Measures of Central Tendency (Mean)

(i) Given,

10, 4, 6, 12, 9

We know that,

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{10 + 4 + 6 + 12 + 9}{5} \\[1em] = \frac{41}{5} \\[1em] = 8.2\end{aligned}$$

Hence, mean of given numbers = 8.2.

(ii) Given,

0.2, 0.02, 2, 2.02, 1.22, 1.02

We know that,

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{0.2 + 0.02 + 2 + 2.02 + 1.22 + 1.02}{6} \\[1em] = \frac{6.48}{6} \\[1em] = 1.08.\end{aligned}$$

Hence, mean of given numbers = 1.08.

(i) Given,

first eight natural numbers = 1, 2, 3, 4, 5, 6, 7, 8

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \frac{36}{8} \\[1em] = 4.5\end{aligned}$$

Hence, mean of given numbers = 4.5.

(ii) Given,

first five prime numbers = 2, 3, 5, 7, 11

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{2 + 3 + 5 + 7 + 11}{5} \\[1em] = \frac{28}{5} \\[1em] = 5.6\end{aligned}$$

Hence, mean of given numbers = 5.6.

(iii) Given,

first six positive even integers; = 2, 4, 6, 8, 10, 12

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{2 + 4 + 6 + 8 + 10 + 12}{6} \\[1em] = \frac{42}{6} \\[1em] = 7\end{aligned}$$

Hence, mean of given numbers = 7.

(iv) Given,

first five positive integral multiples of 3 = 3, 6, 9, 12, 15

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{3 + 6 + 9 + 12 + 15}{5} \\[1em] = \frac{45}{5} \\[1em] = 9.\end{aligned}$$

Hence, mean of given numbers = 9.

(v) Given,

all factors of 20 = 1, 2, 4, 5, 10, 20

Mean = $\frac{\sum x_i}{n}$

Substitute values, we get:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{1 + 2 + 4 + 5 + 10 + 20}{6} \\[1em] = \frac{42}{6} \\[1em] = 7.\end{aligned}$$

Hence, mean of given numbers = 7.

We know that,

Mean = $\frac{\sum x_i}{n}$

We have,

$$\begin{aligned}\Rightarrow \text{Mean of temperatures } = \frac{\text{Sum of temperatures of all days}}{\text{Number of days}} \\[1em] = \frac{35.5 + 30.8 + 28.3 + 31.1 + 23.8 + 29.9 + 32.7}{7} \\[1em] = \frac{212.1}{7} \\[1em] = 30.3 ^{\circ} F\end{aligned}$$

Hence, mean temperature of the week is 30.3 °F.

(i) We know that,

Mean = $\frac{\sum x_i}{n}$

We have,

$$\begin{aligned}\Rightarrow \text{Mean of marks} = \frac{\text{Sum of marks}}{\text{Number of students}} \\[1em] = \frac{38 + 41 + 36 + 31 + 45 + 38 + 27 + 32 + 29 + 39}{10} \\[1em] = \frac{356}{10} \\[1em] = 35.6\end{aligned}$$

Hence, mean of marks = 35.6.

(ii) If 2 marks are added to each student, the mean also increases by 2.

New mean = 35.6 + 2 = 37.6

Hence, mean when marks are increased by 2 = 37.6

(iii) If 1 mark is deducted to each student, the mean also decreases by 1.

New mean = 35.6 - 1 = 34.6

Hence, mean when 1 mark is deducted = 34.6

(iv) If the marks of each student are halved, the mean is also halved.

New mean = $\frac{35.6}{2}$ = 17.8

Hence, mean when marks are halved = 17.8

We know that,

Mean = $\frac{\sum x_i}{n}$

We have,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{11 + 8 + 13 + 10 + x + 9}{6} \\[1em] \Rightarrow 9.5 = \frac{51 + x}{6} \\[1em] \Rightarrow 6(9.5) = 51 + x \\[1em] \Rightarrow 57 = 51 + x \\[1em] \Rightarrow x = 57 - 51 \\[1em] \Rightarrow x = 6.\end{aligned}$$

Hence, value of x is 6.

Mean = $\frac{\text{Sum of terms}}{\text{Number of terms}}$

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of 10 numbers is 13

∴ Sum of 13 terms = 13 × 10 = 130

Sum of 25 terms = 130 + 270 = 400.

Mean = $\frac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

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

$\therefore 52.75 = \frac{\text{Sum of weight of students}}{60}$

Total weight = 60 × 52.75

= 3165 kg.

Mean weight of 25 students among them = 51 kg.

So, the total weight of 25 students = 51 × 25 = 1275 kg.

Remaining students = 60 – 25 = 35

Total weight of remaining 35 students = 3165 – 1275 = 1890 kg

Mean weight of 35 students = $\frac{1890}{35}$ = 54 kg.

Hence, the mean weight of the remaining students is 54 kg.

Given:

Number of observations = 5

Mean = 18

⇒ Sum of all 5 observations = 5 x 18 = 90

On excluding an observation, the mean of the remaining 4 observations = 16

∵ Sum of all remaining 4 observations = 4 x 16 = 64

⇒ Excluded observation = Sum of all 5 observations - Sum of all remaining 4 observations

= 90 - 64

= 26

Hence, the excluded number is 26.

