Statistics

Mean (M) = $\frac{\text{Sum of the all observations}}{\text{No. of observation}}$

Sum of all observations = 8 + 6 + 10 + 12 + 1 + 3 + 4 + 4 = 48.

$M = \frac{48}{8} = 6.$

Hence, mean = 6.

Mean (M) = $\frac{\text{Total social work hrs}}{\text{No. of people}}$

Sum of hours spent by people on social work = 10 + 7 + 13 + 20 + 15 = 65.

$M = \frac{65}{5} = 13.$

Hence, the mean time in a week devoted by people for social work is 13 hours.

Mean (M) = $\frac{\text{Sum of enrollments}}{\text{No. of years}}$

Sum of enrollments = 1620 + 2060 + 2540 + 3250 + 3500 + 3710 = 16680

$M = \frac{16680}{6} = 2780.$

Hence, the mean enrollment = 2780.

Mean (M) = $\frac{\text{Sum of first 12 natural nos.}}{\text{No. of natural nos.}}$

Sum of first 12 natural numbers = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 78.

$M = \frac{78}{12} = 6.5$

Hence, the mean of the first twelve natural numbers = 6.5

First six prime numbers = 2, 3, 5, 7, 11, 13.

Mean (M) = $\frac{\text{Sum of prime numbers}}{\text{No. of prime numbers}}$

Sum of prime numbers = 2 + 3 + 5 + 7 + 11 + 13 = 41.

$M = \frac{41}{6}$.

Hence, the mean of the first six prime numbers = $\frac{41}{6}$.

First seven odd prime numbers = 3, 5, 7, 11, 13, 17, 19.

Mean (M) = $\frac{\text{Sum of first seven odd prime nos.}}{\text{No. of prime nos.}}$

Sum of first seven odd prime numbers = 3 + 5 + 7 + 11 + 13 + 17 + 19 = 75.

$$\begin{aligned}M = \frac{75}{7} \\[1em] = 10\frac{5}{7}.\end{aligned}$$

Hence, the mean of the first seven odd prime numbers = $10\frac{5}{7}$.

Mean marks (M) = $\frac{\text{Total marks}}{\text{No. of students}}$

Total marks = 81 + 72 + 90 + 90 + 85 + 86 + 70 + 93 + 71 = 738.

$M = \frac{738}{9} = 82.$

Hence, mean marks = 82.

Let age of Vijay, Rahul and Rakhi be 4x, 5x and 6x respectively..

Mean age = $\frac{\text{Sum of age of students}}{\text{No. of students}}$

Sum of age of students = 4x + 5x + 6x = 15x.

$$\begin{aligned}\therefore 15 = \frac{15x}{3} \\[1em] \Rightarrow 15 \times 3 = 15x \\[1em] \Rightarrow x = \frac{15 \times 3}{15} \\[1em] \Rightarrow x = 3.\end{aligned}$$

Vijay's age = 4x = 4 × 3 = 12 years,

Rahul's age = 5x = 5 × 3 = 15 years,

Rakhi's age = 6x = 6 × 3 = 18 years.

Hence, age of three students Vijay, Rahul and Rakhi are 12, 15 and 18 years respectively.

Given, mean of 5 numbers = 20.

By formula,

Mean = $\frac{\text{Total sum of observations}}{\text{No. of observations}}$

$$\begin{aligned}\therefore 20 = \frac{\text{Total sum of observations}}{5} \\[1em] \text{Total sum of observations} = 20 \times 5 = 100.\end{aligned}$$

Let number excluded be x.

Given, Mean of the remaining numbers becomes 23.

$$\begin{aligned}\therefore 23 = \frac{100 - x}{4} \\[1em] \Rightarrow 23 \times 4 = 100 - x \\[1em] \Rightarrow 92 = 100 - x \\[1em] \Rightarrow x = 100 - 92 \\[1em] \Rightarrow x = 8.\end{aligned}$$

Hence, excluded number = 8.

Given,

The mean of 25 observations is 27.

By formula,

Mean = $\frac{\text{Total sum of observations}}{\text{No. of observations}}$

$$\begin{aligned}\therefore 27 = \frac{\text{Total sum of observations}}{25} \\[1em] \text{Total sum of observations} = 27 \times 25 = 675.\end{aligned}$$

Given, mean after including new number remains 27.

No. of observations = 25 + 1 = 26.

$$\begin{aligned}\therefore 27 = \frac{\text{New sum of observations}}{26} \\[1em] \text{New sum of observations} = 27 \times 26 = 702.\end{aligned}$$

New number included = New sum of observations - Total sum of observations = 702 - 675 = 27.

Hence, the new included observation = 27.

By formula,

Mean = $\frac{\text{Total sum of observations}}{\text{No. of observations}}$

Given,

The mean of 5 observations = 15

$$\begin{aligned}\therefore 15 = \frac{\text{Total sum of 5 observations}}{5} \\[1em] \Rightarrow \text{Total sum of observations} = 15 \times 5 \\[1em] \Rightarrow \text{Total sum of observations} = 75.\end{aligned}$$

Given,

Mean of first 3 observations = 14

$$\begin{aligned}\therefore 14 = \frac{\text{Sum of first 3 observations}}{3} \\[1em] \Rightarrow \text{Sum of first three observations} = 14 \times 3 \\[1em] \Rightarrow \text{Sum of first three observations} = 42.\end{aligned}$$

Given,

Mean of last 3 observations = 17

$$\begin{aligned}\therefore 17 = \frac{\text{Sum of last 3 observations}}{3} \\[1em] \Rightarrow \text{Sum of last three observations} = 17 \times 3 \\[1em] \Rightarrow \text{Sum of last three observations} = 51.\end{aligned}$$

Third observation = Sum of first three observations + Sum of last three observations - Total sum of observation
= 51 + 42 - 75 = 18.

Hence, the third observation = 18.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

Given,

Mean of 8 variates = 10.5

$$\begin{aligned}\therefore 10.5 = \frac{\text{Sum of eight variates}}{8} \\[1em] \text{Sum of 8 variates} = 10.5 \times 8 = 84.\end{aligned}$$

Given,

Seven variates = 3, 15, 7, 19, 2, 17 and 8.

