Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive

Mean of the observation is given by –

So, adding the given observations, we get

Sum of the given observations = x + (x + 2) + (x + 4) + (x + 6) + (x + 8)

= x + x + 2 + x + 4 + x + 6 + x + 8

= 5x + 20

Total number of observations = 5

Mean = 11 (Given)

⇒

⇒ 55 = 5x + 20

⇒ 5x = 55 – 20 = 35

⇒ x = 7

Thus, x = 7

Mean of the observation is given by –

Total number of observations = 25

Mean of 25 observation = 27 (Given)

We get

⇒ Sum of 25 observations = 27 × 25 = 675

If each observation is decreased by 7, the Sum gets affected.

New Sum = 675 – (25 × 7) = 675 – 175 = 500

Thus, new mean = 20

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 20 & Σf_ix_i = 426

∵ mean is given by

⇒

⇒

Thus, mean is 5.325

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 40 & Σf_ix_i = 1150

∵ mean is given by

⇒

⇒

Thus, mean is 28.75

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 40 & Σf_ix_i = 1980

∵ mean is given by

⇒

⇒

Thus, mean is 49.5

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 50 & Σf_ix_i = 13200

∵ mean is given by

⇒

⇒

Thus, mean is 264

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 80 & Σf_ix_i = 8244

∵ mean is given by

⇒

⇒

Thus, mean is 103.05

Here, the method being used is direct method as it is easy to calculate the mid - points of the class intervals and the rest calculations were simple and easy.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 12 + p & Σf_ix_i = 270 + 25p

∵ mean is given by

⇒

⇒ 288 + 24p = 270 + 25p

⇒ 25p – 24p = 288 – 270

⇒ p = 18

Thus, p is 18

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 44 + f and Σf_ix_i = 752 + 20f

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 18)

⇒ 792 + 18f = 752 + 20f

⇒ 20f – 18f = 792 – 752

⇒ 2f = 40

⇒ f = 20

Thus, f is 20.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 39 + p and Σf_ix_i = 2370 + 30p

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 54)

⇒ 2106 + 54p = 2370 + 30p

⇒ 54p – 30p = 2370 – 2106

⇒ 24p = 264

⇒ p = 11

Thus, p is 11.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 63 + x + y and Σf_ix_i = 2775 + 25x + 45y

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 42)

⇒ 2646 + 42x + 42y = 2775 + 25x + 45y

⇒ 42x – 25x + 42y – 45y = 2775 – 2646

⇒ 17x – 3y = 129 …(i)

As given in the question, frequency(Σf_i) = 100

And as calculated by us, frequency (Σf_i) = 63 + x + y

Equalizing them, we get

63 + x + y = 100

⇒ x + y = 37 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 3 and then add it to eq.(i), we get

(17x – 3y) + [3(x + y)] = 129 + 111

⇒ 17x – 3y + 3x + 3y = 240

⇒ 20x = 240

⇒ x = 12

Substitute x = 12 in equation (ii),

12 + y = 37

⇒ y = 37 – 12

⇒ y = 25

Thus, x = 12 and y = 25.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 35 + f₁ + f₂ and Σf_ix_i = 6150 + 190f₁ + 210f₂

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 188)

⇒ 6580 + 188f₁ + 188f₂ = 6150 + 190f₁ + 210f₂

⇒ 190f₁ – 188f₁ + 210f₂ – 188f₂ = 6580 – 6150

⇒ 2f₁ + 22f₂ = 430 …(i)

As given in the question, frequency(Σf_i) = 100

And as calculated by us, frequency (Σf_i) = 35 + f₁ + f₂

Comparing them, we get

35 + f₁ + f₂ = 100

⇒ f₁ + f₂ = 65 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 2 and then subtracting it from eq.(i), we get

(2f₁ + 22f₂) – [2(f₁ + f₂)] = 430 – 130

⇒ 2f₁ + 22f₂ – 2f₁ – 2f₂ = 300

⇒ 20 f₂ = 300

⇒ f₂ = 15

Substitute f₂ = 15 in equation (ii),

f₁ + 15 = 65

⇒ f₁ = 65 – 15

⇒ f₁ = 50

Thus, f₁ = 50 and f₂ = 15.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 32 + f₁ + f₂ and Σf_ix_i = 1940 + 30f₁ + 70f₂

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 57.6)

⇒ 1843.2 + 57.6f₁ + 57.6f₂ = 1940 + 30f₁ + 70f₂

⇒ 57.6f₁ – 30f₁ + 57.6f₂ – 70f₂ = 1940 – 1843.2

⇒ 27.6f₁ – 12.4f₂ = 96.8

⇒ 69f₁ – 31f₂ = 242 …(i)