Age (x)	Number of students (f)	fx
12	6	72
13	8	104
14	5	70
15	7	105
16	9	144
17	5	85
Total	∑ f = 40	∑ fx = 580

We know that,

n = ∑f = 40.

By formula,

Mean = $\frac{\sum fx}{n} = \frac{580}{40}$ = 14.5 years

Hence, mean age of the group is 14.5 years.

Variate (x)	frequency (f)	fx
5	7	35
6	8	48
7	14	98
8	11	88
9	10	90
Total	∑ f = 50	∑ fx = 359

We know that,

n = ∑f = 50.

By formula,

Mean = $\frac{\sum fx}{n} = \frac{359}{50}$ = 7.18

Hence, mean of the frequency distribution is 7.18.

Number of misprints per page (x)	Number of pages (f)	fx
0	154	0
1	95	95
2	36	72
3	7	21
4	6	24
5	2	10
Total	∑ f = 300	∑ fx = 222

We know that,

n = ∑f = 300.

By formula,

Mean = $\frac{\sum fx}{n} = \frac{222}{300}$ = 0.74 per page.

Hence, average number of misprints per page is 0.74.

Category	Wages in ₹/day (x)	Number of workers (f)	fx
A	250	2	500
B	300	4	1200
C	350	8	2800
D	400	12	4800
E	450	10	4500
F	500	6	3000
G	550	8	4400
Total		∑fi = 50	∑fx= 21200

(i) We know that,

n = ∑f = 50.

By formula,

Mean wage = $\frac{\sum fx}{n} = \frac{21200}{50}$ = ₹ 424

Hence, mean wage is ₹ 424.

(ii) If all frequencies in a distribution are multiplied by a constant, the mean remains unchanged.

n = ∑f = 50 x 2 = 100.

∑fx = 21200 x 2 = 42400

By formula,

Mean wage = $\frac{\sum fx}{n} = \frac{42400}{100}$ = ₹ 424

Hence, new mean wage is ₹ 424.

Variable (x)	Frequency (f)	fx
5	20	100
6	17	102
7	f	7f
8	10	80
9	8	72
10	6	60
11	7	77
12	6	72
Total	∑ f = 74 + f	∑fx = 563 + 7f

We know that,

n = ∑f = 74 + f.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{n} \\[1em] 7.5 = \frac{563 + 7f}{74 + f}.\end{aligned}$$

⇒ 7.5(74 + f) = 563 + 7f

⇒ 555 + 7.5f = 563 + 7f

⇒ 7.5f - 7f = 563 - 555

⇒ 0.5f = 8

⇒ f = $\frac{8}{0.5}$

⇒ f = 16.

Hence, missing frequency f is 16.

Variate (xi)	Frequency (fi)	fi xi
8	12	96
12	16	192
15	20	300
18	p	18p
20	16	320
25	8	200
30	4	120
Total	∑fi = 76 + p	∑ fixi = 1228 + 18p

We know that,

n = ∑f = 76 + p.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{n} \\[1em] 16.6 = \frac{1228 + 18p}{76 + p}.\end{aligned}$$

⇒ 16.6(76 + p) = 1228 + 18p

⇒ 1261.6 + 16.6p = 1228 + 18p

⇒ 1261.6 - 1228 = 18p - 16.6p

⇒ 33.6 = 1.4p

⇒ p = $\frac{33.6}{1.4}$

⇒ p = 24.

Hence, numerical value of p is 24.

Variate (x)	Frequency (f)	fx
5	2	10
8	5	40
10	8	80
12	22	264
x	7	7x
20	4	80
25	2	50
Total	∑ f = 50	∑fx = 524 + 7x

We know that,

n = ∑f = 50.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{n} \\[1em] 12.58 = \frac{524 + 7x}{50}.\end{aligned}$$

⇒ 12.58(50) = 524 + 7x

⇒ 629 = 524 + 7x

⇒ 7x = 629 - 524

⇒ 7x = 105

⇒ x = $\frac{105}{7}$

⇒ x = 15.

Hence, numerical value of x is 15.

Let assumed mean (A) = 62.

Height (x)	Number of plants (f)	d = x - A	fd
58	15	-4	-60
60	14	-2	-28
A = 62	20	0	0
65	18	+3	54
66	8	+4	32
68	5	+6	30
Total	∑ f = 80		∑fd = 28

We know that,

n = ∑f = 80.

By formula,

$$\begin{aligned}\text{Mean} = A + \frac{\sum fd}{n} \\[1em] = 62 + \frac{28}{80} \\[1em] = 62 + 0.35 \\[1em] = 62.35.\end{aligned}$$

Hence, mean height of the plants is 62.35 cm.

Number of match sticks (x)	Number of boxes (f)	d = x - A	fd
40	6	-4	-24
42	7	-2	-14
43	12	-1	-12
A = 44	9	0	0
45	10	1	10
48	6	4	24
Total	∑ f = 50		∑ fd = -16

(i) We know that,

n = ∑f = 50.

