Data Handling

(i) On arranging the given numbers in ascending order of their magnitudes, we get :

52, 63, 63, 68, 70, 71, 73, 75, 75, 76, 77, 82, 83, 90, 92, 93, 99, 103, 105 and 115.

Hence, the numbers in ascending order are 52, 63, 63, 68, 70, 71, 73, 75, 75, 76, 77, 82, 83, 90, 92, 93, 99, 103, 105 and 115.

(ii) From the above arrangement,

Largest number = 115

Smallest number = 52

By formula,

Range of given numbers = Largest number − Smallest number

Substituting the values, we get :

Range = 115 − 52 = 63

Hence, the range of the given numbers is 63.

Writing the numbers from the lowest to the highest, marking the tally marks against each number and then counting the tally marks, we get the required frequency distribution table as shown below :

Numbers	Tally marks	Frequency
15	IIII	5
16	IIII II	7
17	IIII IIII I	11
18	IIII	5
19	IIII I	6
20	III	3
21	III	3
Total		40

Hence, the above table is the required frequency distribution table.

Writing the scores from the lowest to the highest, marking the tally marks against each score and then counting the tally marks, we get the required frequency table as shown below :

Score	Tally marks	Frequency
1	IIII	4
2	IIII	4
3	III	3
4	III	3
5	III	3
6	III	3
Total		20

Hence, the above table is the required frequency table for the scores.

(i) Writing the wages from the lowest to the highest, marking the tally marks against each wage and then counting the tally marks, we get the required frequency distribution table as shown below :

Wages (in ₹)	Tally marks	Frequency
3500	I	1
3750	I	1
4000	IIII	4
4250	III	3
4750	I	1
Total		10

Hence, the above table is the required frequency distribution table.

(ii) From the above table,

Highest wages = ₹ 4750

Lowest wages = ₹ 3500

By formula,

Range of wages = Highest wages − Lowest wages

Substituting the values, we get :

Range = ₹ (4750 − 3500)

= ₹ 1250

Hence, the range of wages is ₹ 1250.

(iii) The maximum wages are ₹ 4750 and the frequency of ₹ 4750 is 1.

Hence, 1 worker is getting the maximum wages.

(i) Writing the marks from the lowest to the highest, marking the tally marks against each mark and then counting the tally marks, we get the required frequency distribution table as shown below :

Marks	Tally marks	No. of students
10	II	2
20	III	3
30	III	3
40	IIII II	7
50	IIII I	6
60	IIII II	7
70	IIII	5
80	IIII	4
90	III	3
Total		40

Hence, the above table is the required frequency distribution table.

(ii) Marks equal to or more than 70 are 70, 80 and 90, whose frequencies are 5, 4 and 3 respectively.

∴ Number of such students = 5 + 4 + 3

= 12

Hence, 12 students have marks equal to or more than 70.

(iii) Marks below 40 are 10, 20 and 30, whose frequencies are 2, 3 and 3 respectively.

∴ Number of such students = 2 + 3 + 3

= 8

Hence, 8 students obtained marks below 40.

On arranging the given data in descending order of their magnitudes, we get :

4.1, 4.0, 3.9, 3.9, 3.8, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2 and 3.1.

Hence, the data in descending order is 4.1, 4.0, 3.9, 3.9, 3.8, 3.8, 3.7, 3.7, 3.6, 3.5, 3.5, 3.4, 3.3, 3.2 and 3.1.

(i) From the above arrangement,

Largest number = 4.1

Smallest number = 3.1

By formula,

Range of given numbers = Largest number − Smallest number

Substituting the values, we get :

Range = 4.1 − 3.1

= 1

Hence, the range of the given data is 1.

(ii) The numbers less than 3.5 are 3.4, 3.3, 3.2 and 3.1.

∴ Number of such numbers = 4

Hence, 4 numbers are less than 3.5.

(iii) The numbers 3.8 or above are 4.1, 4.0, 3.9, 3.9, 3.8 and 3.8.

∴ Number of such numbers = 6

Hence, 6 numbers are 3.8 or above.