As given in the question, frequency(Σf_i) = 50

And as calculated by us, frequency (Σf_i) = 32 + f₁ + f₂

Comparing them, we get

32 + f₁ + f₂ = 50

⇒ f₁ + f₂ = 18 …(ii)

We will now solve equations (i) and (ii), multiply eq.(ii) by 31 and then adding to eq.(i), we get

(69f₁ – 31f₂) + [31(f₁ + f₂)] = 242 + 558

⇒ 69f₁ – 31f₂ + 31f₁ + 31f₂ = 800

⇒ 100f₁ = 800

⇒ f₁ = 8

Substitute f₁ = 8 in equation (ii),

8 + f₂ = 18

⇒ f₂ = 18 – 8

⇒ f₂ = 10

Thus, f₁ = 8 and f₂ = 10.

We will find the mean heartbeats per minute by direct method.

We have got

Σf_i = 30 & Σf_ix_i = 2277

∵ mean is given by

⇒

⇒

Thus, mean is 75.9 heartbeats per minute.

We will find the mean marks per student using Assumed - mean method, where A = Assumed mean.

We have got

A = 25, Σf_i = 100 & Σf_id_i = 300

∵ mean is given by

⇒

⇒

Thus, mean is 28.

We will find the mean of the frequency distribution using Assumed - mean method, where A = Assumed mean.

We have got

A = 150, Σf_i = 80 & Σf_id_i = - 300

∵ mean is given by

⇒

⇒

Thus, mean is 146.25.

We will find the mean of the data using Assumed - mean method, where A = Assumed mean.

We have got

A = 50, Σf_i = 220 & Σf_id_i = 2760

∵ mean is given by

⇒

⇒

Thus, mean is 62.55.

We will solve this using direct method.

We have got

Σf_i = 40 and Σf_ix_i = 2780

∵ mean is given by

⇒

⇒

Thus, mean is 69.5%.

We will find the mean of the frequency distribution using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 25 and h = 10

We have got

A = 25, h = 10, Σf_i = 50 & Σf_iu_i = 4

∵ mean is given by

⇒

⇒

Thus, mean is 25.8

We will find the mean of the data using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 40 and h = 10

We have got

A = 40, h = 10, Σf_i = 86 & Σf_iu_i = 7

∵ mean is given by

⇒

⇒

Thus, mean is 40.81

We will find the mean weight of packet using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 202.5 and h = 1

We have got

A = 202.5, h = 1, Σf_i = 70 & Σf_iu_i = - 38

∵ mean is given by

⇒

⇒

Thus, mean is 201.96 g.

We will find the mean of the frequency distribution using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 45 and h = 10

We have got

A = 45, h = 10, Σf_i = 150 & Σf_iu_i = - 37

∵ mean is given by

⇒

⇒

Thus, mean is 42.53.

We will find the mean marks using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 37.5 and h = 15

We have got

A = 37.5, h = 15, Σf_i = 50 & Σf_iu_i = 60

∵ mean is given by

⇒

⇒

Thus, mean marks are 55.5.

We will find the mean of the frequency distribution using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 33 and h = 6

We have got

A = 33, h = 6, Σf_i = 40 & Σf_iu_i = 2

∵ mean is given by

⇒

⇒

Thus, mean age is 33.3 years.

We will find the mean of the frequency distribution using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 550 and h = 20

We have got

A = 550, h = 20, Σf_i = 40 & Σf_iu_i = - 12

∵ mean is given by

⇒

⇒

Thus, mean is 544.

We will find the mean age using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 42 and h = 5

Since, the class intervals are inclusive type, we’ll first convert it into exclusive type by extending the class interval from both the ends.

We have got

A = 42, h = 5, Σf_i = 70 & Σf_iu_i = - 37

∵ mean is given by

⇒

⇒

Thus, mean age is 544 years.

We will find the average age using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 29.5 and h = 10

Since, the class intervals are inclusive type, we’ll first convert it into exclusive type by extending the class interval from both the ends.

We have got

A = 29.5, h = 10, Σf_i = 80 & Σf_iu_i = 43

∵ mean is given by

⇒

⇒

Thus, mean age is 34.88 years.

We will find the mean weight using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 92 and h = 5

Since, the class intervals are inclusive type, we’ll first convert it into exclusive type by extending the class interval from both the ends.