By formula,

$$\begin{aligned}\text{Mean} = A + \frac{\sum fd}{n} \\[1em] = 44 + \frac{-16}{50} \\[1em] = 44 - 0.32 \\[1em] = 43.68\end{aligned}$$

Hence, mean number of match sticks per box is 43.68

(ii) Total number of sticks = Mean × Total Boxes

= 43.68 × 50

= 2184 sticks

Number of match sticks to be added = (50 × 45) − 2184 = 66.

Hence, extra match sticks to be added = 66.

Marks	Number of students (f)	class marks (x)	Deviation d = x - A	fd
0 – 10	3	5	-20	-60
10 – 20	8	15	-10	-80
20 – 30	14	A = 25	0	0
30 – 40	9	35	10	90
40 – 50	4	45	20	80
50 – 60	2	55	30	60
Total	∑f = 40			∑fd = 90

By formula,

$$\begin{aligned}\text{Mean} = A + \frac{\sum fd}{\sum f} \\[1em] = 25 + \frac{90}{40} \\[1em] = 25 + 2.25 \\[1em] = 27.25.\end{aligned}$$

Hence, required mean = 27.25

Construct the table as under, taking assumed mean as 45:

Class	Class Mark (yi)	Deviation (di = yi - a)	Frequency(fi)	fidi
20 - 30	25	-20	6	-120
30 - 40	35	-10	9	-90
40 - 50	45	0	14	0
50 - 60	55	10	10	100
60 - 70	65	20	7	140
70 - 80	75	30	4	120
Total			Σfi = 50	Σfidi = 150

By formula,

$\text{Mean} = a + \frac{\sum f_id_i}{\sum f_i}$

$= 45 + \frac{150}{50}$

= 45 + 3

= 48.

Hence, required mean = 48.

Marks	No. of students (f)	Class mark (y)	fy
0 – 10	7	5	35
10 – 20	a	15	15a
20 – 30	8	25	200
30 – 40	10	35	350
40 – 50	5	45	225
Total	∑ f = 30 + a		∑ fy = 810 + 15a

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum f_iy_i}{\sum f_i} \\[1em] 24 = \frac{810 + 15a}{30 + a} \\[1em] 24(30 + a) = 810 + 15a \\[1em] 720 + 24a = 810 + 15a \\[1em] 24a - 15a = 810 - 720 \\[1em] 9a = 90 \\[1em] a = 10\end{aligned}$$

Hence, the value of a = 10.

We construct the following table, taking assumed mean a = 25. Here, c (width of each class) = 10.

Marks	No.of students	Class mark (yi)	ui = (yi - a)/c	No. of students (fi)	fi ui
0 – 10	10	5	-2	10	-20
10 – 20	9	15	-1	9	-9
20 – 30	25	a = 25	0	25	0
30 – 40	30	35	1	30	30
40 – 50	16	45	2	16	32
50 – 60	10	55	3	10	30
Total				∑ fi = 100	∑ fi ui = 63

By formula,

$$\begin{aligned}\text{Mean} = a + c \times \frac{\sum f_iu_i}{\sum f_i} \\[1em] = 25 + 10 \times \frac{63}{100} \\[1em] = 25 + \frac{630}{100} \\[1em] = 25 + 6.3 \\[1em] = 31.3\end{aligned}$$

Hence, mean of the following distribution is 31.3

In the given table i is the class interval which is equal to 15.

Class	Class mark (x)	d = (x - A)	u = d/i	Frequency (f)	fu
0-15	7.5	-45	-3	3	-9
15-30	22.5	-30	-2	4	-8
30-45	37.5	-15	-1	7	-7
45-60	A = 52.5	0	0	6	0
60-75	67.5	15	1	8	8
75-90	82.5	30	2	2	4
Total				Σf = 30	Σfu = -12

Mean = A + $\frac{Σfu}{Σf} \times i = 52.5 + \frac{-12}{30} \times 15$

= $52.5 - \frac{180}{30}$

= 52.5 - 6

= 46.50

Hence, mean = 46.50

We construct the following table, taking assumed mean a = 97.5. Here, c (width of each class) = 5.

Weight in grams	No.of apples (fi)	Class mark (yi)	ui = (yi - a)/c	fiui
80 – 85	5	82.5	-3	-15
85 – 90	8	87.5	-2	-16
90 – 95	10	92.5	-1	-10
95 – 100	12	a=97.5	0	0
100 – 105	8	102.5	1	8
105 – 110	4	107.5	2	8
110 – 115	3	112.5	3	9
Total	∑ fi = 50			∑ fi ui = -16

By formula,

$$\begin{aligned}\text{Mean} = a + c \times \frac{\sum f_iu_i}{\sum f_i} \\[1em] = 97.5 + 5 \times \frac{-16}{50} \\[1em] = 97.5 - \frac{16}{10} \\[1em] = 97.5 - 1.6 \\[1em] = 95.9 \approx 96\end{aligned}$$

Hence, mean weight of the apples is 96 g.

Since, class are discontinuous we will first convert them into continuous class intervals.

Adjustment factor

= $\frac{\text{Lower limit of a class -Upper limit of previous class}}{2} = \frac{80 - 79}{2} = \frac{1}{2}$ = 0.5

Adding the adjustment factor to upper limit and subtracting from lower limit we get the continuous class intervals.

We construct the following table, taking assumed mean a = 92. Here, c (width of each class) = 5.