Sum of all observations = 53 + 61 + 60 + 67 + 64

= 305

and, number of observations = 5

By formula,

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

Substituting the values, we get :

Required mean = $\frac{305}{5}$

= 61

Hence, the mean of the given observations is 61.

First six natural numbers are 1, 2, 3, 4, 5 and 6.

Sum of all observations = 1 + 2 + 3 + 4 + 5 + 6

= 21

and, number of observations = 6

By formula,

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

Substituting the values, we get :

Required mean = $\frac{21}{6}$

= 3.5

Hence, the mean of first six natural numbers is 3.5.

First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17 and 19.

Sum of all observations = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

= 100

and, number of observations = 10

By formula,

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

Substituting the values, we get :

Required mean = $\frac{100}{10}$

= 10

Hence, the mean of first ten odd natural numbers is 10.

All the factors of 10 are 1, 2, 5 and 10.

Sum of all observations = 1 + 2 + 5 + 10

= 18

and, number of observations = 4

By formula,

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

Substituting the values, we get :

Required mean = $\frac{18}{4}$

= 4.5

Hence, the mean of all factors of 10 is 4.5.

Sum of all observations = $(x + 3) + (x + 5) + (x + 7) + (x + 9) + (x + 11)$

= $5x + 35$

and, number of observations = 5

By formula,

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

Substituting the values, we get :

$$\begin{aligned}\text{Required mean} = \frac{5x + 35}{5} \\[1em] = \frac{5(x + 7)}{5} \\[1em] = x + 7\end{aligned}$$

Hence, the mean of the given observations is $\bm{x + 7}$.

Sum of all observations = 19.8 + 15.4 + 13.7 + 11.71 + 11.8 + 12.6 + 12.8 + 18.6 + 20.5 + 21.1

= 158.01

and, number of observations = 10

By formula,

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

Substituting the values, we get :

Required mean = $\frac{158.01}{10}$

= 15.801

Hence, the mean of the given values of variable $\bm{x}$ is 15.801.

Given,

Mean of the observations = 32

(i) If each observation is increased by 3, the new mean will also be increased by 3.

∴ New mean = 32 + 3

= 35

Hence, the resulting mean is 35.

(ii) If each observation is decreased by 7, the new mean will also be decreased by 7.

∴ New mean = 32 − 7

= 25

Hence, the resulting mean is 25.

(iii) If each observation is multiplied by 2, the new mean will also be multiplied by 2.

∴ New mean = 32 × 2

= 64

Hence, the resulting mean is 64.

(iv) If each observation is divided by 0.5, the new mean will also be divided by 0.5.

∴ New mean = 32 ÷ 0.5

= $\frac{32}{0.5}$

= 64

Hence, the resulting mean is 64.

(v) If each observation is increased by 60%, the new mean will also be increased by 60%.

∴ New mean = 32 + 60% of 32

= 32 + $\frac{60}{100} \times 32$

= 32 + 19.2

= 51.2

Hence, the resulting mean is 51.2.

(vi) If each observation is decreased by 20%, the new mean will also be decreased by 20%.

∴ New mean = 32 − 20% of 32

= 32 − $\frac{20}{100} \times 32$

= 32 − 6.4

= 25.6

Hence, the resulting mean is 25.6.

Sum of all observations = ₹ (85.40 + 88.00 + 86.50 + 84.75 + 82.60 + 87.25)

= ₹ 514.50

and, number of observations = 6

By formula,

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

Substituting the values, we get :

Mean pocket expenses = ₹ $\frac{514.50}{6}$

= ₹ 85.75

Hence, the mean pocket expenses of Anuj per day are ₹ 85.75.

Given,

$$\begin{aligned}\frac{8 + 10 + 7 + (x + 2) + 6}{5} = 9 \\[1em] \Rightarrow \frac{33 + x}{5} = 9 \\[1em] \Rightarrow 33 + x = 9 \times 5 \\[1em] \Rightarrow 33 + x = 45 \\[1em] \Rightarrow x = 45 - 33 \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, the value of $\bm{x = 12}$.

First six multiples of 3 are 3, 6, 9, 12, 15 and 18.

