Mean & Median of Ungrouped Data & Frequency Polygon

Weights of 7 boys (in kg) :

52, 57, 55, 60, 54, 59, 55

Number of boys (n) = 7

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

Sum of observations = 52 + 57 + 55 + 60 + 54 + 59 + 55 = 392

Total number of observations = 7

Mean = $\frac{392}{7}$ = 56.

Hence, mean weight of the group is 56 kg.

Marks of 7 students:

340, 180, 260, 164, 56, 275, 307

Number of students (n) = 7

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

Sum of observations = 340 + 180 + 260 + 164 + 56 + 275 + 307 = 1582

Total number of observations = 7

Mean = $\frac{ 1582}{7}$ = 226.

Hence, mean marks per student is 226.

First six prime numbers are:

2, 3, 5, 7, 11, 13

Mean = $\frac{\text{Sum of prime numbers}}{\text{Total number of prime numbers}}$

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

Mean = $\frac{41}{6} \approx$ 6.83

Hence, mean of first six prime numbers is 6.83.

First ten odd numbers are:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Sum of observations = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

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

= $\frac{100}{10}$ = 10.

Hence, mean of first ten odd numbers is 10.

Factors of 20 are the numbers that divide 20 exactly.

The factors are :

1, 2, 4, 5, 10, 20

Total number of observations = 6

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

= $\frac{1 + 2 + 4 + 5 + 10 + 20}{6}$

= $\frac{42}{6}$ = 7.

Hence, mean of all the factors of 20 is 7.

We know that,

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

= $\frac{35.2 + 31.1 + 27.6 + 31.8 + 29.3 + 23.8}{6}$

= $\frac{178.8}{6}$

= 29.8 °F

Hence, mean temperature = 29.8 °F.

Numbers: 6, 8, 9, x, 13

Mean = 10

Number of observations (n) = 5

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

⇒ 10 = $\frac{6 + 8 + 9 + x + 13}{5}$

⇒ 50 = 36 + x

⇒ x = 50 - 36 = 14.

Hence, x = 14.

Mean height of 6 girls = 148 cm

Number of girls (n) = 6

Heights of 5 girls:

142 cm, 154 cm, 146 cm, 145 cm, 150 cm

Let height of the sixth girl be x cm.

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

⇒ 148 = $\frac{142 + 154 + 146 + 145 + 150 + x}{6}$

⇒ 148 × 6 = 737 + x

⇒ 888 = 737 + x

⇒ x = 888 - 737 = 151.

Hence, height of the sixth girl = 151 cm.

$\text{Mean} = \frac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (weight)

Weight (x)	Number of workers (f)	fx
60	4	240
63	5	315
66	3	198
72	1	72
75	2	150
Total	15	975

So, ∑fx = 975 and ∑f = 15

Therefore,

$\text{Mean} = \frac{\sum fx}{\sum f}$

= $\frac{975}{15}$ = 65.

Hence, mean weight = 65 kg.

$\text{Mean} = \frac{\sum fx}{\sum f}$

where

f = frequency (Number of workers)

x = observation (Daily wages)

Daily Wages (₹) (x)	Number of Workers (f)	fx
90	12	1080
110	14	1540
120	13	1560
130	11	1430
150	10	1500
Total	60	7110

so, ∑fx = 7110 and ∑f = 60

Mean = $\frac{7110}{60}$ = 118.5.

Hence, mean of daily wages of 60 workers is ₹118.5

$\text{Mean} = \frac{\sum fx}{\sum f}$

where

f = frequency (Number of plants)

x = observation (Height)

Height (x)	Number of Plants (f)	fx
58	20	1160
60	25	1500
62	15	930
64	8	512
66	12	792
74	10	740
Total	90	5634

∑fx = 5634 and ∑f = 90

Mean = $\frac{5634}{90}$ = 62.6

Hence, mean height = 62.6 cm.

Mean = $\frac{\sum f_ix_i}{\sum f_i}$

xi	fi	fixi
6	5	30
12	4	48
18	p	18p
24	6	144
30	4	120
36	6	216

∑f_i = 5 + 4 + p + 6 + 4 + 6 = 25 + p

∑f_ix_i​ = 30 + 48 + 18p + 144 + 120 + 216 = 558 + 18p

⇒ 21.6 = $\frac{558+18p}{25+p}$

⇒ 21.6(25 + p) = 558 + 18p

⇒ 540 + 21.6p = 558 + 18p

⇒ 21.6p − 18p = 558 − 540

⇒ 3.6p = 18

⇒ p = $\frac{18}{3.6}$ = 5

Hence, p = 5.