Weights (in gms)	Class interval	Number of eggs (fi)	Class mark (yi)	ui = (yi - a)/c	fiui
75 – 79	74.5 - 79.5	4	77	-3	-12
80 – 84	79.5 - 84.5	9	82	-2	-18
85 – 89	84.5 - 89.5	13	87	-1	-13
90 – 94	89.5 - 94.5	17	a = 92	0	0
95 – 99	94.5 - 99.5	12	97	1	12
100 – 104	99.5 - 104.5	3	102	2	6
105 – 109	104.5 - 109.5	2	107	3	6
Total		∑ fi= 60			∑ fi ui = -19

By formula,

$$\begin{aligned}\text{Mean} = a + c \times \frac{\sum f_iu_i}{\sum f_i} \\[1em] = 92 + 5 \times \frac{-19}{60} \\[1em] = 92 - \frac{19}{12} \\[1em] = 92 - 1.583 \\[1em] = 90.416 \approx 90\end{aligned}$$

Hence, mean weight of the eggs is 90 g.

We construct the following table, taking assumed mean a = 17.5. Here, c (width of each class) = 5.

Marks	Frequency (fi)	Class mark (yi)	ui = (yi - a)/c	fiui
0-5	3	2.5	-3	-9
5-10	7	7.5	-2	-14
10-15	15	12.5	-1	-15
15-20	24	a = 17.5	0	0
20-25	16	22.5	1	16
25-30	8	27.5	2	16
30-35	5	32.5	3	15
35-40	2	37.5	4	8
Total	∑ fi = 80			∑fi ui = 17

By formula,

$$\begin{aligned}\text{Mean} = a + c \times \frac{\sum f_iu_i}{\sum f_i} \\[1em] = 17.5 + 5 \times \frac{17}{80} \\[1em] = 17.5 + \frac{85}{80} \\[1em] = 17.5 + 1.0625 \\[1em] = 18.5625 \approx 18.56\end{aligned}$$

Hence, mean marks scored by the students is 18.56

We construct the following table, taking assumed mean a = 45.

Number of patients	Number of days (fi)	Class mark (xi)	di = xi - A	fidi
10 – 20	5	15	-30	-150
20 – 30	2	25	-20	-40
30 – 40	7	35	-10	-70
40 – 50	9	a=45	0	0
50 – 60	2	55	10	20
60 – 70	5	65	20	100
Total	∑ fi = 30			∑fi di = -140

By formula,

$$\begin{aligned}\text{Mean} = a + \frac{\sum f_id_i}{\sum f_i} \\[1em] = 45 + \frac{-140}{30} \\[1em] = 45 - \frac{14}{3} \\[1em] = 45 - 4.6666 \\[1em] = 40.3333 \approx 40.33\end{aligned}$$

Hence, average number of patients attending the hospital per month is 40.33.

We construct the following table, taking assumed mean a = 30. Here, c (width of each class) = 10.

Class-interval	Frequency (fi)	Class mark (yi)	ui = (yi - a)/c	fiui
5 – 15	2	10	-2	-4
15 – 25	6	20	-1	-6
25 – 35	4	a = 30	0	0
35 – 45	8	40	1	8
45 – 55	4	50	2	8
Total	∑ fi= 24			∑ fiui = 6

By formula,

$$\begin{aligned}\text{Mean} = a + c \times \frac{\sum f_iu_i}{\sum f_i} \\[1em] = 30 + 10 \times \frac{6}{24} \\[1em] = 30 + \frac{10}{4} \\[1em] = 30 + 2.5 \\[1em] = 32.5\end{aligned}$$

Hence, mean of the frequency distribution is 32.5.

Range is a measure of dispersion, not central tendency, it describes how spread out the data is.

Range is the difference between the highest and lowest values of data.

Hence, option 3 is the correct option.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum x_i}{n} \\[1em] = \frac{34 + 89 + 37 + 144 + 78 + 240 + 128 + 98}{8} \\[1em] = \frac{848}{8} \\[1em] = 106.\end{aligned}$$

Hence, option 3 is the correct option.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum x_i}{n} \\[1em] \Rightarrow 10 = \frac{7 + 5 + 13 + x + 9}{5} \\[1em] \Rightarrow 10 \times 5 = 34 + x \\[1em] \Rightarrow 50 = 34 + x \\[1em] \Rightarrow x = 50 - 34 \\[1em] \Rightarrow x = 16.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Number of observations = 4

By formula,

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

Substituting values we get :

$$\begin{aligned}\text{Mean }= \frac{1^2 + 2^2 + 3^3 + 4^4}{4} \\[1em] = \frac{1 + 4 + 27 + 256}{4} \\[1em] = \frac{288}{4} \\[1em] = 72.\end{aligned}$$

Hence, option 2 is the correct option.

Number (x)	Frequency (f)	fx
4	20	80
5	40	200
6	30	180
7	10	70
Total	∑ f = 100	∑ fx = 530

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{\sum f} \\[1em] = \frac{530}{100} \\[1em] = 5.3.\end{aligned}$$ Hence, option 1 is the correct option.

Given,

36a + 36b = 576

36(a + b) = 576

a + b = $\frac{576}{36}$

a + b = 16

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{a + b}{2} \\[1em] = \frac{16}{2} \\[1em] = 8.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Mean = 43

When each observation is increased by 7, mean will also increase by 7, thus new mean = 43 + 7 = 50.