Sum of all observations = 3 + 6 + 9 + 12 + 15 + 18

= 63

and, number of observations = 6

By formula,

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

Substituting the values, we get :

Required mean = $\frac{63}{6}$

= 10.5

Hence, the mean of first six multiples of 3 is 10.5.

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

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

= 28

and, number of observations = 5

By formula,

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

Substituting the values, we get :

Required mean = $\frac{28}{5}$

= 5.6

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

Given,

$$\begin{aligned}\frac{(x - 5) + (x - 1) + x + (x + 2) + (x + 4) + (x + 12)}{6} = 15 \\[1em] \Rightarrow \frac{6x + 12}{6} = 15 \\[1em] \Rightarrow 6x + 12 = 15 \times 6 \\[1em] \Rightarrow 6x + 12 = 90 \\[1em] \Rightarrow 6x = 90 - 12 \\[1em] \Rightarrow 6x = 78 \\[1em] \Rightarrow x = \frac{78}{6} \\[1em] \Rightarrow x = 13\end{aligned}$$

Since the first four numbers are

$x - 5, x - 1, x$ and $x + 2$

= $13 - 5, 13 - 1$, $13$ and $13 + 2$

= 8, 12, 13 and 15

$$\begin{aligned}\therefore \text{Required mean} = \frac{8 + 12 + 13 + 15}{4} \\[1em] = \frac{48}{4} \\[1em] = 12\end{aligned}$$

Hence, the mean of first four numbers is 12.

First five whole numbers are 0, 1, 2, 3 and 4.

So, the squares of first five whole numbers are 0, 1, 4, 9 and 16.

Sum of all observations = 0 + 1 + 4 + 9 + 16

= 30

and, number of observations = 5

By formula,

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

Substituting the values, we get :

Required mean = $\frac{30}{5}$

= 6

Hence, the mean of squares of first five whole numbers is 6.

Given,

$$\begin{aligned}\frac{6 + 4 + 7 + p + 10}{5} = 8 \\[1em] \Rightarrow \frac{27 + p}{5} = 8 \\[1em] \Rightarrow 27 + p = 8 \times 5 \\[1em] \Rightarrow 27 + p = 40 \\[1em] \Rightarrow p = 40 - 27 \\[1em] \Rightarrow p = 13\end{aligned}$$

Hence, the value of $\bm{p = 13}$.

First six multiples of 5 are 5, 10, 15, 20, 25 and 30.

Sum of all observations = 5 + 10 + 15 + 20 + 25 + 30

= 105

and, number of observations = 6

By formula,

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

Substituting the values, we get :

Required mean = $\frac{105}{6}$

= 17.5

Hence, the mean of first six multiples of 5 is 17.5.

Total rainfall for the week = (0.5 + 2.7 + 2.6 + 0.5 + 2 + 5.8 + 1.5) mm

= 15.6 mm

and, number of days = 7

By formula,

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

Substituting the values, we get :

Average (mean) rainfall = $\frac{15.6}{7}$ mm

= 2.23 mm (approx.)

Hence, the total rainfall for the week is 15.6 mm and the average (mean) rainfall is 2.23 mm (approx.).

Given,

Mean of 100 students = 40

Number of students = 100

Incorrectly recorded score = 83

Actual score = 53

By formula,

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

Therefore,

Incorrect sum of marks = 40 × 100

= 4000

Correct sum of marks = Incorrect sum − Incorrectly recorded score + Actual score

= 4000 − 83 + 53

= 3970

Now,

Correct mean = $\frac{\text{Correct sum of marks}}{\text{Number of students}}$

= $\frac{3970}{100}$

= 39.7

Hence, the correct mean of marks scored by 100 students is 39.7.

Given,

Mean of 5 numbers = 27

Mean of remaining 4 numbers (after excluding one number) = 25

By formula,

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

Therefore,

Sum of 5 numbers = 27 × 5

= 135

Sum of remaining 4 numbers = 25 × 4

= 100

Now,

Excluded number = Sum of 5 numbers − Sum of remaining 4 numbers

= 135 − 100

= 35

Hence, the excluded number is 35.