Sum of seven variates = (3 + 15 + 7 + 19 + 2 + 17 + 8) = 71.

8th variate = Sum of 8 variates - Sum of 7 variates = 84 - 71 = 13.

Hence, 8th variate = 13.

By formula,

Mean = $\frac{\text{Total weight of students}}{\text{No. of students}}$

Given,

$$\begin{aligned}\therefore 45.5 = \frac{\text{Total weight of eight students}}{8} \\[1em] \text{Total weight of eight students} = 45.5 \times 8 = 364.\end{aligned}$$

Weight of two more students are 41.7 kg and 53.3 kg

Now,

The total weight of 10 students = 364 + 41.7 + 53.3 = 459 kg.

New mean weight = $\frac{459}{10}$ = 45.9 kg.

Hence, new mean weight = 45.9 kg.

By formula,

Mean = $\frac{\text{Total sum of observations}}{\text{No. of observations}}$

$$\begin{aligned}\therefore 35 = \frac{\text{Total sum of observations}}{9} \\[1em] \text{Total sum of observations} = 35 \times 9 = 315.\end{aligned}$$

Since, 81 was misread as 18.

So, Actual sum of observation = Total sum of observations - 18 + 81 = 315 - 18 + 81 = 378.

Correct mean = $\frac{378}{9}$ = 42.

Hence, the correct mean = 42.

Given,

Marks scored in 11 questions of a question paper by the student are:

7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6

Arranging it in ascending order, we have

1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8

Here, n = 11 which is odd.

∴ Median = $\frac{(n + 1)}{2}$ th term

= $\frac{11 + 1}{2} = \frac{12}{2}$ = 6th term.

6th term = 5.

Hence, median marks = 5.

By formula,

$\text{Mean } = \frac{\text{Sum of numbers}}{\text{No. of numbers}}$

Sum of numbers = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24

$$\begin{aligned}\therefore \text{Mean } = \frac{24}{10} \\[1em] = 2.4\end{aligned}$$

Given,

Numbers : 2, 3, 4, 3, 0, 5, 1, 1, 3, 2.

First arrange the numbers in ascending order :

0, 1, 1, 2, 2, 3, 3, 3, 4, 5

Here, n = 10 which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{ th observation} + \Big(\frac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2}\text{ th observation} + \Big(\frac{10}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \frac{5 \text{ th observation} + 6\text{ th observation}}{2} \\[1em] = \frac{2 + 3}{2} \\[1em] = \frac{5}{2} \\[1em] = 2.5\end{aligned}$$

Hence, mean = 2.4 and median = 2.5

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{Total minutes}}{\text{No. of students}} \\[1em] = \frac{24 + 30 + 28 + 17 + 22 + 36 + 30 + 19 + 32 + 18 + 20 + 24}{12} \\[1em] = \frac{300}{12} \\[1em] = 25.\end{aligned}$$

Arranging the data in ascending order :

17, 18, 19, 20, 22, 24, 24, 28, 30, 30, 32, 36

Here, n = 12 which is even

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{ th observation} + \Big(\frac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{\frac{12}{2}\text{ th observation} + \Big(\frac{12}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \frac{6 \text{ th observation} + 7\text{ th observation}}{2} \\[1em] = \frac{24 + 24}{2} \\[1em] = \frac{48}{2} \\[1em] = 24\end{aligned}$$

Hence, mean time = 25 minutes and median time = 24 minutes.

By formula,

$\text{Mean} = \frac{\text{Total marks}}{\text{No. of students}}$

Total marks = 41 + 39 + 52 + 48 + 54 + 62 + 46 + 52 + 40 + 96 + 42 + 40 + 98 + 60 + 52
= 822

$$\begin{aligned}\text{Mean} = \frac{822}{15} \\[1em] = 54.8\end{aligned}$$

Given,

Student's marks : 41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.

On arranging the marks obtained by the students in ascending order, we have :

39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98

Here, n = 15 which is odd.

By formula,

$$\begin{aligned}\text{Median } = \frac{n + 1}{2} \text{ th observation} \\[1em] = \frac{15 + 1}{2} \text{ th observation} \\[1em] = \frac{16}{2} \text{ th observation} \\[1em] = 8 \text{ th observation} \\[1em] = 52.\end{aligned}$$

Hence, mean = 54.8 and median = 52.

Given,

Points scored by kabaddi team : 7, 17, 2, 5, 27, 15, 8, 14, 10, 48, 10, 7, 24, 8, 28, 18.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of points}}{\text{No. of matches}} \\[1em] = \frac{7 + 17 + 2 + 5 + 27+ 15 + 8 + 14 + 10 + 48 + 10 + 7 + 24 + 8 + 28 + 18}{16} \\[1em] = \frac{248}{16} \\[1em] = 15.5\end{aligned}$$

Let’s arrange the given data in ascending order:

2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48.

Here, n = 16 when is even

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{ th observation} + \Big(\frac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{\frac{16}{2}\text{ th observation} + \Big(\frac{16}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \frac{8 \text{ th observation} + 9\text{ th observation}}{2} \\[1em] = \frac{10 + 14}{2} \\[1em] = \frac{24}{2} \\[1em] = 12\end{aligned}$$

Hence, the mean and the median of the points scored by the Kabaddi team are 15.5 and 12 respectively.

Given data,

17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93.

Here, n = 14 which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{ th observation} + \Big(\frac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{\frac{14}{2}\text{ th observation} + \Big(\frac{14}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \frac{7 \text{ th observation} + 8\text{ th observation}}{2} \\[1em] = \frac{x + 50}{2}.\end{aligned}$$

Given, median = 47.5

⇒ $\frac{x + 50}{2}$ = 47.5

⇒ x + 50 = 95

⇒ x = 95 - 50

⇒ x = 45.

Hence, the value of x is 45.