Mean = 18.75

xi	fi	fixi
10	5	50
15	10	150
p	7	7p
25	8	200
30	2	60

∑f_i​ = 5 + 10 + 7 + 8 + 2 = 32

∑f_ix_i​ = 50 + 150 + 7p + 200 + 60 = 460 + 7p

Mean = $\frac{\sum f_ix_i}{\sum f_i}$

⇒ 18.75 = $\frac{460 + 7p}{32}$

⇒ 600 = 460 + 7p

⇒ 7p = 600 - 460

⇒ 7p = 140

⇒ p = $\frac{140}{7}$ = 20.

Hence, p = 20.

Let missing frequencies be:

Frequency at age 16 = a

Frequency at age 19 = b

By formula,

Mean = $\frac{\sum f_ix_i}{\sum f_i}$

where

x_i = Age (in years)

f_i = Number of students

Age in years (xi)	Number of students (fi)	fixi
15	3	45
16	a	16a
17	9	153
18	11	198
19	b	19b
20	3	60

∑f_i = 3 + a + 9 + 11 + b + 3 = 40

⇒ a + b + 26 = 40

⇒ a + b = 14

⇒ b = 14 - a ........(1)

Mean = $\frac{\sum f_ix_i}{\sum f_i}$

⇒ 17.45 = $\frac{45 + 16a + 153 + 198 + 19b + 60}{40}$

⇒ 17.45 × 40 = 456 + 16a + 19b

⇒ 698 = 456 + 16a + 19b

⇒ 242 = 16a + 19b ..........(2)

Substitute equation 1 in equation 2

⇒ 16a + 19(14 − a) = 242

⇒ 16a + 266 − 19a = 242

⇒ −3a + 266 = 242

⇒ −3a = −24

⇒ a = 8.

⇒ Put a = 8 in equation 1

⇒ b = 14 - 8

⇒ b = 6.

Hence, the missing frequencies are 8 & 6.

By using assumed mean method,

Let A = 1100

Weekly wages in ₹ (xi)	Number of workers (fi)	di = xi - A	fidi
950	24	-150	-3600
1000	18	-100	-1800
1050	13	-50	-650
A = 1100	15	0	0
1250	20	150	3000
1500	11	400	4400
1600	9	500	4500
Total	110		5850

By formula,

Mean = A + $\frac{\sum f_id_i}{\sum f_i}$

= 1100 + $\frac{5850}{110}$

= 1100 + 53.18

= 1153.18.

Hence, mean weekly wage is ₹ 1153.18.

We construct the following table, taking assumed mean A = 21

Here c (width of the class) = 1

xi	fi	ui = $\frac{(x_i - 21)}{c}$	fiui
18	170	-3	-510
19	320	-2	-640
20	530	-1	-530
A = 21	700	0	0
22	230	1	230
23	140	2	280
24	110	3	330
Total	∑fi = 2200		∑fiui = -840

Mean = A + c × $\frac{\sum f_iu_i}{\sum f_i}$

= 21 + 1 × $\frac{-840}{2200}$

= 21 + $\frac{-840}{2200}$

= 21 - 0.382 = 20.62.

Hence, mean of following distribution is 20.62.

(i) By arranging data in ascending order, we get :

6, 8, 9, 15, 16, 18, 21, 22, 25

Number of observations, n = 9, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th observation

⇒ $\frac{9 + 1}{2}$ th observation

⇒ $\frac{10}{2}$th observation

⇒ 5th observation

∴ Median = 16.

Hence, median is 16.

(ii) By arranging data in ascending order, we get :

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81

Number of observations, n = 12, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{12}{2}\right)^{\text{th}} \text{term} + \left(\frac{12}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{6th term} + \text{7th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{15} + \text{17}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{32}}{2} \\[1em]\end{aligned}$$

∴ Median = 16.

Hence, median = 16.

(iii) By arranging data in ascending order, we get :

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

Number of observations, n = 10, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{10}{2}\right)^{\text{th}} \text{term} + \left(\frac{10}{2} + 1\right)^{\text{th}} \text{term}}{2}\\[1em] \Rightarrow \text{Median} = \frac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{60} + \text{63}}{2}\\[1em] \Rightarrow \text{Median} = \frac{\text{123}}{2}\end{aligned}$$

∴ Median = 61.5.

