1em] ⇒ 7 \times 6 = 34 + x \\[1em] ⇒ 42 = 34 + x \\[1em] ⇒ x = 42 - 34 \\[1em] ⇒ x = 8\end{aligned}$$

Hence, the value of x is 8.

Arranging the given data in ascending order, we get :

0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4

The frequency distribution table is :

Number of children (xi)	Tally Marks	Number of families (Frequency fi)
0	||	2
1	|||| |	6
2	|||| ||||	9
3	||||	5
4	|||	3
Total		25

Arranging the given data in ascending order, we get :

1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6

The frequency distribution table is :

Outcome (xi)	Tally Marks	Frequency (fi)
1	||||	4
2	|||| |	6
3	|||| ||||	9
4	|||| |	6
5	||	2
6	|||	3
Total		30

We calculate the mean as follows:

Weight (in kg) xi	Frequency fi	fixi
50	6	300
52	8	416
54	15	810
56	14	784
60	7	420
Total	∑fi = 50	∑fixi = 2730

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{2730}{50}$ kg

= 54.6 kg

Hence, the mean weight is 54.6 kg.

We calculate the mean as follows:

Height (in cm) xi	Number of plants fi	fixi
58	20	1160
60	25	1500
62	15	930
64	8	512
66	12	792
74	10	740
Total	∑fi = 90	∑fixi = 5634

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{5634}{90}$ cm

= 62.6 cm

Hence, the mean height is 62.6 cm.

We calculate the mean as follows:

Height (in cm) xi	Number of boys fi	fixi
142	10	1420
144	13	1872
146	12	1752
148	7	1036
150	17	2550
152	6	912
Total	∑fi = 65	∑fixi = 9542

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{9542}{65}$ cm

= 146.8 cm

Hence, the mean height is 146.8 cm.

For calculating the mean, we prepare the following table:

Variable (xi)	Frequency (fi)	fixi
10	13	130
12	p	12p
14	15	210
16	32	512
18	28	504
Total	∑fi = (88 + p)	∑fixi = (1356 + 12p)

Mean = $\frac{∑f_ix_i}{∑f_i}$

⇒ 15 = $\frac{1356 + 12p}{88 + p}$

⇒ 15(88 + p) = 1356 + 12p

⇒ 1320 + 15p = 1356 + 12p

⇒ 15p - 12p = 1356 - 1320

⇒ 3p = 36

⇒ p = $\frac{36}{3}$

⇒ p = 12

Hence, the value of p is 12.

(i) On arranging the given set of data in ascending order, we get :

0, 8, 18, 25, 31, 34, 39, 46, 53, 65, 72

Number of observations, n = 11 (odd)

Median = $\Big[\frac{n+1}{2}\Big]^{\text{th}}$ term

$$\begin{aligned}= \Big[\frac{11+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{12}{2}\Big]^{th} \text{ term} \\[1em] = 6^{th} \text{ term} \\[1em] = 34\end{aligned}$$

Hence, the median is 34.

(ii) On arranging the given set of data in ascending order, we get :

6, 8, 9, 11, 13, 15, 16, 17, 18, 20, 21, 22, 25

Number of observations, n = 13 (odd)

Median = $\Big[\frac{n+1}{2}\Big]^{th}$ term

$$\begin{aligned}= \Big[\frac{13+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{14}{2}\Big]^{th} \text{ term} \\[1em] = 7^{th} \text{ term} \\[1em] = 16\end{aligned}$$

Hence, the median is 16.

(i) On arranging the given set of data in ascending order, we get :

9, 10, 17, 19, 21, 22, 32, 35

Number of observations, n = 8 (even)

Median = $\frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{ term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{ term}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{th} \text{ term} + \Big(\frac{8}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(4)^{th} \text{ term} + (5)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[19 + 21\Big] \\[1em] = \frac{1}{2}\Big[40\Big] \\[1em] = 20\end{aligned}$$

Hence, the median is 20.