Given observations in ascending order :

3, 6, 7, 10, x, x + 4, 19, 20, 25, 28

Here, n = 10 which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{ th observation} + \Big(\frac{n}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2}\text{ th observation} + \Big(\frac{10}{2} + 1\Big)\text{ th observation}}{2} \\[1em] = \frac{5 \text{ th observation} + 6\text{ th observation}}{2} \\[1em] = \frac{x + x + 4}{2} \\[1em] = \frac{2x + 4}{2} \\[1em] = x + 2.\end{aligned}$$

Given, median = 13

⇒ x + 2 = 13

⇒ x = 13 - 2

⇒ x = 11.

Hence, the value of x is 11.

(i) Discrete

(ii) Continuous

(iii) Discrete

(iv) Continuous

(v) Continuous

The frequency distribution table for given grouped data is :

Classes	Tally Marks	Frequency
0 - 4	II	2
5 - 9	IIII II	7
10 - 14	IIII III	8
15 - 19	IIII I	6
20 - 24	II	2

(i) The frequency distribution table for given grouped data is :

Classes	Tally Marks	Frequency
1 - 10	IIII III	8
11 - 20	IIII II	7
21 - 30	IIII I	6
31 - 40	IIII I	6

(ii) Range = Highest mark - Lowest mark = 40 - 2 = 38.

Hence, range of marks = 38.

(iii) Class mark = $\frac{\text{Lower limit + Upper limit}}{2}$

Third class of frequency table = 21-30.

Substituting values we get :

Class mark = $\frac{21 + 30}{2} = \frac{51}{2}$ = 25.5

Hence, class mark = 25.5

(i) Variant — A particular value of a variable is called variate.

(ii) Class size — The difference between the actual upper limit and the actual lower limit of a class is called its class size.

(iii) Class mark — The class mark of a class is the value midway between its actual lower limit and actual upper limit.

(iv) Class limits — In the frequency table the class interval is called class limits.

(v) True class limits — In a continuous distribution, the class limits are called true or actual class limits.

(vi) Frequency of a class — The number of tally marks opposite to a variate is its frequency and it is written in the next column opposite to tally marks of the variate.

(vii) Cumulative frequency of a class — The sum of frequency of all previous classes and that particular class is called the cumulative frequency of the class.

(i) The number of observations in a particular class is called frequency of the class.

(ii) The difference between the class marks of two consecutive classes is the size of the class.

(iii) Range = Highest value - Lowest value

= 25 - 11 = 14.

The range of the data 16, 19, 23, 13, 11, 25, 18 is 14.

(iv) The mid-point of the class interval is called its class marks.

(v) Class mark = $\frac{4 + 9}{2} = \frac{13}{2}$ = 6.5

The class mark of the class 4 – 9 is 6.5.

(i) The frequency distribution table for given grouped data is :

Classes	Tally Marks	Frequency
1 - 10	IIII II	7
11 - 20	IIII III	8
21 - 30	IIII II	7
31 - 40	IIII IIII	10
41 - 50	IIII III	8

(ii) Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

$= \frac{11 - 10}{2} = \frac{1}{2}$ = 0.5

Subtracting the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for the given data is :

Classes before adjustment	Classes after adjustment	Frequency
1 - 10	0.5 - 10.5	7
11 - 20	10.5 - 20.5	8
21 - 30	20.5 - 30.5	7
31 - 40	30.5 - 40.5	10
41 - 50	40.5 - 50.5	8

(iii) Third class : 20.5 - 30.5

Hence, lower limit : 20.5 and upper limit : 30.5

(iv) Class mark = $\frac{\text{Lower limit + Upper limit}}{2}$

= $\frac{30.5 + 40.5}{2} = \frac{71}{2}$ = 35.5

Hence, class mark of fourth class = 35.5

(i) Fifth class : 48 - 52

Hence, upper limit = 52 and lower limit = 48.

(ii) Adjustment factor = (Lower limit of class - Upper limit of previous class) / 2

= $\frac{48 - 47}{2} = \frac{1}{2}$ = 0.5

True lower limit = 48 - 0.5 = 47.5

True upper limit = 52 + 0.5 = 52.5

Hence, true class limits of fifth class = 47.5 and 52.5

(iii) Third class : 38 - 42

True lower limit = 38 - 0.5 = 37.5

True upper limit = 42 + 0.5 = 42.5

Hence, class boundaries of third class = 37.5 and 42.5

(iv) Fourth class : 43 - 47

Class mark = $\frac{\text{Lower class limit + Upper class limit}}{2}$

= $\frac{43 + 47}{2} = \frac{90}{2}$

= 45.

Hence, class mark of fourth class = 45.

(v) Fifth class : 53 - 57

True lower limit = 53 - 0.5 = 52.5

True upper limit = 57 + 0.5 = 57.5

Hence, width = 5.

The cumulative frequency table is shown below :

Marks %	Frequency	Cumulative frequency
10 - 19	7	7
20 - 29	11	18 (11 + 7)
30 - 39	20	38 (20 + 18)
40 - 49	46	84 (46 + 38)
50 - 59	57	141 (57 + 84)
60 - 69	37	178 (37 + 141)
70 - 79	15	193 (15 + 178)
80 - 89	7	200 (7 + 193)

The cumulative frequency distribution table is given below:

Classes	Frequency	Cumulative frequency
0 - 10	5	5
10 - 20	5	10 (5 + 5)
20 - 30	7	17 (7 + 10)
30 - 40	10	27 (10 + 17)
40 - 50	8	35 (8 + 27)

From table,

The no. of students obtaining below 20 marks = 10.

(i) The cumulative frequency table for given data is :

Classes	Tally Marks	Frequency	Cumulative frequency
0 - 10	III	3	3
10 - 20	II	2	5 (2 + 3)
20 - 30	IIII IIII I	11	16 (11 + 5)
30 - 40	IIII IIII IIII III	18	34 (18 + 16)
40 - 50	-	0	34 (0 + 34)
50 - 60	III	3	37 (3 + 34)
60 - 70	II	2	39 (2 + 37)
70 - 80	III	3	42 (3 + 39)
80 - 90	IIII	5	47 (5 + 42)
90 - 100	III	3	50 (3 + 47)

(ii) From table,

Class 30 - 40 has the largest frequency distribution table.