Hence, median is 61.5.

By arranging data in ascending order, we get :

0, 7, 18, 26, 32, 35, 38, 47, 53, 66, 72

Number of observations, n = 11, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th observation

⇒ $\frac{11 + 1}{2}$ th observation

⇒ $\frac{12}{2}$th observation

⇒ 6th observation

∴ Median = 35.

Hence, median is 35 runs.

By arranging data in ascending order, we get :

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Number of observations, n = 9, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th observation

⇒ $\frac{9 + 1}{2}$ th observation

⇒ $\frac{10}{2}$th observation

⇒ 5th observation

∴ Median = 147.3.

Hence, median is 147.3 cm.

By arranging data in ascending order, we get :

31, 34, 36, 37, 40, 43, 46, 50, 52, 53

Number of observations, n = 10, which is even.

By formula,

$$\begin{aligned}\text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow\text{Median} = \frac{\left(\frac{10}{2}\right)^{\text{th}} \text{term} + \left(\frac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{40} + \text{43}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{83}}{2} \\[1em]\end{aligned}$$

∴ Median = 41.5.

Hence, median is 41.5 years.

By arranging data in ascending order, we get :

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Number of observations, n = 8, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{8}{2}\right)^{\text{th}} \text{term} + \left(\frac{8}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{4th term} + \text{5th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{13.4} + \text{14.3}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{27.7}}{2} \\[1em]\end{aligned}$$

∴ Median = 13.85.

Hence, median is 13.85 kg.

Cumulative frequency table :

Weight (in kg)	Number of boys (Frequency)	Cumulative Frequency (cf)
45	8	8
46	5	13
48	6	19
50	9	28
52	7	35
54	4	39
55	2	41

Total number of observations = 41, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th observation

⇒ $\frac{41 + 1}{2}$ th observation

⇒ $\frac{42}{2}$th observation

⇒ 21st observation

From the table,

From 20th to 28th position, the weight = 50 kg

∴ Median = 50 kg.

Hence, median weight is 50 kg.

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Variate	frequency	cumulative frequency
3	3	3
6	4	7
7	13	20
10	2	22
12	8	30
15	10	40

Total number of observations = 40, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{40}{2}\right)^{\text{th}} \text{term} + \left(\frac{40}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{20th term} + \text{21st term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{7} + \text{10}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{17}}{2} \\[1em]\end{aligned}$$

∴ Median = 8.5.

Hence, median is 8.5.

Cumulative frequency table :

Diameter of heart (in mm)	frequency	cumulative frequency
120	7	7
121	9	16
122	15	31
123	12	43
124	6	49
125	11	60

Total number of observations = 60, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{60}{2}\right)^{\text{th}} \text{term} + \left(\frac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{122} + \text{122}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{244}}{2} \\[1em]\end{aligned}$$

∴ Median = 122.

Hence, median = 122 mm.

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Variate	Frequency	Cumulative frequency
16	9	9
18	8	17
20	13	30
23	4	34
25	4	38
26	6	44
28	11	55
30	5	60

Total number of observations = 60, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{60}{2}\right)^{\text{th}} \text{term} + \left(\frac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{20} + \text{23}}{2} \\[1em] \Rightarrow\text{Median} = \frac{\text{43}}{2} \\[1em]\end{aligned}$$

∴ Median = 21.5.

Hence, median is 21.5.

Frequency distribution table :

Marks	Class mark	Number of students (Frequency)
0 - 10	5	15
10 - 20	15	32
20 - 30	25	55
30 - 40	35	35
40 - 50	45	13

Steps to draw frequency polygon :

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

2. Take 2 cm along y-axis = 10 students.

3. Find the mid-point 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 (-10) - 0 with zero frequency and join the other end with mid-point of class 50 - 60 with zero frequency.

The required frequency polygon is shown below:

Frequency distribution table :

Class - interval	Class mark	Frequency
0 - 9	4.5	9
9 - 18	13.5	15
18 - 27	22.5	6
27 - 36	31.5	12
36 - 45	40.5	18

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = 9 units.

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

3. Find the mid-point 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 (-9) - 0 with zero frequency and join the other end with mid-point of class 45 - 54 with zero frequency.