(ii) On arranging the given set of data in ascending order, we get :

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of observations, n = 10 (even)

Median = $\frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{ term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{ term} \Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{th} \text{ term} + \Big(\frac{10}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(5)^{th} \text{ term} + (6)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[60 + 63\Big] \\[1em] = \frac{1}{2}\Big[123\Big] \\[1em] = 61.5\end{aligned}$$

Hence, the median is 61.5.

On arranging the given set of data in ascending order, we get :

31, 34, 35, 40, 41, 42, 44, 47, 48, 52, 53, 61, 65

Number of observations, n = 13 (odd)

Median = $\Big[\frac{n+1}{2}\Big]^{th}$ term

$$\begin{aligned}= \Big[\frac{13+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{14}{2}\Big]^{th} \text{ term} \\[1em] = 7^{th} \text{ term} \\[1em] = 44\end{aligned}$$

Hence, the median weight is 44 kg.

On arranging the given set of data in ascending order, we get :

31, 34, 36, 37, 40, 42, 43, 46, 47, 50

Number of observations, n = 10 (even)

Median = $\frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{ term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{ term}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{th} \text{ term} + \Big(\frac{10}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(5)^{th} \text{ term} + (6)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[40 + 42\Big] \\[1em] = \frac{1}{2}\Big[82\Big] \\[1em] = 41\end{aligned}$$

Hence, the median marks is 41.

The first 15 odd numbers are :

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Number of observations, n = 15 (odd)

Median = $\Big[\frac{n+1}{2}\Big]^{th}$ term

$$\begin{aligned}= \Big[\frac{15+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{16}{2}\Big]^{th} \text{ term} \\[1em] = 8^{th} \text{ term} \\[1em] = 15\end{aligned}$$

Hence, the median of first 15 odd numbers is 15.

The first 50 whole numbers are :

0, 1, 2, 3, ..., 49

Number of observations, n = 50 (even)

$$\begin{aligned}Median = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{ term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{50}{2}\Big)^{th} \text{ term} + \Big(\frac{50}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(25)^{th} \text{ term} + (26)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[24 + 25\Big] \\[1em] = \frac{1}{2}\Big[49\Big] \\[1em] = 24.5\end{aligned}$$

Hence, the median of first 50 whole numbers is 24.5.

Arranging the terms in ascending order, we get :

Daily wages (in ₹) xi	280	300	320	360	380	400	420
Number of labourers fi	10	8	27	17	4	16	18

Now, we prepare the cumulative frequency table :

Daily wages (in ₹) xi	Number of labourers fi	Cumulative Frequency
280	10	10
300	8	18
320	27	45
360	17	62
380	4	66
400	16	82
420	18	100

Total number of terms = ∑f_i = N = 100, which is even.

Median = $\frac{1}{2}\Big[\Big(\frac{N}{2}\Big)^{th} + \Big(\frac{N}{2} + 1\Big)^{th}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{100}{2}\Big)^{th} \text{ term} + \Big(\frac{100}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(50)^{th} \text{ term} + (51)^{st} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[360 + 360\Big] \\[1em] = \frac{1}{2}\Big[720\Big] \\[1em] = 360\end{aligned}$$

[Note that each one of 46th, 47th, ......, 62nd labourer earns ₹ 360.]

Hence, the median wages is ₹ 360.

Arranging the terms in ascending order, we get :

Height (in cm) xi	152	153	154	155	156	157
Number of students fi	4	3	5	9	7	7

Now, we prepare the cumulative frequency table :

Height (in cm) xi	Number of students fi	Cumulative Frequency
152	4	4
153	3	7
154	5	12
155	9	21
156	7	28
157	7	35

Total number of terms = ∑f_i = N = 35, which is odd.

Median = $\Big[\frac{N+1}{2}\Big]^{th}$

$$\begin{aligned}= \Big[\frac{35+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{36}{2}\Big]^{th} \text{ term} \\[1em] = 18^{th} \text{ term} \\[1em] = 155\end{aligned}$$

[Note that each one of 13th, 14th, ......, 21st student has height 155 cm.]