Given,

Mean of 5 numbers = 27

Mean of 6 numbers (after including one new number) = 25

By formula,

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

Therefore,

Sum of 5 numbers = 27 × 5

= 135

Sum of 6 numbers = 25 × 6

= 150

Now,

Included number = Sum of 6 numbers − Sum of 5 numbers

= 150 − 135

= 15

Hence, the included number is 15.

Given,

Mean of first 5 numbers = 20

Mean of other 5 numbers = 30

By formula,

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

Therefore,

Sum of first 5 numbers = 20 × 5

= 100

Sum of other 5 numbers = 30 × 5

= 150

Sum of all the 10 numbers = 100 + 150

= 250

Now,

Mean of all the 10 numbers = $\frac{\text{Sum of all the 10 numbers}}{\text{Number of observations}}$

= $\frac{250}{10}$

= 25

Hence, the mean of all the 10 numbers taken together is 25.

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

2, 5, 7, 9, 11, 15, 17, 19, 23

Since, the number of terms ($n$) is 9 and 9 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{9 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 5^{\text{th}}\text{ term} \\[1em] = 11\end{aligned}$$

Hence, the median of the given data is 11.

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

0, 3, 7, 9, 10, 13, 20

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 9\end{aligned}$$

Hence, the median of the given data is 9.

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

17, 18, 19, 19, 20, 20, 21, 22, 23, 25

As the number of terms ($n$) is 10 and 10 is even :

$$\begin{aligned}\therefore \text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(20 + 20) \\[1em] = \frac{1}{2} \times 40 \\[1em] = 20\end{aligned}$$

Hence, the median of the given data is 20.

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

1.9, 2.7, 3.6, 3.8, 5.6, 6.5, 7.6, 8.9, 9.4, 10.8, 15.6

Since, the number of terms ($n$) is 11 and 11 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{11 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 6^{\text{th}}\text{ term} \\[1em] = 6.5\end{aligned}$$

Hence, the median of the given data is 6.5.

Making the table as shown below :

Term (x)	Frequency (f)	f × x
18	3	54
22	5	110
26	10	260
30	2	60
34	8	272
38	2	76
Total	30	832

Sum of all observations = 832

and, number of observations = 30

By formula,

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

Substituting the values, we get :

Required mean = $\frac{832}{30}$

= 27.73 (approx.)

Since, the frequency of the term 26 is maximum, i.e. 10,

∴ Mode = 26

Hence, the mean of the given data is 27.73 (approx.) and the mode is 26.

In the given data, the number 6 appears maximum times, i.e. 5 times.

∴ Mode = 6

Hence, the mode of the given data is 6.

In the given data, the number 10 appears maximum times, i.e. 3 times.

∴ Mode = 10

Hence, the mode of the given data is 10.

Since, the frequency of the term 18 is maximum, i.e. 13,

∴ Mode = 18

Hence, the mode of the given data is 18.

Since, the number of plants having height 41 cm is maximum, i.e. 153,

∴ Mode = 41

Hence, the mode of the given data is 41.

Sum of all observations = (140 + 142 + 135 + 133 + 137 + 150 + 148 + 138) cm

= 1123 cm

and, number of observations = 8

By formula,

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

Substituting the values, we get :

Mean height = $\frac{1123}{8}$ cm

= 140.375 cm

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

133, 135, 137, 138, 140, 142, 148, 150

As the number of terms ($n$) is 8 and 8 is even :

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{8}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{8}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[4^{\text{th}}\text{ term} + 5^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(138 + 140) \\[1em] = \frac{1}{2} \times 278 \\[1em] = 139 \text{ cm}\end{aligned}$$

Hence, the mean height of the girls is 140.375 cm and their median height is 139 cm.

Sum of all observations = 12 + 24 + 24 + 12 + 30 + 12

= 114

and, number of observations = 6

By formula,

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

Substituting the values, we get :

Required mean = $\frac{114}{6}$

= 19

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

12, 12, 12, 24, 24, 30

As the number of terms ($n$) is 6 and 6 is even :

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{6}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{6}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[3^{\text{rd}}\text{ term} + 4^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(12 + 24) \\[1em] = \frac{1}{2} \times 36 \\[1em] = 18\end{aligned}$$

In the given data, the number 12 appears maximum times, i.e. 3 times.