(iii) From table,

34 students score less than 40 marks.

(iv) Total marks = 100

60% marks = $\frac{60}{100} \times 100$ = 60.

From tables, students scoring less than 60 = 37

Total students = 50.

Students scoring more than 60% = 50 - 37 = 13.

Hence, 13 students scored more than 60% marks.

The frequency distribution table for given data is :

Class	Cumulative frequency	Frequency
0 - 4	7	7
4 - 7	38	31 (38 - 7)
7 - 10	175	137 (175 - 38)
10 - 13	248	73 (248 - 175)
13 - 16	300	52 (300 - 248)

From table,

Frequency of class 10 - 13 = 73.

Hence, the no. of children in the age group 10 - 13 are 73.

The frequency distribution table for given data is :

Class	Cumulative frequency	Frequency
0 - 10	2	2
11 - 20	7	5 (7 - 2)
21 - 30	18	11 (18 - 7)
31 - 40	32	14 (32 - 18)
41 - 50	43	11 (43 - 32)
51 - 60	50	7 (50 - 43)

Here, maximum = 43.2 and minimum = 23.6.

Range = 43.2 - 23.6 = 19.6

Let us form 5 classes of each size 4.

Since, we want to score 43.2 in last class and 43.5 is the upper limit of the last class, so the lower limit of first class is 23.5

The frequency distribution table of the given data is as follows :

Classes	Tally marks	Frequency
23.5 - 27.5	IIII	4
27.5 - 31.5	IIII II	7
31.5 - 35.5	IIII	4
35.5 - 39.5	IIII	4
39.5 - 43.5	IIII IIII I	11

Class size = Difference between two successive class marks = 104 - 94 = 10.

Class mark of fourth class = 124.

Lower limit of class = Class mark - (Class size/2)

= 124 - $\frac{10}{2}$

= 124 - 5

= 119.

Upper class limit = Class mark + (Class size/2)

= 124 + $\frac{10}{2}$

= 124 + 5

= 129.

Hence, class size = 10, lower limit = 119 and upper limit = 129.

Class size = Difference between two successive class marks = 16.5 - 9.5 = 7.

Class mark of third class = 23.5

Lower limit of class = Class mark - (Class size/2)

= 23.5 - $\frac{7}{2}$

= 23.5 - 3.5

= 20.

Upper class limit = Class mark + (Class size/2)

= 23.5 + $\frac{7}{2}$

= 23.5 + 3.5

= 27.

Hence, class size = 7, lower limit = 20 and upper limit = 27.

Steps of construction :

1. Take states along x-axis.

2. Take 1 cm along y-axis = 20 lac hectares.

3. Construct rectangles corresponding to the above distribution table.

The required bar graph is shown in the adjoining figure.

Steps of construction :

1. Take days along x-axis.

2. Take 1 cm along y-axis = 50 books.

3. Construct rectangles corresponding to the above distribution table.

The required bar graph is shown in the adjoining figure.

Steps of construction :

1. Take year along x-axis.

2. Take 1 cm along y-axis = 10 %.

3. Construct rectangles corresponding to the above distribution table.

The required bar graph is shown in the adjoining figure.

Steps of construction :

1. Take country along y-axis.

2. Take birth rate along x-axis.

3. Construct rectangles corresponding to the above distribution table.

The required bar graph is shown in the adjoining figure.

Steps of construction :

1. Take class along x-axis.

2. Take 1 cm along y-axis = 5 students.

3. Construct rectangles corresponding to the above distribution table.

The required bar graph is shown in the adjoining figure.

Steps of construction of histogram :

1. Take 2 cm along x-axis = 10 marks.

2. Take 1 cm along y-axis = 1 student.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of histogram :

1. Since the scale on x-axis starts at 850, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 850.

2. Take 2 cm along x-axis = 100 rupees.

3. Take 1 cm along y-axis = 10 workers.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

We know that class mark is the mid-point of class. So, frequency distribution table for above data is :

Class marks	Class	Frequency
12.5	10 - 15	7
17.5	15 - 20	12
22.5	20 - 25	20
27.5	25 - 30	28
32.5	30 - 35	8
37.5	35 - 40	11

Steps of construction of histogram :

1. Since the scale on x-axis starts at 10, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 10.

2. Take 2 cm along x-axis = 5 units.

3. Take 1 cm along y-axis = 5 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Frequency distribution table for above data is :

Age (in years)	No. of children (Cumulative frequency)	Frequency
0 - 2	12	12
2 - 4	15	3 (15 - 12)
4 - 6	36	21 (36 - 15)
6 - 8	45	9 (45 - 36)
8 - 10	72	27 (72 - 45)
10 - 12	90	18 (90 - 72)

Steps of construction of histogram :

1. Take 2 cm along x-axis = 2 years.

2. Take 1 cm along y-axis = 3 children.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{66 - 65}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustment	Classes after adjustment	Frequency
59 - 65	58.5 - 65.5	10
66 - 72	65.5 - 72.5	5
73 - 79	72.5 - 79.5	25
80 - 86	79.5 - 86.5	15
87 - 93	86.5 - 93.5	30
94 - 100	93.5 - 100.5	10

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 58.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 58.5

2. Take 2 cm along x-axis = 7 units.

3. Take 1 cm along y-axis = 5 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Frequency distribution table :

Class intervals	Class marks	Frequency
40 - 50	45	15
50 - 60	55	28
60 - 70	65	45
70 - 80	75	32
80 - 90	85	41
90 - 100	95	18

Steps to draw frequency polygon :

1. Since, the scale on x-axis starts at 30, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 30.

2. Take 1 cm along x-axis = 10 units.

3. Take 1 cm along y-axis = 5 units.

4. Find the mid-points of class-intervals.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 30 - 40 with zero frequency and join the other end with mid-point of class 100 - 110 with zero frequency.

The required frequency polygon is shown below:

Frequency distribution table :

Marks	Class marks	Students
10 - 20	15	10
20 - 30	25	25
30 - 40	35	12
40 - 50	45	08
50 - 60	55	05

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 10 marks.