The required frequency polygon is shown below:

Frequency distribution table :

Height (in cm)	Class mark	Frequency
140 - 145	142.5	24
145 - 150	147.5	32
150 - 155	152.5	16
155 - 160	157.5	20
160 - 165	162.5	44

Steps to draw frequency polygon :

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

2. Take 2 cm along y-axis = 10 boys.

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 140.

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 135 - 140 with zero frequency and join the other end with mid-point of class 165 - 170 with zero frequency.

The required frequency polygon is shown below :

Frequency distribution table :

Weight (in kg)	Class mark	Frequency
35 - 40	37.5	6
40 - 45	42.5	17
45 - 50	47.5	30
50 - 55	52.5	8
55 - 60	57.5	3

Steps to draw frequency polygon :

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

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

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 35.

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 - 35 with zero frequency and join the other end with mid-point of class 60 - 65 with zero frequency.

The required frequency polygon is shown below :

Frequency distribution table :

Weekly wages (in ₹)	Class mark	Frequency
750 - 850	800	52
850 - 950	900	41
950 - 1050	1000	65
1050 - 1150	1100	54
1150 - 1250	1200	38

Steps to draw frequency polygon :

1. Take 2 cm along x-axis = ₹ 100.

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

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 750.

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 650 - 750 with zero frequency and join the other end with mid-point of class 1250 - 1350 with zero frequency.

The required frequency polygon is shown below:

The following frequency distribution table is discontinuous. Convert it into continuous frequency distribution.

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

= $\frac{6 - 5}{2} = \frac{1}{2}$ = 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 table :

Class-interval	Class mark	Frequency
0.5 - 5.5	3	5
5.5 - 10.5	8	8
10.5 - 15.5	13	12
15.5 - 20.5	18	7
20.5 - 25.5	23	4

Steps to draw frequency polygon :

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

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

3. Find the mid-point of class intervals.

4. A kink is drawn near x-axis to show that the scale begins at 0.5.

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 -5.5 - 0.5 with zero frequency and join the other end with mid-point of class 25.5 - 30.5 with zero frequency.

The required frequency polygon is shown below:

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

= $\frac{7 + 9 + 4 + 6 + 5}{5}$

= $\frac{31}{5}$ = 6.2

Hence, option 3 is the correct option.

The data is already arranged in the ascending order.

Number of observations, n = 12, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{12}{2}\right)^{\text{th}} \text{term} + \left(\frac{12}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{6th term} + \text{7th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{22} + \text{24}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{46}}{2} \\[1em]\end{aligned}$$

∴ Median = 23.

Hence, option 2 is the correct option.

Here n = 10, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \frac{\left(\frac{10}{2}\right)^{\text{th}} \text{term} + \left(\frac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{63} = \frac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{63} = \frac{\text{x} + \text{x+2}}{2} \\[1em]\end{aligned}$$

⇒ 126 = 2x + 2

⇒ 2x = 124

⇒ x = 62.

Hence, option 2 is the correct option.

By arranging data in ascending order we get :

10, 12, 13, 17, 18, 20

Number of observations (n) = 6

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

= $\frac{10+12+13+17+18+20}{6}$

= $\frac{90}{6}$ = 15.

Hence, option 2 is the correct option.

First 8 prime numbers are :

2, 3, 5, 7, 11, 13, 17, 19

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

= $\frac{2+3+5+7+11+13+17+19}{8}$

= $\frac{77}{8}$ = 9.625.

Hence, option 1 is the correct option.

Mean = 17

Number of observations = 6

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

⇒ 17 = $\frac{10+12+16+20+p+26}{6}$

⇒ 102 = 84 + p

⇒ p = 102 - 84

⇒ p = 18.

Hence, option 2 is the correct option.

Mean of 10 observations = 20

Mean of 15 observations = 16

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

Sum of 10 observations = 10 × 20 = 200

Sum of 15 observations = 15 × 16 = 240

Total Sum = 200 + 240 = 440

Total number of observations = 25

Mean = $\frac{440}{25}$ = 17.6.

Hence, option 4 is the correct option.

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

= $\frac{102+107+82+103+97+111+88+94}{8}$

= $\frac{784}{8}$ = 98 cm.

Hence, option 4 is the correct option.