∴ Mode = 12

Hence, the mean is 19, the median is 18 and the mode is 12.

Sum of all observations = 21 + 24 + 21 + 6 + 15 + 18 + 21 + 45 + 9 + 6 + 27 + 15

= 228

and, number of observations = 12

By formula,

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

Substituting the values, we get :

Required mean = $\frac{228}{12}$

= 19

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

6, 6, 9, 15, 15, 18, 21, 21, 21, 24, 27, 45

As the number of terms ($n$) is 12 and 12 is even :

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{12}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{12}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[6^{\text{th}}\text{ term} + 7^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(18 + 21) \\[1em] = \frac{1}{2} \times 39 \\[1em] = 19.5\end{aligned}$$

In the given data, the number 21 appears maximum times, i.e. 3 times.

∴ Mode = 21

Hence, the mean is 19, the median is 19.5 and the mode is 21.

Steps:

1. On a graph paper, draw two mutually perpendicular lines OX and OY which intersect each other at point O. The line OX is taken horizontal and is called the x-axis, whereas the line OY is taken vertical and is called the y-axis.

2. On the x-axis, starting from O, mark points at equal distances. At these points write the names of the different brands of soap given. In this question, the brands are A, B, C, D and E.

3. Along y-axis, mark the heights of the bars (rectangles) in proportion to the given data (number of buyers). Along the y-axis, take 1 cm = 6 buyers.

The required bar graph is as given below:

Steps:

1. On a graph paper, draw two mutually perpendicular lines OX and OY which intersect each other at point O. The line OX is taken horizontal and is called the x-axis, whereas the line OY is taken vertical and is called the y-axis.

2. On the x-axis, starting from O, mark points at equal distances. At these points write the names of the different countries given. In this question, the countries are India, U.K., China, Germany and Sweden.

3. Along y-axis, mark the heights of the bars (rectangles) in proportion to the given data (birth rate per thousand). Along the y-axis, take 1 cm = 5 units.

The required bar graph is as given below:

First five odd natural numbers are 1, 3, 5, 7 and 9.

Sum of all observations = 1 + 3 + 5 + 7 + 9

= 25

and, number of observations = 5

By formula,

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

Substituting the values, we get :

Required mean = $\frac{25}{5}$ = 5

Hence, option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{6 + 8 + 5 + x + 4}{5} = 7 \\[1em] \Rightarrow \frac{23 + x}{5} = 7 \\[1em] \Rightarrow 23 + x = 7 \times 5 \\[1em] \Rightarrow 23 + x = 35 \\[1em] \Rightarrow x = 35 - 23 \\[1em] \Rightarrow x = 12\end{aligned}$$

Hence, option 1 is the correct option.

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

7, 8, 14, 15, 25, 26, 36

Since, the number of terms ($n$) is 7 and 7 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{7 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 4^{\text{th}}\text{ term} \\[1em] = 15\end{aligned}$$

Hence, option 2 is the correct option.

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

2, 4, 4, 6, 7, 8, 9, 10, 12, 14

As the number of terms ($n$) is 10 and 10 is even :

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(7 + 8) \\[1em] = \frac{1}{2} \times 15 \\[1em] = 7.5\end{aligned}$$

Hence, option 4 is the correct option.

In the given data, the number 12 appears maximum times, i.e. 4 times.

∴ Mode = 12

Hence, option 3 is the correct option.

From the given data,

Largest number = 78

Smallest number = 18

By formula,

Range of given numbers = Largest number − Smallest number

Substituting the values, we get :

Range = 78 − 18

= 60

Hence, option 2 is the correct option.

Given,

Mean weight of 10 oranges = 120 g

If the weight of each orange is reduced by 10 g, the new mean weight will also be reduced by 10 g.

∴ New mean weight = (120 − 10) g

= 110 g

Hence, option 2 is the correct option.