2. Take 1 cm along y-axis = 5 students.

3. Find the mid-points of class-intervals.

4. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

5. Join consecutive points by line segments.

6. Join first end point with mid-point of class 0 - 10 with zero frequency and join the other end with mid-point of class 60 - 70 with zero frequency.

The required frequency polygon is shown below:

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{21 - 20}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustment	Classes after adjustment	Class mark	Frequency
16 - 20	15.5 - 20.5	18	4
21 - 25	20.5 - 25.5	23	12
26 - 30	25.5 - 30.5	28	18
31 - 35	30.5 - 35.5	33	26
36 - 40	35.5 - 40.5	38	14
41 - 45	40.5 - 45.5	43	10
46 - 50	45.5 - 50.5	48	6

Steps to draw frequency polygon :

1. Since, the scale on x-axis starts at 10.5, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 10.5.

2. Take 1 cm along x-axis = 5 marks.

3. Take 1 cm along y-axis = 5 students.

4. Find the mid-points of class-intervals.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 10.5 - 15.5 with zero frequency and join the other end with mid-point of class 50.5 - 55.5 with zero frequency.

The required frequency polygon is shown alongside.

Frequency distribution table :

Cost of living index	Class marks	Number of weeks
140 - 150	145	5
150 - 160	155	10
160 - 170	165	20
170 - 180	175	9
180 - 190	185	6
190 - 200	195	2

Steps to draw frequency polygon :

1. Since, the scale on x-axis starts at 130, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 130.

2. Take 2 cm along x-axis = 10 units (cost of living index).

3. Take 2 cm along y-axis = 5 weeks.

4. Find the mid-points of class-intervals.

5. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

6. Join consecutive points by line segments.

7. Join first end point with mid-point of class 130 - 140 with zero frequency and join the other end with mid-point of class 200 - 210 with zero frequency.

The required frequency polygon is shown alongside.

Frequency distribution table :

Weekly earnings	Class marks	No. of workers
150 - 165	157.5	8
165 - 180	172.5	14
180 - 195	187.5	22
195 - 210	202.5	12
210 - 225	217.5	15
225 - 240	232.5	6

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 135, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 135.

2. Take 2 cm along x-axis = 15 rupees.

3. Take 2 cm along y-axis = 5 workers.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 135 - 150 with zero frequency, and join the other end point with the mid-point of class 240 - 255 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

Frequency distribution table :

Age (in years)	Class marks	No. of patients
10 - 20	15	3
20 - 30	25	8
30 - 40	35	30
40 - 50	45	36
50 - 60	55	27
60 - 70	65	15
70 - 80	75	6

Steps of construction of histogram :

1. Take 1 cm along x-axis = 10 years.

2. Take 1 cm along y-axis = 6 patients.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 0 - 10 with zero frequency, and join the other end point with the mid-point of class 80 - 90 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

First represent the given data in the form of a frequency distribution table.

Here, class intervals represent water bills class and frequency represents no. of houses.

Class Intervals	Class marks	Tally Marks	Frequency
30 - 40	35	III	3
40 - 50	45	I	1
50 - 60	55	IIII	4
60 - 70	65	IIII	5
70 - 80	75	IIII IIII	9
80 - 90	85	III	3
90 - 100	95	III	3
100 - 110	105	IIII	4

Steps of construction of histogram :

1. Take 1 cm along x-axis = 10 units.

2. Take 1 cm along y-axis = 1 unit.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 20 - 30 with zero frequency, and join the other end point with the mid-point of class 110 - 120 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

First represent the given data in a frequency distribution table as shown below:

Class Intervals	Class marks	Tally Marks	Frequency
40 - 42	41	II	2
42 - 44	43	IIII IIII II	12
44 - 46	45	IIII II	7
46 - 48	47	IIII I	6
48 - 50	49	IIII II	7
50 - 52	51	IIII	4
52 - 54	53	II	2

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 38, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 38.

2. Take 1 cm along x-axis = 2 units.

3. Take 1 cm along y-axis = 2 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 38 - 40 with zero frequency, and join the other end point with the mid-point of class 54 - 56 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

(i) From graph,

The frequency of class 400 - 425 is 18.

(ii) From graph,

Minimum frequency of a class = 4.

Hence, the class having minimum frequency is 475 - 500.

(iii) Cumulative frequency of class 425 - 450 = Sum of frequency of class 425 - 450 and previous classes

= 10 + 18 + 6 = 34.

Hence, cumulative frequency of class 425 - 450 = 34.

(iv) The cumulative frequency distribution table for given distribution is shown below :

Classes	Frequency	Cumulative frequency
375 - 400	6	6
400 - 425	18	24 (6 + 18)
425 - 450	10	34 (24 + 10)
450 - 475	20	54 (34 + 20)
475 - 500	4	58 (54 + 4)

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

Adjustment factor = $\frac{\text{(Lower limit of one class - Upper limit of previous class)}}{2}$

= $\frac{\text{(6 - 5)}}{2} = \frac{1}{2} = 0.5$

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustment	Classes after adjustment	Class marks	No. of students of 10A	No. of students of 10B
1 - 5	0.5 - 5.5	3	1	2
6 - 10	5.5 - 10.5	8	3	6
11 - 15	10.5 - 15.5	13	8	3
16 - 20	15.5 - 20.5	18	9	10
21 - 25	20.5 - 25.5	23	4	7
26 - 30	25.5 - 30.5	28	5	6
31 - 35	30.5 - 35.5	33	6	4
36 - 40	35.5 - 40.5	38	4	2

Steps to draw frequency polygon :

1. Take 1 cm along x-axis = 4 units.

2. Take 1 cm along y-axis = 1 units.

3. Find the mid-points of class-intervals.

4. Find points corresponding to given frequencies of classes and the mid-points of class-intervals, and plot them.

5. Join consecutive points by line segments.

6. Join first end point with mid-point of class -5.5 - 0.5 with zero frequency and join the other end with mid-point of class 40.5 - 45.5 with zero frequency.

The required frequency polygon is shown alongside.

Range = Upper limit - Lower limit = 100 - 46 = 54.