Hence, option 4 is the correct option.

To calculate the mean, we need a single representative value for each class. We assume that the data points in that interval are balanced around the middle, known as the Class Mark.

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

Hence, option 2 is the correct option.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum x_i}{n} \\[1em] \Rightarrow 11 = \frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8)}{5} \\[1em] \Rightarrow 11 \times 5 = 5x + 20 \\[1em] \Rightarrow 55 = 5x + 20 \\[1em] \Rightarrow 55 - 20 = 5x \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \frac{35}{5} \\[1em] \Rightarrow x = 7.\end{aligned}$$

Mean of first three observations, x = 7, x + 2 = 7 + 2 = 9, x + 4 = 7 + 4 = 11

$$\begin{aligned}\text{Mean} = \frac{\sum x_i}{n} \\[1em] = \frac{7 + 9 + 11}{3} \\[1em] = \frac{27}{3} \\[1em] =9.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

Five consecutive odd numbers :

A = x - 4,

B = x - 2,

C = x,

D = x + 2,

E = x + 4.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum x_i}{n} \\[1em] \Rightarrow 37 = \frac{(x - 4) + (x - 2) + x + (x + 2) + (x + 4)}{5} \\[1em] \Rightarrow 37 = \frac{5x}{5} \\[1em] \Rightarrow x = 37.\end{aligned}$$

C = x = 37

B = 37 - 2 = 35

D = 37 + 2 = 39

B × D = 35 × 39 = 1365.

Hence, option 1 is the correct option.

Five consecutive odd numbers = x - 4, x - 2, x, x + 2, x + 4.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum x_i}{n} \\[1em] \Rightarrow 37 = \frac{(x - 4) + (x - 2) + x + (x + 2) + (x + 4)}{5} \\[1em] \Rightarrow 37 \times 5 = 5x \\[1em] \Rightarrow x = 37.\end{aligned}$$

Set A = 33, 35, 37, 39, 41

Smallest number of Set B = 41 + 13 = 54

Set B = 54, 56, 58, 60

$$\begin{aligned}\text{Average} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{54 + 56 + 58 + 60}{4} \\[1em] = \frac{228}{4} \\[1em] = 57.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

The mean of four observations is 20

When a constant c is added to each observation, total increase is 4(c)

New sum = 80 + 4c

New mean = $\frac{80 + 4c}{4} = 22$ 80 + 4c = 22(4)

4c = 88 - 80

c = $\frac{8}{4}$

c = 2.

Hence, option 2 is the correct option.

The original mean =

$\bar{x} = \frac{x_1 + x_2 + x_3 + \dots + x_n}{n}$

The new mean is the sum of the new observations divided by :

$$\begin{aligned}\text{New Mean} = \frac{(x_1 - a) + (x_2 - a) + (x_3 - a) + \dots + (x_n - a)}{n} \\[1em] = \frac{(x_1 + x_2 + \dots + x_n) - (a + a + \dots + a)}{n}\end{aligned}$$

Since there are n terms of a, the sum of the a is na:

$$\begin{aligned}\text{New Mean} = \frac{(x_1 + x_2 + \dots + x_n) - na}{n}\\[1em] = \frac{x_1 + x_2 + \dots + x_n}{n} - \frac{na}{n} \\[1em] = \bar{x} - a.\end{aligned}$$

Hence, option 3 is the correct option.

Let the original observations be x₁, x₂, ......., x_k with a mean of $\bar{x}$.

The original mean is:

$\bar{x} = \frac{\sum x_i}{k}$

If every observation is multiplied by m, the new observations are, mx₁, mx₂,......., mx_k.

The sum of these new values is

$\sum mx_i = m(\sum x_i)$.

Adding "n" to every term increases mean by "n":

$\text{Mean} = \frac{m(\sum x_i)}{k} = m \bar{x} + n$

Hence, option 2 is the correct option.

Sum of observations when observations were misread = Number of observations × Initial Mean

= 100(50)

= 5000

To find the correct sum We need to subtract the wrong value and add the correct value,

Sum = 5000 - 50 + 40

= 4990

∴ Mean = $\frac{4990}{100}$ = 49.9

Hence, option 2 is the correct option.

x	f	fx
5	3	15
15	p	15p
25	3	75
35	6	210
45	2	90
Total	∑ f = 14 + p	∑ fx = 390 + 15p

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum fx}{\sum f} \\[1em] \Rightarrow 25 = \frac{390 + 15p}{14 + p} \\[1em] \Rightarrow 25(14 + p) = 390 + 15p \\[1em] 350 + 25p = 390 + 15p \\[1em] \Rightarrow 25p - 15p = 390 - 350 \\[1em] 10p = 40 \\[1em] \Rightarrow p = \frac{40}{10} \\[1em] \Rightarrow p = 4.\end{aligned}$$

Hence, option 3 is the correct option.

Marks	Number of students (f)	Class mark (x)	fx
0 – 10	12	5	60
10 – 20	18	15	270
20 – 30	27	25	675
30 – 40	20	35	700
40 – 50	17	45	765
50 – 60	6	55	330
Total	∑ f = 100		∑ fx = 2800

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{\sum f} \\[1em] = \frac{2800}{100} \\[1em] = 28.\end{aligned}$$

Hence, option 4 is the correct option.