Hence, the median height is 155 cm.

(i) Arranging the given data in ascending order, we get :

5, 7, 8, 8, 9, 9, 9, 11, 12, 16

Since 9 occurs most frequently in the given data, the mode is 9.

Hence, the mode is 9.

(ii) Arranging the given data in ascending order, we get :

16, 19, 21, 25, 25, 25, 25, 29, 34, 37, 38

Since 25 occurs most frequently in the given data, the mode is 25.

Hence, the mode is 25.

From the given frequency distribution, it is clear that the maximum number of pairs sold is 8 which corresponds to shoe size 5.

Hence, the modal shoe-size is 5.

Arranging the terms in ascending order and preparing the cumulative frequency table :

Height (in cm) xi	Number of students fi	Cumulative Frequency	fixi
147	6	6	882
148	3	9	444
149	12	21	1788
150	4	25	600
151	10	35	1510
152	8	43	1216
153	7	50	1071
Total	N = ∑fi = 50		∑fixi = 7511

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{7511}{50}$

= 150.22

We have, N = 50, which is even.

Median = $\frac{1}{2}\Big[\Big(\frac{N}{2}\Big)^{th} \text{ term} + \Big(\frac{N}{2} + 1\Big)^{th} \text{ term}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{50}{2}\Big)^{th} \text{ term} + \Big(\frac{50}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(25)^{th} \text{ term} + (26)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[150 + 151\Big] \\[1em] = \frac{1}{2}\Big[301\Big] \\[1em] = 150.5\end{aligned}$$

Now, Mode = 3(Median) - 2(Mean)

= 3 × 150.5 - 2 × 150.22

= 451.5 - 300.44

= 151.06

Hence, Mean = 150.22, Median = 150.5 and Mode = 151.06.

Arranging the terms in ascending order and preparing the cumulative frequency table:

Marks obtained xi	Number of students fi	Cumulative Frequency	fixi
1	0	0	0
2	0	0	0
3	1	1	3
4	6	7	24
5	4	11	20
6	9	20	54
7	7	27	49
8	1	28	8
9	5	33	45
10	2	35	20
Total	N = ∑fi = 35		∑fixi = 223

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{223}{35}$

= 6.37 (approx.)

We have, N = 35, which is odd.

Median = $\Big[\frac{N+1}{2}\Big]^{th}$ term

= $\Big[\frac{35+1}{2}\Big]^{th}$ term

= $\Big[\frac{36}{2}\Big]^{th}$ term

= $18^{th}$ term

= 6

Now, Mode = 3(Median) - 2(Mean)

= 3 × 6 - 2 × 6.37

= 18 - 12.74

= 5.26

Hence, Mean = 6.37, Median = 6 and Mode = 5.26.

Arranging the terms in ascending order and preparing the cumulative frequency table :

Weight (in grams) xi	Number of boxes fi	Cumulative Frequency	fixi
350	5	5	1750
375	4	9	1500
400	3	12	1200
425	3	15	1275
450	7	22	3150
475	2	24	950
500	6	30	3000
Total	N = ∑fi = 30		∑fixi = 12825

Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{12825}{30}$

= 427.5

We have, N = 30, which is even.

Median = $\frac{1}{2}\Big[\Big(\frac{N}{2}\Big)^{th} \text{ term} + \Big(\frac{N}{2} + 1\Big)^{th} \text{ term}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{30}{2}\Big)^{th} \text{ term} + \Big(\frac{30}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(15)^{th} \text{ term} + (16)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[425 + 425\Big] \\[1em] = \frac{1}{2}\Big[850\Big] \\[1em] = 425\end{aligned}$$

Now, Mode = 3(Median) - 2(Mean)

= 3 × 425 - 2 × 427.5

= 1275 - 855

= 420

Hence, Mean = 427.5 g, Median = 425 g and Mode = 420 g.