Hence, Option 2 is the correct option.

Class mark = $\frac{\text{Upper limit + Lower limit}}{2}$

= $\frac{90 + 120}{2} = \frac{210}{2}$

= 105.

Hence, Option 2 is the correct option.

Lower limit of class = Mid value - $\frac{\text{Width}}{2}$

= 10 - $\frac{6}{2}$

= 10 - 3

= 7.

Hence, Option 2 is the correct option.

Let x and y be the upper and lower class limit of frequency distribution.

Given,

Width of class = 5

∴ x - y = 5

⇒ x - 10 = 5

⇒ x = 5 + 10 = 15.

Upper class limit of highest class = No. of continuous class × Width + Lower class limit of lowest class

= 5 × 5 + 10

= 25 + 10

= 35.

Hence, Option 3 is the correct option.

As the class marks are 15, 20, 25, ......... which are at equal gaps, so the classes are of equal size.

∴ size of class = difference between two consecutive class marks
= 20 - 15 = 5.

Half of class size = $\frac{5}{2}$ = 2.5

∴ Lower limit of class corresponding to the class mark 20 = class mark - half of class size
= 20 - 2.5 = 17.5

∴ Upper limit of class corresponding to the class mark 20 = class mark + half of class size
= 20 + 2.5 = 22.5

∴ Class corresponding to the class mark 20 is 17.5 - 22.5

Hence, Option 2 is the correct option.

The number 20 will be included in 20 - 30 interval.

Hence, Option 2 is the correct option.

Elements from above data in the class 310 - 330 are :

310, 310, 320, 319, 318, 316.

Hence, the frequency of class 310 - 330 is 6.

Hence, Option 3 is the correct option.

By formula,

$\text{Mean } = \frac{\text{Sum of observations}}{\text{No. of observations}}$

Sum of observations = (x - 1) + (x + 1) + (x + 3) + (x + 5) = 4x + 8.

$$\begin{aligned}\text{Mean } = \frac{4x + 8}{4} \\[1em] = \frac{4(x + 2)}{4} \\[1em] = x + 2.\end{aligned}$$

Hence, Option 2 is the correct option.

By formula,

$$\begin{aligned}\text{Mean } = \frac{\text{Sum of observations}}{\text{No. of observations}} \\[1em] 30 = \frac{\text{Sum of observations}}{5} \\[1em] \text{Sum of observations} = 150\end{aligned}$$

Let excluded number be x.

Given, on excluding x the mean is 28.

$$\begin{aligned}\therefore \frac{150 - x}{4} = 28 \\[1em] \Rightarrow 150 - x = 112 \\[1em] \Rightarrow x = 150 - 112 \\[1em] \Rightarrow x = 38.\end{aligned}$$

Hence, Option 4 is the correct option.

By formula,

$\text{Mean} = \frac{\text{Sum of observations}}{\text{No. of observations}}$

Given, Mean of x₁, x₂ = 7.5

$$\begin{aligned}\therefore 7.5 = \frac{x_1 + x_2}{2} \\[1em] \Rightarrow x_1 + x_2 = 15 ...........(i)\end{aligned}$$

Given, Mean of x₁, x₂ and x₃ = 8

$$\begin{aligned}\therefore 8 = \frac{x_1 + x_2 + x_3}{3} \\[1em] \Rightarrow x_1 + x_2 + x_3 = 24 ...........(ii)\end{aligned}$$

Subtracting equation (i) from (ii), we get :

⇒ x₁ + x₂ + x₃ - (x₁ + x₂) = 24 - 15

⇒ x₁ - x₁ + x₂ - x₂ + x₃ = 9

⇒ x₃ = 9.

Hence, Option 1 is the correct option.

Let x₁, x₂, x₃ be the observations.

So, their mean (M) = $\frac{x_1 + x_2 + x_3}{3}$

Since, each observations is increased by 5, so observations will be

x₁ + 5, x₂ + 5, x₃ + 5.

$$\begin{aligned}\text{Mean (M}_1) = \frac{x_1 + 5 + x_2 + 5 + x_3 + 5}{3} \\[1em] = \frac{x_1 + x_2 + x_3 + 15}{3} \\[1em] = \frac{x_1 + x_2 + x_3}{3} + \frac{15}{3} \\[1em] = \frac{x_1 + x_2 + x_3}{3} + 5.\end{aligned}$$

So, M₁ = M + 5.

Hence, Option 4 is the correct option.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

Given,

Mean of 100 observations is 50.

$$\begin{aligned}\therefore 50 = \frac{\text{Sum of observations}}{100} \\[1em] \text{Sum of observations} = 50 \times 100 = 5000.\end{aligned}$$

Since, observation which was 50 is replaced by 150.

So,

Sum of observation = 5000 - 50 + 150 = 5100.

New mean = $\frac{5100}{100}$ = 51.

Hence, Option 2 is the correct option.

We know, class marks are the mean of lower and upper limit of the class intervals.

Hence, Option 3 is the correct option.

Arranging given data in ascending order :

3, 4, 4, 5, 6, 7, 7, 7, 12.

No. of observation (n) = 9.

Here, n is odd.

By formula,

$$\begin{aligned}\text{Median} = \frac{(n + 1)}{2} \text{th observation} \\[1em] = \frac{9 + 1}{2} \text{th observation} \\[1em] = \frac{10}{2} \text{th observation} \\[1em] = 5 \text{th observation} \\[1em] = 6.\end{aligned}$$

Hence, Option 3 is the correct option.

Arranging given data in ascending order :

22, 34, 39, 45, 54, 54, 56, 68, 78, 84.

No. of observation (n) = 10.

Here, n is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2} \text{th observation} + \Big(\frac{n}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \Big(\frac{10}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \frac{\text{5th observation + 6th observation}}{2} \\[1em] = \frac{54 + 54}{2} \\[1em] = \frac{108}{2} \\[1em] = 54.\end{aligned}$$

Hence, Option 3 is the correct option.