Class	Frequency (f)	Class mark(x)	fx
0 – 10	8	5	40
10 – 20	p	15	15p
20 – 30	12	25	300
30 – 40	13	35	455
40 – 50	10	45	450
Total	∑ f = 43 + p		∑ fx = 1245 + 15p

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum fx}{\sum f} \\[1em] \Rightarrow 27 = \frac{1245 + 15p}{43 + p} \\[1em] \Rightarrow 27(43 + p) = 1245 + 15p \\[1em] 1161 + 27p = 1245 + 15p \\[1em] \Rightarrow 27p - 15p = 1245 - 1161 \\[1em] 12p = 84 \\[1em] \Rightarrow p = \frac{84}{12} \\[1em] \Rightarrow p = 7.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

Mean of 5 numbers = 27

Sum = Number of observations × Initial Mean

= 5(27)

= 135

When one number is excluded, 4 numbers remain, and their new mean is 25.

Sum of numbers(when one number is excluded) = Number of observations × new mean

= 4(25)

= 100

The difference between the two sums is the value of the number that was removed = 135 - 100 = 35

Hence, option 4 is the correct option.

Observations (x)	Frequency (f)	fx
2x -5	2	4x- 10
x + 3	3	3x + 9
7 - x	4	28 - 4x
5 - x	6	30 - 6x
x + 9	1	x + 9
Total	∑ f = 16	∑ fx = 66 - 2x

By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{\sum f} \\[1em] 4 = \frac{66 - 2x}{16} \\[1em] 2x = 66 - 64 \\[1em] 2x = 2 \\[1em] x = 1.\end{aligned}$$

Hence, option 1 is the correct option.

Let the n observations x₁, x₂, ......., x_n. The mean is $\bar x$.

$$\begin{aligned}\bar{x} = \frac{x_1 + x_2 + ... + x_n}{n} \\[1em] \sum x = n\bar x\end{aligned}$$

Given,

The first observation is increased by 1, second by 2 and so on.

New sum of observations = (x₁ + 1)+ (x₂ + 2)+ .......+ (x_n + n)

= (x₁+ x₂+ .......+ x_n) + (1 + 2 + ..... + n)

= ∑x + (1 + 2 + 3 + ... + n)

$= n\bar x + \frac{n(n + 1)}{2}$

$$\begin{aligned}\text{ Mean }= \frac{\text{ Sum of observations}}{\text{ Number of observations}} \\[1em] = \frac{n\bar x + \frac{n(n + 1)}{2}}{n} \\[1em] = \bar x + \frac{n + 1}{2}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

$\bar x = a + \Big(\frac{\sum f_i d_i}{\sum f_i}\Big)$

the deviation d_i are calculated as:

d_i = x_i - a

where x_i are the class marks.

So, the deviations are taken from mid-points of the classes.

Hence, option 3 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \sum f_i \bar x \\[1em] \Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \bar x \sum f_i \text{ .....(1)}\end{aligned}$$

We know that,

$$\begin{aligned}\bar x = \frac{\sum f_i x_i}{\sum f_i} \\[1em] \bar x \sum f_i = \sum f_i x_i\end{aligned}$$

Substituting the values in equation (1), we get :

$\therefore \sum f_i(x_i - \bar x) =\bar x \sum f_i - \bar x \sum f_i = 0$

Hence, option 2 is the correct option.

In the Step-Deviation Method formula $\bar x = a + h \Big(\frac{\sum f_i u_i}{\sum f_i}\Big)$,

the term u_i represents the step-deviation.

The step-deviation is calculated as:

$u_i = \frac{x_i - a}{h}$

Where :

x_i : The class mark (mid-point).

a : The assumed mean.

h : The class size.

Hence, option 3 is the correct option.

Sum of marks obtained in class 1 = 100(30) = 3000

Sum of marks obtained in class 2 = 50(60) = 3000

Total marks obtained in class 1 and class 2 = 6000

Total number of students in class 1 and class 2 = 100 + 50 = 150

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{6000}{150} \\[1em] = 40.\end{aligned}$$

Hence, option 1 is the correct option.

Mean (xi)	fi	fi xi
2	45	90
2.5	40	100
2	15	30
Total	∑ fi = 100	∑ fi xi = 220

$$\begin{aligned}\text{Mean} = \frac{\sum f_ix_i}{\sum f_i} \\[1em] = \frac{220}{100} \\[1em] = 2.2\end{aligned}$$

Hence, option 2 is the correct option.

Total number of items in three groups = 2 + 3 + 5 = 10

Total sum = Number of observations × Mean

= 10 × 12 = 120

Total number of items in first two groups = 2 + 3 = 5

Sum of first two groups = Number of observations × Mean

= 3 × 5 = 15

The sum of the third group is the difference between the total sum and the sum of the first two groups = 120 - 15 = 105.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{105}{5} \\[1em] = 21.\end{aligned}$$

Hence, option 4 is the correct option.

Total weight of 17 boxes = Mean × Number of boxes

= 92 × 17 = 1564 kg.