To represent the sports preferences of the student body, we take the names of the sports along the x-axis and the number of students along the y-axis.

Using a scale of 1 cm = 5 students, we may draw the column graph for the given data as shown below.

We represent the subjects on the x-axis and the marks obtained on the y-axis.

Taking a scale of 1 cm = 5 marks, the bar graph illustrating the student's performance in various subjects is drawn as follows.

The countries are represented along the x-axis and the birth-rate per thousand is taken along the y-axis.

Following the given hint to take 2 cm = 7 infants on the y-axis, we draw the bar graph for the data as shown below.

Cities are represented along the x-axis and their respective distances from Delhi along the y-axis.

Taking a scale of 1 cm = 100 km. The bar graph representing the air distances (in km) is provided below.

From the given bar graph, the marks obtained in different subjects are :

Hindi = 55, Mathematics = 95, Science = 80, English = 65, Social Science = 70, Computer Science = 90.

(i) What information is represented by the bar graph?

The bar graph shows the marks obtained by a student in various subjects in an examination.

(ii) How many marks were obtained in English?

The marks obtained in English are 65.

(iii) In which subject least marks were obtained?

Least marks were obtained in Hindi.

(iv) What was the total marks obtained in all the subjects together?

Total marks obtained in all subjects together

= 55 + 95 + 80 + 65 + 70 + 90

= 455

Hence, the total marks obtained in all subjects together is 455.

From the given bar graph, the wheat production (in million tons) in different years are :

2008 = 90, 2009 = 70, 2010 = 85, 2011 = 100, 2012 = 130, 2013 = 110.

(i) What information is given by the bar graph?

The bar graph represents quantity of wheat produced (in million tons, i.e., in MT) by a state in various years from 2008 to 2013.

(ii) What was the wheat production in 2010?

The wheat production in 2010 was 85 MT.

(iii) In which year the production was minimum?

The production was minimum in the year 2009.

(iv) In which year there was maximum increase in production of wheat?

Increase in production :

From 2008 to 2009 = 70 - 90 = -20 million tons (a decrease)

From 2009 to 2010 = 85 - 70 = 15 million tons

From 2010 to 2011 = 100 - 85 = 15 million tons

From 2011 to 2012 = 130 - 100 = 30 million tons

Clearly, the maximum increase in production was in the year 2012.

The maximum increase in production of wheat was in the year 2012.

(v) Find the percentage increase in wheat production in the year 2012.

Percentage increase in wheat production in 2012

$$\begin{aligned}= \frac{\text{Increase in production}}{\text{Production in previous year}} \times 100 \\[1em] = \frac{130 - 100}{100} \times 100 \\[1em] = \frac{30}{100} \times 100 \\[1em]\end{aligned}$$

= 30%

Hence, the percentage increase in wheat production in the year 2012 is 30%.

Range = Highest observation - Lowest observation

= 56 - 5

= 51

Hence, option 3 is the correct option.

The mean of n observations x₁, x₂, ..............., x_n is given by :

Mean = $\frac{x_1 + x_2 + ........ + x_n}{n} = \frac{∑x_i}{n}$

Hence, option 4 is the correct option.

When n is odd, then :

Median = value of $\Big(\frac{n + 1}{2}\Big)^{th}$ observation.

Hence, option 2 is the correct option.

The Empirical formula for calculating mode is :

Mode = 3(Median) - 2(Mean)

Hence, option 2 is the correct option.

The first 49 natural numbers are :

1, 2, 3, 4, ............, 49

Number of observations, n = 49 (odd)

Median = $\Big[\frac{n+1}{2}\Big]^{th}$ term

= $\Big[\frac{49+1}{2}\Big]^{th}$ term

= $\Big[\frac{50}{2}\Big]^{th}$ term

= $25^{th}$ term

= 25

Hence, option 3 is the correct option.

(i) A collection of information in the form of numerical figures, is called data.