Given,

Mean of 5 numbers = 30

Mean of remaining 4 numbers (after excluding one number) = 30 − 3 = 27

By formula,

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

Therefore,

Sum of 5 numbers = 30 × 5

= 150

Sum of remaining 4 numbers = 27 × 4

= 108

Now,

Excluded number = Sum of 5 numbers − Sum of remaining 4 numbers

= 150 − 108

= 42

Hence, option 3 is the correct option.

Given,

Mean of 8 numbers = 25

Mean of 9 numbers (after including one new number) = 25 + 2 = 27

By formula,

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

Therefore,

Sum of 8 numbers = 25 × 8

= 200

Sum of 9 numbers = 27 × 9

= 243

Now,

Included number = Sum of 9 numbers − Sum of 8 numbers

= 243 − 200

= 43

Hence, option 3 is the correct option.

The given data is already in ascending order of their magnitudes :

53, 56, 58, 62, 63, 65, 68, 70, 71, 84

As the number of terms ($n$) is 10 and 10 is even :

$$\begin{aligned}\text{Median} = \frac{1}{2}\Big[\Big(\frac{n}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{n}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}\Big[\Big(\frac{10}{2}\Big)^{\text{th}}\text{ term} + \Big(\frac{10}{2} + 1\Big)^{\text{th}}\text{ term}\Big] \\[1em] = \frac{1}{2}[5^{\text{th}}\text{ term} + 6^{\text{th}}\text{ term}] \\[1em] = \frac{1}{2}(63 + 65) \\[1em] = \frac{1}{2} \times 128 \\[1em] = 64\end{aligned}$$

∴ 100 − median = 100 − 64

= 36

Hence, option 2 is the correct option.

The mean of a group of observations is given by

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

So, Statement 1 is true.

The value of the mean is not necessarily from the observations under consideration.

For example, for the observations 26, 25, 40, 36, 50 and 45,

$$\begin{aligned}\text{Mean} = \frac{26 + 25 + 40 + 36 + 50 + 45}{6} \\[1em] = \frac{222}{6} \\[1em] = 37\end{aligned}$$

which is not one of the given observations.

So, Statement 2 is false.

∴ Statement 1 is true, and Statement 2 is false.

Hence, option 3 is the correct option.

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

1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 6, 7, 7

Since, the number of terms ($n$) is 13 and 13 is odd :

$$\begin{aligned}\text{Median} = \Big(\frac{n + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = \Big(\frac{13 + 1}{2}\Big)^{\text{th}}\text{ term} \\[1em] = 7^{\text{th}}\text{ term} \\[1em] = 4\end{aligned}$$

So, Assertion (A) is true.

If $x_1, x_2, ......., x_n$ is a set of $n$ observations arranged in ascending order and $n$ is odd, then

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

So, Reason (R) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.

In the given data, the number 4 appears maximum times, i.e. 5 times.

∴ Mode = 4

So, Assertion (A) is true.

Mode is the value of the observation that occurs most frequently. The value which divides the arranged data into two equal parts, with half of the observations below it and the other half above it, is the median and not the mode.

So, Reason (R) is false.

∴ A is true, R is false.

Hence, option 1 is the correct option.

First 5 odd natural numbers are 1, 3, 5, 7 and 9.

$$\begin{aligned}\text{Mean} = \frac{1 + 3 + 5 + 7 + 9}{5} \\[1em] = \frac{25}{5} \\[1em] = 5.\end{aligned}$$

5 is an odd number and not an even number.

So, Assertion (A) is false.

The value of the mean is not necessarily from the observations under consideration.

For example, the mean of 26, 25, 40, 36, 50 and 45 is 37, which is not one of these observations.

So, Reason (R) is true.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Given,

Mean of 8 observations = 64

If 4 is subtracted from each observation, the new mean will also be decreased by 4.

∴ New mean = 64 − 4

= 60

So, Assertion (A) is true.

If the mean of $n$ observations is $M$ and each observation is decreased by $k$, then the mean of the resulting observations will be $M - k$.

So, Reason (R) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.