If 18 new boxes are added, mean increases by 3 kg.

So, new mean = 92 + 3 = 95 kg.

Total weight of 35 boxes = New mean × Number of boxes

= 95 × 35

= 3325 kg

The weight added by the 18 new boxes = Total weight of 35 boxes - Total weight of 17 boxes

= 3325 - 1564

= 1761 kg.

By formula,

$$\begin{aligned}\text{Mean of 18 boxes} = \frac{\text{Weight of 18 boxes}}{\text{ Number of boxes}} \\[1em] = \frac{1761}{18} \\[1em] = 97.8 \text{ kg.}\end{aligned}$$

Hence, option 3 is the correct option.

Class	Frequency (fi)	Class mark (xi)	fixi
0 – 20	17	10	170
20 – 40	28	30	840
40 – 60	32	50	1600
60 – 80	f	70	70f
80 – 100	19	90	1710
Total	∑ fi = 96 + f		∑ fixi = 4320 + 70f

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum f_i x_i}{\sum f_i} \\[1em] \Rightarrow 50 = \frac{4320 + 70f}{96 + f} \\[1em] \Rightarrow 50(96 + f) = 4320 + 70f \\[1em] 4800 + 50f = 4320 + 70f \\[1em] \Rightarrow 4800 - 4320 = 70f - 50f \\[1em] \Rightarrow 480 = 20f \\[1em] \Rightarrow f = \frac{480}{20} \\[1em] \Rightarrow f = 24.\end{aligned}$$

Hence, option 1 is the correct option.

The mean is an algebraic measure that cannot be determined graphically.

To calculate mean, we must sum all observations and divide by the total number. Since it depends on the exact value of every single observation rather than their position or frequency alone there is no standard geometric construction or curve that can pinpoint the mean on a graph.

Hence, option 1 is the correct option.

Directions : The marks obtained by 10 students in a class-test were as follows :

36, 64, 48, 52, 57, 73, 26, 39, 78, 67

31. The mean marks of the whole class is :

(a) 49.1
(b) 53.7
(c) 54
(d) 60

32. If the maximum marks in the test were 80, the mean percentage of marks obtained by the students is :

(a) 65%
(b) 67.5%
(c) 68%
(d) 72%

33. The mean marks of the top 5 scorers in the class is :

(a) 65.4
(b) 66.8
(c) 67.2
(d) 67.8

34. As per the Board’s instruction each student who obtained less than 50 marks was awarded 3 grace marks. The new mean of the marks thus obtained increases by :

(a) 0.9
(b) 1.2
(c) 1.5
(d) 1.8

31. By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{36 + 64 + 48 + 52 + 57 + 73 + 26 + 39 + 78 + 67}{10} \\[1em] = \frac{540}{10} \\[1em] = 54.\end{aligned}$$

Hence, option (c) is the correct option.

32. By formula,

$$\begin{aligned}\text{Percentage of marks} = \frac{\text{Mean of marks obtained}}{\text{Maximum marks}} \times 100 \\[1em] = \frac{54}{80} \times 100 \\[1em] = 0.675 \times 100 \\[1em] = 67.5 %\end{aligned}$$

Hence, option (b) is the correct option.

33. Top five marks in class are 78, 73, 67, 64, 57.

By formula,

$$\begin{aligned}\text{Mean} = \frac{\text{ Sum of all observations}}{\text{ Number of observations}} \\[1em] = \frac{78 + 73 + 67 + 64 + 57}{5} \\[1em] = \frac{339}{5} \\[1em] = 67.8\end{aligned}$$

Hence, option (d) is the correct option.

34. The students who scored less than 50 are: 36, 48, 26, 39

Since each of these 4 students gets 3 marks,

Total marks increased = 4(3) = 12 marks

The new mean of the marks thus obtained increases by :

$$\begin{aligned}\frac{\text{Total marks increased}}{\text{Number of students}} \\[1em] = \frac{12}{10} \\[1em] = 1.2\end{aligned}$$

Hence, option (b) is the correct option.

Directions :
At a courier company, a daily report of parcels received for dispatch is prepared every evening, which classifies the parcels on the basis of their weights. The report of a certain day is as under :

Weight of parcel (in grams) (x)	Number of parcels (f)
Below 600	60
Below 500	58
Below 400	54
Below 300	35
Below 200	22
Below 100	10

35.How many parcels have weights in the range of 300 – 400 grams?
(a) 4
(b) 12
(c) 13
(d) 19

36.How many parcels have weights in the range of 200 – 300 grams?
(a) 10
(b) 12
(c) 13
(d) 19

37.In which of the following weight ranges, the number of parcels is the lowest?
(a) 100 – 200
(b) 200 – 300
(c) 400 – 500
(d) 500 – 600

38.The mean weight of parcels received on that particular day is :
(a) 226.78 g
(b) 251.67 g
(c) 284.28 g
(d) 302.16 g

Table :

Class interval (x)	Number of parcels (f)	Class mark (x)	fx
0-100	10	50	500
100-200	12	150	1800
200-300	13	250	3250
300 - 400	19	350	6650
400 - 500	4	450	1800
500 - 600	2	550	1100
Total	∑ f = 60		∑ fx = 15100

35. From table,

The number of parcels for the 300 – 400 range = 19 (54 - 35).