Since, number of observations (n) = 10, which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2} \text{th observation} + \Big(\frac{n}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \Big(\frac{10}{2} + 1\Big) \text{th observation}}{2} \\[1em] = \frac{\text{5th observation + 6th observation}}{2}\end{aligned}$$

Since, 5th and 6th observation is not changed at all, so there is no change in median.

Hence, Option 1 is the correct option.

In a histogram, there are no gaps between the bars, indicating the continuous nature of the data.

The width of each bar (its base) represents the class interval or class size.

The height of each bar represents the frequency (or count) of observations falling within that specific class interval.

∴ Statement 1 is true.

In a bar graph, the bars have arbitrary width and does not represent any numerical quantity.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.

Range = Maximum Value - Minimum Value.

∴ Reason (R) is true.

For Batsman A:

Runs: 55, 60, 30, 80

Minimum value (Min) = 30

Maximum value (Max) = 80

Range of A = Max − Min = 80 − 30 = 50.

For Batsman B:

Runs: 81, 87, 76, 92

Minimum value (Min) = 76

Maximum value (Max) = 92

Range of B = Max − Min = 92 − 76 = 16.

Thus, B does not have higher range than A.

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

The first 5 odd natural numbers are: 1, 3, 5, 7, 9.

The mean (arithmetic average) is calculated as the sum of all observations divided by the number of observations.

Sum = 1 + 3 + 5 + 7 + 9 = 25

Number of observations = 5

Mean = $\frac{\text{Sum of observations}}{\text{Number of observations}} = \frac{25}{5}$ = 5

∴ Assertion (A) is true.

The mean is the arithmetic average (sum of values divided by the count).

The median is the middle value of a set of observations when they are arranged in ascending or descending order.

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

According to assertion: if 6 students score 78, 62, 91, 37, 80 and 66 marks in a subject, then median score is 72.

Arrange the scores in ascending order: 37, 62, 66, 78, 80, 91

There are 6 scores, so n = 6. This is an even number.

If number of observations is even, then

Median = $\frac{\frac{n}{2}\text{th observation} + \Big(\frac{n}{2} + 1\Big)\text{th observation}}{2}$, after putting all observations in ascending or descending order.

∴ Reason (R) is true.

Substituting the value,

$$\begin{aligned}\text{Median }= \frac{\frac{6}{2}\text{th observation} + \Big(\frac{6}{2} + 1\Big)\text{th observation}}{2}\\[1em] = \frac{3\text{rd observation} + (3 + 1)\text{th observation}}{2}\\[1em] = \frac{3\text{th observation} + 4\text{th observation}}{2}\\[1em] = \frac{66 + 78}{2}\\[1em] = \frac{144}{2}\\[1em] = 72.\end{aligned}$$

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

Sum of observations = 8 + 0 + 5 + 3 + 2 + 9 + 1 + 5 + 4 + 7 + 2 + 5 = 51.

$\text{Mean} = \frac{51}{12} = 4.25$

By arranging data in ascending order, we get :

0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9

Here, n = 12 which is even.

By formula,

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

Substituting the values we get,

$$\begin{aligned}\text{Median} = \frac{\frac{12}{2}\text{th observation} + \Big(\frac{12}{2} + 1\Big)\text{th observation}}{2} \\[1em] = \frac{\text{6th observation + 7th observation}}{2} \\[1em] = \frac{4 + 5}{2} \\[1em] = \frac{9}{2} \\[1em] = 4.5\end{aligned}$$

Hence, mean = 4.25 and median = 4.5

Positive factors of 48 :

1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

By formula,

$\text{Mean} = \frac{\text{Sum of positive factors of 48}}{\text{No. of factors}}$

Sum of positive factors of 48 = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 + 48 = 124.

Mean $= \frac{124}{10} = 12.4$

Here, n = 10 which is even.

By formula,

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

Substituting the values we get,

$$\begin{aligned}\text{Median} = \frac{\frac{10}{2}\text{th observation} + \Big(\frac{10}{2} + 1\Big)\text{th observation}}{2} \\[1em] = \frac{\text{5th observation + 6th observation}}{2} \\[1em] = \frac{6 + 8}{2} \\[1em] = \frac{14}{2} \\[1em] = 7\end{aligned}$$

Hence, mean = 12.4 and median = 7.

By formula,

Mean = $\frac{\text{Sum of weight of students}}{\text{No. of students}}$

Given,

Mean weight of 60 students of a class = 52.75 kg

$$\begin{aligned}\therefore 52.75 = \frac{\text{Total weight}}{60} \\[1em] \text{Total weight} = 60 \times 52.75 \\[1em] = 3165 \text{ kg}.\end{aligned}$$

Mean weight of 35 students among them = 54 kg.

So, the total weight of 35 students = 54 × 35 = 1890 kg.

Remaining students = 60 – 35 = 25

Total weight of 25 students = 3165 – 1890 = 1275 kg

Mean weight of 25 students = $\frac{1275}{25}$ = 51 kg.

Hence, the mean weight of the remaining students is 51 kg.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{Total age}}{\text{No. of students}} \\[1em] \therefore 14.5 = \frac{\text{Total age}}{18} \\[1em] \text{Total age} = 18 \times 14.5 = 261.\end{aligned}$$

Total age of 20 students = 261 + 15 + 16 = 292 years.

So, mean age of 20 students = $\frac{292}{20}$ = 14.6 years.

Hence, the new mean age is 14.6 years.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

Sum of observations = x + 1 + x + 3 + x + 5 + 2x + 2 + 3x + 3 = 8x + 14.

Given,

Mean of the five observations = 14.

$$\begin{aligned}\Rightarrow \frac{8x + 14}{5} = 14 \\[1em] \Rightarrow 8x + 14 = 70 \\[1em] \Rightarrow 8x = 56 \\[1em] \Rightarrow x = \frac{56}{8} = 7.\end{aligned}$$

First three observations are

⇒ x + 1 = 7 + 1 = 8,

⇒ x + 3 = 7 + 3 = 10,

⇒ x + 5 = 7 + 5 = 12.

Mean = $\frac{8 + 10 + 12}{3} = \frac{30}{3}$ = 10.

Hence, the mean of first three observations is 10.