Mean of x₁, x₂, x₃, ....., x_n is 'a'

So,

$\frac{x_1 + x_2 + \dots + x_n}{n} = a$

x₁ + x₂ + ..... + x_n = na

∴ $\sum x_i = na$

New observation is :

x₁ + a, x₂ + a, ...., x_n + a

$$\begin{aligned}\Rightarrow \text{New Mean} = \frac{(x_1 + a) + (x_2 + a) + \dots + (x_n + a)}{n} \\[1em] \Rightarrow \text{New Mean} = \frac{(x_1 + x_2 + \dots + x_n) + (a + a + \dots + n \text{ times})}{n} \\[1em] \Rightarrow \text{New Mean} = \frac{\sum x_i + na}{n}\end{aligned}$$

Since, $\sum x_i = na$

∴ $\text{New Mean} = \frac{na + na}{n} = \frac{2na}{n}$ = 2a.

Hence, option 2 is the correct option.

Mean of y₁, y₂, y₃, ...., y_n is 102

So,

$\frac{y_1 + y_2 + \dots + y_n}{n} = 102$

New observation is :

5y₁, 5y₂, 5y₃, ...., 5y_n

$$\begin{aligned}\text{New Mean} = \frac{5y_1 + 5y_2 + \dots + 5y_n}{n} \\[1em] = \frac{5(y_1 + y_2 + \dots + y_n)}{n} \\[1em] = 5 \times \left( \frac{y_1 + y_2 + \dots + y_n}{n} \right)\end{aligned}$$

Since,

$\frac{y_1 + y_2 + \dots + y_n}{n} = 102$

∴ $\text{New Mean} = 5 \times 102$ = 510.

Hence, option 3 is the correct option.

Average score of 5 innings = 42

⇒ Total score = 42 × 5 = 210

Total score of first 4 innings = 38 + 41 + 43 + 47 = 169

Fifth innings score = 210 - 169 = 41.

Hence, option 2 is the correct option.

Given

Wrong mean = 44

Total observations = 25

So,

Total wrong sum = 25 × 44 = 1100

Two numbers were wrongly copied:

Wrong numbers → 28 and 42

Correct numbers → 34 and 46

⇒ 34 - 28 = 6

⇒ 46 - 42 = 4

⇒ Increased sum = 6 + 4 = 10

So, the correct sum = 1100 + 10 = 1110

∴ Correct mean = $\frac{1110}{25}$ = 44.4

Hence, option 3 is the correct option.

Given,

Median = 24

Set of numbers arranged in ascending order,

11, 12, 14, 18, (x + 4), 30, 32, 35, 41

Number of observations (n) = 9, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th observation

⇒ 24 = $\frac{9 + 1}{2}$ th observation

⇒ 24 = $\frac{10}{2}$th observation

⇒ 24 = 5th observation

⇒ 24 = x + 4

⇒ x = 24 - 4

⇒ x = 20.

Hence, option 2 is the correct option.

Given,

Mean = 3.5

Total number of observations = 6

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

⇒ 3.5 = $\frac{x - 3 + x - 1 + 7 + x + 2x - 1 + 3x - 5}{6}$

⇒ 3.5 = $\frac{8x - 3}{6}$

⇒ 21 = 8x - 3

⇒ 8x = 24

⇒ x = 3.

∴ The observations are :

0, 2, 7, 3, 5, 4

By arranging the data in ascending order, we get :

0, 2, 3, 4, 5, 7

Number of observations (n) = 6, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{6}{2}\right)^{\text{th}} \text{term} + \left(\frac{6}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{3rd term} + \text{4th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{3} + \text{4}}{2} \\[1em] \Rightarrow \text{Median} = \frac{7}{2} \\[1em]\end{aligned}$$

∴ Median = 3.5.

Hence, option 2 is the correct option.

Total innings = 20

Arranging the table into Runs and frequency :

Runs	Frequency
0	1
5	3
17	6
32	5
61	3
70	2

0 × 1 = 0

5 × 3 = 15

17 × 6 = 102

32 × 5 = 160

61 × 3 = 183

70 × 2 = 140

⇒ Total runs = 0 + 15 + 102 + 160 + 183 + 140 = 600

⇒ Mean = $\frac{\text{Total Runs}}{\text{Total innings}}$

= $\frac{600}{20}$ = 30.

Hence, option 4 is the correct option.

1. Range = Highest value - Lowest value

= 57 - 36 = 21 kg.

Hence, option (b) is the correct option.

2. Mean = $\frac{\text{Total Weight}}{\text{Total number of observations}}$

= $\frac{1350}{30}$ = 45 kg.