(ii) When information is collected and presented, randomly, then it is called raw data.

(iii) A particular observation that gives the value of a variable is called variate.

(iv) An arrangement of the numerical figures of a data in ascending or descending order is called an array.

(v) The difference between the highest and the lowest values of observations in a data is called the range of the data.

(vi) The number of times a particular observation occurs in a data is called the frequency of the observation.

(vii) The representation of data in a tabular form showing the frequency of each observation is called frequency distribution.

(viii) The arithmetic mean of some observations is equal to the ratio of the sum of the observations to the number of observations.

(ix) When an ungrouped data is arranged in an array, then the value of the middle-most observation is called the median of the data.

(x) The value of the variable which occurs most frequently is called the mode of the data.

(i) True
Reason — Each numerical figure in a data is called an observation.

(ii) False
Reason — The difference between the highest and the lowest values of observations gives the range of the data, not the frequency.

(iii) False
Reason — The first 4 multiples of 2 are 2, 4, 6 and 8.

Mean = $\frac{2 + 4 + 6 + 8}{4}$

= $\frac{20}{4}$

= 5

So, the mean of first 4 multiples of 2 is 5, not 4.

(iv) True
Reason — By definition, mode is the value of the variable which occurs most frequently in a given data.

(v) True
Reason — The first 10 natural numbers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Number of observations, n = 10 (even)

Median = $\frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{th} \text{ term} + \Big(\frac{n}{2} + 1\Big)^{th} \text{ term}\Big]$ term

$$\begin{aligned}= \frac{1}{2}\Big[(5)^{th} \text{ term} + (6)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[5 + 6\Big] \\[1em] = \frac{11}{2} \\[1em] = 5.5\end{aligned}$$

Arranging the given data in ascending order, we get :

0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4

We prepare the following table :

Number of children (xi)	Number of families (fi)	Cumulative Frequency	fixi
0	2	2	0
1	6	8	6
2	9	17	18
3	5	22	15
4	3	25	12
Total	N = ∑fi = 25		∑fixi = 51

(1) From the table, the number of families having 3 children = 5.

Hence, option 3 is the correct option.

(2) From the table, the cumulative frequency of families having 3 children = 22.

Hence, option 4 is the correct option.

(3) Mean = $\frac{∑f_ix_i}{∑f_i}$

= $\frac{51}{25}$

= 2.04

Hence, option 2 is the correct option.

(4) We have, N = 25, which is odd.

Median = $\Big[\frac{N+1}{2}\Big]^{th}$ term

$$\begin{aligned}= \Big[\frac{25+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\frac{26}{2}\Big]^{th} \text{ term} \\[1em] = 13^{th} \text{ term} \\[1em] = 2\end{aligned}$$

[Note that each one of 9th, 10th, ......, 17th term is 2.]

Hence, option 2 is the correct option.

From the given bar graph, the number of air conditioners sold in different months are :

April = 510, May = 460, June = 640, July = 720, August = 680, September = 550.

(1) The highest number of air conditioners sold = 720, which is in the month of July.

Hence, option 2 is the correct option.

(2) Difference between the number of air conditioners sold in June and August

= 680 - 640

= 40

Hence, option 1 is the correct option.

(3) Total number of air conditioners sold in six months

= 510 + 460 + 640 + 720 + 680 + 550

= 3560

Hence, option 2 is the correct option.

(4) Highest number of air conditioners sold in a month = 720 (in July)

Lowest number of air conditioners sold in a month = 460 (in May)

Required ratio = 720 : 460

= $\frac{720}{460}$

= $\frac{36}{23}$

= 36 : 23

Hence, option 2 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

The median is the value of the middle-most observation when the data is arranged in ascending order. If we remove one extreme observation from each end of the data, the middle-most observation remains the same. So, the median does not get affected.

Hence, Assertion is true.

By definition, mode is the value of the variable which occurs most frequently in a given data.

Hence, Reason is true.