By formula,

Mean = $\frac{\text{Total height}}{\text{No. of students}}$

Given,

Mean height of 36 students of a class = 150.5 cm

$$\begin{aligned}\therefore 150.5 = \frac{\text{Total height}}{36} \\[1em] \text{Total height} = 150.5 \times 36 \\[1em] = 5418 \text{ cm}.\end{aligned}$$

As, height of one student was wrongly copied as 165 cm instead of 156 cm.

∴ Actual total height = 5418 – 165 + 156 = 5409 cm.

Actual Mean height = $\frac{5409}{36}$ = 150.25

Hence, the correct mean height is 150.25 cm.

By formula,

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

Given,

Mean of 40 items = 35

$$\begin{aligned}\therefore 35 = \frac{\text{Sum of observations}}{40} \\[1em] \text{Sum of observations} = 40 \times 35 = 1400.\end{aligned}$$

Since,

Sum of two items were misread as 36 and 29 instead of 63 and 22.

Actual sum of observation = 1400 - 36 - 29 + 63 + 22

= 1420.

Correct mean = $\frac{1420}{40}$ = 35.5

Hence, the correct mean is 35.5

Here n = 10, which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\frac{n}{2}\text{th observation} + \Big(\frac{n}{2} + 1\Big)\text{th observation}}{2} \\[1em] \therefore 63 = \frac{\frac{10}{2}\text{th observation} + \Big(\frac{10}{2} + 1\Big)\text{th observation}}{2} \\[1em] \Rightarrow 126 = \text{5th observation + 6th observation} \\[1em] \Rightarrow 126 = x + x + 2 \\[1em] \Rightarrow 2x = 126 - 2 \\[1em] \Rightarrow 2x = 124 \\[1em] \Rightarrow x = 62.\end{aligned}$$

Hence, the value of x = 62.

Steps of construction of histogram :

1. Take 2 cm along x-axis = 10 marks.

2. Take 1 cm along y-axis = 5 students.

3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 8, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 8.

2. Take 1 cm along x-axis = 10 marks.

3. Take 1 cm along y-axis = 5 students.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 8 - 14 with zero frequency, and join the other end point with the mid-point of class 62 - 68 wirh zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{101 - 100}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustment	Classes after adjustment	Class mark	Frequency
91 - 100	90.5 - 100.5	95.5	16
101 - 110	100.5 - 110.5	105.5	28
111 - 120	110.5 - 120.5	115.5	44
121 - 130	120.5 - 130.5	125.5	20
131 - 140	130.5 - 140.5	135.5	32
141 - 150	140.5 - 150.5	145.5	12
151 - 160	150.5 - 160.5	155.5	4

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 80.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 80.5

2. Take 1 cm along x-axis = 10 units.

3. Take 1 cm along y-axis = 5 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 80.5 - 90.5 with zero frequency, and join the other end point with the mid-point of class 160.5 - 170.5 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

From data,

Least term = 52 and Greatest term = 129.

Range = Greatest term - Least term = 129 – 52 = 77.

Construct a frequency distribution table:

Class interval	Class mark	Tally numbers	Frequency
50 - 60	55	II	2
60 - 70	65	IIII I	6
70 - 80	75	IIII	5
80 - 90	85	IIII III	8
90 - 100	95	IIII	5
100 - 110	105	IIII II	7
110 - 120	115	III	3
120 - 130	125	IIII	4
Total		40

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 40, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 40.

2. Take 1 cm along x-axis = 10 units.

3. Take 1 cm along y-axis = 2 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 40 - 50 with zero frequency, and join the other end point with the mid-point of class 130 - 140 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

From above data :

Least mark = 19

Greatest marks = 30

Range = 30 – 19 = 11.

Given,

We need to take class intervals as 19 - 20, 21 - 22 etc.

The following frequency distribution is discontinuous, to convert it into continuous frequency distribution,

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\frac{21 - 20}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Continuous frequency distribution for given data is :

Classes before adjustment	Classes after adjustment	Class mark	Frequency
19 - 20	18.5 - 20.5	19.5	3
21 - 22	20.5 - 22.5	21.5	5
23 - 24	22.5 - 24.5	23.5	10
25 - 26	24.5 - 26.5	25.5	10
27 - 28	26.5 - 28.5	27.5	5
29 - 30	28.5 - 30.5	29.5	2

Steps of construction of histogram :

1. Since, the scale on x-axis starts at 16.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 16.5.

2. Take 2 cm along x-axis = 2 units.

3. Take 1 cm along y-axis = 2 units.

4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the adjoining figure.

Steps of construction of frequency polygon :

1. Mark the mid-points of upper bases of rectangles of the histogram.

2. Join the consecutive mid-points by line segments.

3. Join the first end point with the mid-point of class 16.5 - 18.5 with zero frequency, and join the other end point with the mid-point of class 30.5 - 32.5 with zero frequency.

The required frequency polygon is shown by thick line segments in the diagram.

(i) From graph,

Class marks of two successive class are 36 and 30.

Class size = 36 - 30 = 6.

Hence, the class size of each class is 6.

(ii) Lower limit of class = Class mark - $\frac{\text{Class size}}{2}$

= 48 - $\frac{6}{2}$

= 48 - 3

= 45.

Upper limit of class = Class mark + $\frac{\text{Class size}}{2}$

= 48 + $\frac{6}{2}$

= 48 + 3

= 51.

Hence, class whose class mark is 48 is 45 - 51.

(iii) From table,

Class with class mark = 54 has greatest frequency.

Lower limit of class = Class mark - $\frac{\text{Class size}}{2}$

= 54 - $\frac{6}{2}$

= 54 - 3

= 51.

Upper limit of class = Class mark + $\frac{\text{Class size}}{2}$

= 54 + $\frac{6}{2}$

= 54 + 3

= 57.

Hence, class 51 - 57 has highest frequency.

(iv) Frequency table for the given distribution :

Class mark	Class	Frequency
30	27 - 33	4
36	33 - 39	12
42	39 - 45	18
48	45 - 51	6
54	51 - 57	20
60	57 - 63	8