Hence, option (a) is the correct option.

3. By arranging data in the ascending order, we get :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 30, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\left(\frac{n}{2}\right)^{\text{th}} \text{term} + \left(\frac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\left(\frac{30}{2}\right)^{\text{th}} \text{term} + \left(\frac{30}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{15th term} + \text{16th term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\text{45} + \text{45}}{2} \\[1em] \Rightarrow \text{Median} = \frac{90}{2} \\[1em]\end{aligned}$$

∴ Median = 45 kg.

Hence, option (b) is the correct option.

4. Earlier total sum = 1350

After adding the new student (weighing 45 kg) total sum = 1350 + 45 = 1395

∴ New mean = $\frac{1395}{31}$ = 45 kg.

Hence, option (d) is the correct option.

5. After adding the absentee ( weighing 45 kg) the observations are :

36, 38, 38, 39, 40, 40, 40, 40, 41, 41, 41, 41, 42, 44, 45, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50, 50, 51, 51, 51, 52, 57

Number of observations (n) = 31, which is odd.

Median = $\frac{n + 1}{2}$ th observation

= $\frac{31 + 1}{2}$ th observation

= $\frac{32}{2}$th observation

= 16th observation

Median = 45 kg.

Hence, option (a) is the correct option.

Given,

Mean of 19 number = 38

⇒ Total sum = 19 × 38 = 722

Mean of first 10 numbers = 36

⇒ Total sum = 10 × 36 = 360

Mean of last 10 numbers = 40

⇒ Total sum = 10 × 40 = 400

The 10th number is included in both groups:

⇒ First 10 numbers → includes 10th

⇒ Last 10 numbers → also includes 10th

So,

⇒ 10th number = (Sum of first 10 numbers + Sum of last 10 numbers)- Sum of all the 19 numbers

= (360 + 400) - 722

= 760 - 722 = 38.

∴ Assertion (A) is true.

We know that,

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

∴ Reason (R) is true.

Both Assertion (A) and Reason (R) are true.

Hence, option 3 is the correct option.

Given,

Wrong mean = 21

Total observations = 15

So,

Total wrong sum = 21 × 15 = 315

One value 15 was wrongly copied as 18

∴ 18 - 15 = 3

New total sum = 315 - 3 = 312

New mean = $\frac{312}{15}$

= 20.8.

But given mean = 20.

∴ Assertion (A) is false.

If given observation is increased by p, then

$$\begin{aligned}\Rightarrow \text{New Mean} = \frac{(x_1 + p) + (x_2 + p) + \dots + (x_n + p)}{n} \\[1em] \Rightarrow \text{New Mean} = \frac{(x_1 + x_2 + \dots + x_n) + np}{n} \\[1em] \Rightarrow \text{New Mean} = \bar{x} + p \\[1em]\end{aligned}$$

∴ Reason (R) is true.

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

Hence option 2 is the correct option.

(i) If we have a set of numbers, the average cannot be lower than the lowest value because there is no data pulling the average down past that point.

∴ Statement (i) is wrong.

(ii) Mean of a data cannot be bigger than the larger number because the average represents a "middle" value. So, their average cannot mathematically go beyond that value.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

Average Mean marks = 70

Total marks = 70 × Number of students

(i) This can be possible.

E.g : Suppose 3 students scored :

120, 50, 40

Mean = $\frac{210}{3}$ = 70.

So, scoring more than 100 is possible.

∴ Statement (i) is wrong.

(ii) Scoring less than 10 marks is also possible.

E.g : Suppose a student scored :

0, 110, 100

Mean = $\frac{210}{3}$ = 70.

So, scoring less than 10 marks is also possible.

∴ Statement (ii) is wrong.

Both are wrong.

Hence, option 4 is the correct option.

Total people = 10,625

Since, number of observations is odd

∴ Median = $\frac{n + 1}{2}$

= $\frac{10625 + 1}{2}$

= $\frac{10626}{2}$ = 5313.

So median is the 5313th person.

Number of economically poor people are = 5313 - 1 = 5312.

This means that number of rich people are = 10625 - 5312 = 5313.

∴ Number of economically poor people is less than the number of rich people.

Hence, option 1 is the correct option.