However, Reason is about mode, while Assertion is about median. So, Reason is not the correct explanation of Assertion.

Hence, option 2 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

The mean of given observations always lies between the lowest and the highest observations. So, the mean cannot be less than each of the given observations.

Hence, Assertion is false.

The mean of n observations x₁, x₂, x₃, ............... x_n is given by $\frac{x_1 + x_2 + ........ + x_n}{n}$.

Hence, Reason is true.

Hence, option 4 is the correct option.

One day's temperature is much lower than the others, so it is an outlier.

The mean gets affected by extreme values.
The median represents the middle value and is not much affected by outliers.
The mode is the most frequent value.
The range measures spread, not the central value.

Therefore, the best statistic to find the central value is median.

Hence, option 2 is the correct option.

Mode is the value (here, the rank) which occurs most frequently in the data.

From the given information, the number of times Anupam secured each rank is :

• First rank → 2 times
• Third rank → 3 times
• Fourth rank → 1 time
• Fifth rank → 6 times

The fifth rank occurs the most number of times (6 times), so the mode of the data is the fifth rank.

Thus, the mode tells us the rank which Anupam secured the most number of times. So, the question "Which rank did Anupam secure the most?" can be answered using the mode of the data.

Hence, option 1 is the correct option.

Given,

Arithmetic mean of runs scored in 11 matches = 49

Number of matches (observations) = 11

Runs scored in the 11th match = x

Sum of all observations = 84 + 80 + 0 + 10 + 60 + 75 + 50 + 100 + 20 + 40 + x

= 519 + x

Mean = $\frac{\text{Sum of all observations}}{\text{Number of all observations}}$

⇒ 49 = $\frac{519 + x}{11}$

⇒ 49 × 11 = 519 + x

⇒ 539 = 519 + x

⇒ x = 539 - 519

⇒ x = 20

So, the runs scored by Sunil in the 11th match is 20.

Hence, option 1 is the correct option.

Statement-1 : Mode is the most frequently occurring value in a given data — this part is correct. But the mode is not always less than the median. The mode may be less than, equal to or greater than the median, depending on the data.

So, Statement-1 is false.

Statement-2 : The given data is :

1, 2, 2, 3, 2, 3, 4, 2, 3, 2, 2, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3

Arranging the data in ascending order, we get :

1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4

We prepare the following frequency table :

Observation (xi)	Frequency (fi)	Cumulative Frequency
1	2	2
2	11	13
3	10	23
4	1	24
Total	N = ∑fi = 24

Mode = 2, since 2 occurs the most number of times (11 times).

Number of observations, N = 24, which is even.

Median = $\frac{1}{2}\Big[\Big(\frac{N}{2}\Big)^{th} \text{ term} + \Big(\frac{N}{2} + 1\Big)^{th} \text{ term}\Big]$

$$\begin{aligned}= \frac{1}{2}\Big[\Big(\frac{24}{2}\Big)^{th} \text{ term} + \Big(\frac{24}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[(12)^{th} \text{ term} + (13)^{th} \text{ term}\Big] \\[1em] = \frac{1}{2}\Big[2 + 2\Big] \\[1em] = \frac{4}{2} \\[1em] = 2\end{aligned}$$

So, Median = 2 and Mode = 2, i.e., the median is equal to the mode and not less than the mode.

So, Statement-2 is false.

Since both Statement-1 and Statement-2 are false,

Hence, option 3 is the correct option.

From the given bar graph, the heights of the bars (in units) for the different departments are :

Department A = 4 units, Department B = 12 units, Department C = 10 units, Department D = 2 unit.

Difference in the heights of the bars for Department B and Department D

= 12 units - 2 unit

= 10 units

Given, the difference in the average salary of an employee in Department B and Department D = ₹40,000.

So, 10 units = ₹40,000

⇒ 1 unit = $\frac{₹40,000}{10}$

⇒ 1 unit = ₹4,000

Hence, option 4 is the correct option.