Hence, option (d) is the correct option.

36.From table,

The number of parcels for the 200 – 300 range = 13 (35 - 22).

Hence, option (c) is the correct option.

37.The lowest frequency is 2, which occurs in the 500 – 600 range.

Hence, option (d) is the correct option.

38. By formula,

$$\begin{aligned}\text{Mean} = \frac{\sum fx}{\sum f} \\[1em] = \frac{15100}{60} \\[1em] = 251.67 \text{ g.}\end{aligned}$$

Hence, option (b) is the correct option.

First 9 natural numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9

By formula,

$$\begin{aligned}\Rightarrow \text{Mean } = \frac{\text{Sum of all observations}}{\text{Total number of observations}}\\[1em] = \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9}{9}\\[1em] = \frac{45}{9}\\[1em] = 5.\end{aligned}$$

∴ (A) is false, (R) is true.

Hence, option 4 is the correct option.

The standard formula to calculate the mean ($\bar{x}$) using the step-deviation method is:

$\bar{x} = A + \Big( \frac{\sum ft}{\sum f} \Big) \times h$

∴ Assertion (A) is true.

The step-deviation is defined as the difference between the class mark (x) and the assumed mean (A), divided by the class size (h):

$t = \frac{x - A}{h}$

∴ Reason (R) is true.

Reason (R) defines the step-deviation t that appears in the formula stated in Assertion (A), so R is the correct explanation of A.

Hence, option 1 is the correct option.

The expression $\sum f_i(x_i - \bar{x})$ represents the sum of the deviations of all observations from their mean, weighted by their frequencies.

$$\begin{aligned}\Rightarrow \sum f_i(x_i - \bar x) = \sum f_ix_i - \sum f_i \bar x \\[1em] = \sum f_ix_i - \bar x \sum f_i \text{ .....(1)}\end{aligned}$$

We know that,

$$\begin{aligned}\bar x = \frac{\sum f_i x_i}{\sum f_i} \\[1em] \bar x \sum f_i = \sum f_i x_i\end{aligned}$$

Substituting the values in equation (1),

$\therefore \sum f_i(x_i - \bar x) = \bar x \sum f_i - \bar x \sum f_i = 0$

Since the value is 0 (not 1),

∴ Assertion (A) is false.

The sum of the deviations from the mean is 0. This is a fundamental and correct property of the arithmetic mean.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

Mean = $\frac{\text{Sum of terms}}{\text{Number of terms}}$

∴ Sum of terms = Mean × Number of terms

Given, mean of 15 numbers is 18

∴ Sum of 15 terms = 18 × 15 = 270

Given, mean of remaining 10 numbers is 13

∴ Sum of remaining terms = 13 × 10 = 130

Sum of 25 terms = 270 + 130 = 400.

Mean = $\frac{400}{25}$ = 16

Since the mean of the 25 numbers is 16 (not 14),

∴ Assertion (A) is false.

The mean of variates having corresponding frequencies is correctly given by $\bar x = \frac{\sum f_ix_i}{\sum f_i}$.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

(a) Class marks : 13, 15, 17, 19, 21, 23.

Difference between two consecutive class marks = 15 - 13 = 2.

Lower limit of class = Class mark - 1

Upper limit of class = Class mark + 1

Class	Frequency
12 - 14	8
14 - 16	2
16 - 18	3
18 - 20	4
20 - 22	5
22 - 24	6

(b)

Class	Class mark (x)	Frequency (f)	fx
12 - 14	13	8	104
14 - 16	15	2	30
16 - 18	17	3	51
18 - 20	19	4	76
20 - 22	21	5	105
22 - 24	23	6	138
Total		Σf = 28	Σfx = 504

Mean = $\frac{Σfx}{Σf} = \frac{504}{28}$ = 18.

Hence, mean = 18.

Given,

Mean of 5, 7, 8, 4 and n is o.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum x_i}{n} \\[1em] \Rightarrow o = \frac{5 + 7 + 8 + 4 + n}{5} \\[1em] \Rightarrow 5o = 24 + n \\[1em] \Rightarrow o = \frac{24 + n}{5} \text{ .....(1)}\end{aligned}$$

Mean of 5, 7, 8, 4, n and o is n.

$$\begin{aligned}\Rightarrow n = \frac{5 + 7 + 8 + 4 + n + o}{6} \\[1em] \Rightarrow 6n = 24 + n + o \\[1em] \Rightarrow 5n = 24 + o \text{ .....(2)}\end{aligned}$$

From equation (1), o = $\frac{24 + n}{5}$.

Substituting in equation (2), we get:

$$\begin{aligned}\Rightarrow 5n = 24 + \frac{24 + n}{5} \\[1em] \Rightarrow 5n = \frac{120 + 24 + n}{5} \\[1em] \Rightarrow 25n = 120 + 24 + n \\[1em] \Rightarrow 24n = 144 \\[1em] \Rightarrow n = 6.\end{aligned}$$

Substituting n = 6 in equation (1), we get:

$$\begin{aligned}\Rightarrow o = \frac{24 + 6}{5} \\[1em] \Rightarrow o = \frac{30}{5} = 6.\end{aligned}$$

Hence, n = 6 and o = 6.