Mean of a and $\frac{1}{a}$ is x

$\Rightarrow \frac{a + \frac{1}{a}}{2} = x$

$\Rightarrow a + \frac{1}{a} = 2x$

We know that,

(a + b)³ = a³ + b³ + 3ab(a + b)

Substituting $b = \frac{1}{a}$ :

$$\begin{aligned}\Rightarrow \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3(a)\left(\frac{1}{a}\right)\left(a + \frac{1}{a}\right) \\[1em] \Rightarrow \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \\[1em]\end{aligned}$$

Substituting $a + \frac{1}{a} = 2x$ into the identity:

$$\begin{aligned}\Rightarrow (2x)^3 = a^3 + \frac{1}{a^3} + 3(2x) \\[1em] \Rightarrow 8x^3 = a^3 + \frac{1}{a^3} + 6x \\[1em] \Rightarrow a^3 + \frac{1}{a^3} = 8x^3 - 6x \\[1em]\end{aligned}$$

The mean of a³ and $\frac{1}{a^3}$ is:

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{a^3 + \frac{1}{a^3}}{2} \\[1em] \Rightarrow \text{Mean} = \frac{8x^3 - 6x}{2} \\[1em] \Rightarrow \text{Mean} = 4x^3 - 3x \\[1em] \Rightarrow \text{Mean} = x(4x^2 - 3)\end{aligned}$$

Hence, option 1 is the correct option.

Let the original numbers be :

x₁, x₂, x₃, ....., x₅₀

And their mean be 'a'

So,

$$\begin{aligned}\Rightarrow a = \frac{x_1 + x_2 + \dots + x_{50}}{50} \\[1em] \Rightarrow x_1 + x_2 + \dots + x_{50} = 50a \\[1em]\end{aligned}$$

Each number is subtracted from 53.

∴ The new numbers are :

$(53 - x_1), (53 - x_2), (53 - x_3), \dots, (53 - x_{50})$

Sum of new numbers is :

⇒ $(53 - x_1) + (53 - x_2) + (53 - x_3) + \dots + (53 - x_{50})$

⇒ 53 + 53 + ..... + 53 - (x₁ + x₂ + x₃ + ..... + x₅₀)

⇒ 50 × 53 - 50a

⇒ 2650 - 50a

Mean = $\frac{\text{Total Sum}}{\text{Total observations}}$

⇒ -3.5 = $\frac{2650 - 50a}{50}$

⇒ -175 = 2650 - 50a

⇒ 50a = 2650 + 175

⇒ a = $\frac{2825}{50}$

⇒ a = 56.5

∴ Mean of original numbers = 56.5.

Hence, option 4 is the correct option.

Total observations = 10

First 5 numbers increased by 5

∴ Total increase in sum = 5 × 5 = 25

Next 5 numbers decreased by 5

∴ Total decrease in sum = 5 × 5 = 25

So, net change = 25 - 25 = 0

So new total sum = original total sum.

Hence, there is no change in the mean of a data.

Total numbers = 5

For 5 numbers :

Median = $\frac{n + 1}{2}$ observation

⇒ 3rd observation.

For median to be 12

The arrangement should be :

3, 4, 12, 13, x or 3, 4, 12, x, 13.

In order to maximum mean, the value of x should be greatest, thus arrangement will be :

3, 4, 12, 13, x

Since, one of the values must is repeated, so value of x will be equal to 13.

3, 4, 12, 13, 13

Mean = $\frac{3 + 4 + 12 + 13 + 13}{5} = \frac{45}{5}$ = 9.

Hence, maximum mean value = 9.

Mean = ₹1,00,000

Median (mid value) = ₹80,000

This means:

50% employees earn ≤ ₹80,000

50% employees earn ≥ ₹80,000

(i) To be a Grade A officer, an employee must earn at least ₹85,000.

Since the income (₹85,000) is higher than the median (₹80,000), So fewer than 50% of the employees can qualify for Grade A.

∴ There will be more non Grade A officers than the Grade A officers.

Hence, statement (i) is false.

(ii) Given data shows that Mean (₹1,00,000) is significantly higher than the Median (₹80,000). This means there is a small group of very high earners pulling the average up.

The overall mean (₹1,00,000) includes non Grade A officers who earn less than ₹85,000. Therefore there group mean will definitely be below ₹85,000.

To pull this anchor of the non-Grade A officers up from less than ₹85,000 to a total average of ₹1,00,000, the Grade A officers must have a mean significantly higher than ₹1,00,000.

∴ Mean salaries of Grade A officers is more than ₹1,00,000.

Hence, statement (ii) is true.