We have got

A = 92, h = 5, Σf_i = 60 & Σf_iu_i = - 19

∵ mean is given by

⇒

⇒

Thus, mean weight is 90 g.

We will find the mean marks using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 17.5 and h = 5

Since, the class intervals are less - than type, we’ll first convert it into exclusive type.

We have got

A = 17.5, h = 5, Σf_i = 80 & Σf_iu_i = 17

∵ mean is given by

⇒

⇒

Thus, mean marks correct to 2 decimal places are 18.56.

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 140

⇒ N/2 = 140/2 = 70

The cumulative frequency just greater than (N/2 = ) 70 is 115, so the corresponding median class is 45 - 60 and accordingly we get C_f = 65(cumulative frequency before the median class).

Now, since median class is 45 - 60.

∴ l = 45, h = 15, f = 50, N/2 = 70 and C_f = 65

Median is given by,

⇒

= 45 + 1.5

= 46.5

Thus, median age is 46.5 years.

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 41, so the corresponding median class is 35 - 42 and accordingly we get C_f = 25(cumulative frequency before the median class).

Now, since median class is 35 - 42.

∴ l = 35, h = 7, f = 16, N/2 = 25 and C_f = 25

Median is given by,

⇒

= 35 + 0

= 35

Thus, median marks are 35.

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 150

⇒ N/2 = 150/2 = 75

The cumulative frequency just greater than (N/2 = ) 75 is 120, so the corresponding median class is 200 - 300 and accordingly we get C_f = 72(cumulative frequency before the median class).

Now, since median class is 200 - 300.

∴ l = 200, h = 100, f = 48, N/2 = 75 and C_f = 72

Median is given by,

⇒

= 200 + 6.25

= 206.25

Thus, median wage is Rs. 206.25.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 49

⇒ N/2 = 49/2 = 24.5

The cumulative frequency just greater than (N/2 = ) 24.5 is 25, so the corresponding median class is 15 - 20 and accordingly we get C_f = 11(cumulative frequency before the median class).

Now, since median class is 15 - 20.

∴ l = 15, h = 5, f = 15, N/2 = 24.5 and C_f = 11

Median is given by,

⇒

= 15 + 4.5

= 19.5

Thus, median is 19.5.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 67

⇒ N/2 = 67/2 = 33.5

The cumulative frequency just greater than (N/2 = ) 33.5 is 42, so the corresponding median class is 125 - 145 and accordingly we get C_f = 22(cumulative frequency before the median class).

Now, since median class is 125 - 145.

∴ l = 125, h = 20, f = 20, N/2 = 33.5 and C_f = 22

Median is given by,

⇒

= 125 + 11.5

= 136.5

Thus, median is 136.5.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 100

⇒ N/2 = 100/2 = 50

The cumulative frequency just greater than (N/2 = ) 50 is 56, so the corresponding median class is 150 - 155 and accordingly we get C_f = 34(cumulative frequency before the median class).

Now, since median class is 150 - 155.

∴ l = 150, h = 5, f = 22, N/2 = 50 and C_f = 34

Median is given by,

⇒

= 150 + 3.636

= 153.64

Thus, median is 153.64 cm.

Median(given) = 24, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x is the unknown frequency.

Median = 24 (as already mentioned in the question)

24 lies between 20 - 30 ⇒ Median class = 20 - 30

∴ l = 20, h = 10, f = x, N/2 = (55 + x)/2 and C_f = 30

Median is given by,

⇒

⇒

⇒ 24 – 20 = (10x – 50)/2x

⇒ (4)(2x) = 10x – 50

⇒ 8x = 10x – 50

⇒ 10x – 8x = 50

⇒ 2x = 50

⇒ x = 25

Thus, the unknown frequency is 25.

Given: Median = 16 & N = 70

Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x is the unknown frequency.

Median = 16 (as already mentioned in the question)

16 lies between 15 - 20 ⇒ Median class = 15 - 20

∴ l = 15, h = 5, f = 15, N/2 = (55 + a + b)/2 and C_f = 24 + a

Median is given by,

⇒

⇒

⇒ 16 – 15 = (7 – a + b)/6

⇒ 6 = 7 – a + b

⇒ a – b = 1 …(i)

And given that N = 70

⇒ 55 + a + b = 70

⇒ a + b = 15 …(ii)

Solving equations (i) & (ii), we get

(a – b) + (a + b) = 1 + 15

⇒ 2a = 16

⇒ a = 8

Substituting a = 8 in eq.(i),

8 – b = 1

⇒ b = 7

Thus, the unknown frequencies are a = 8 and b = 7.

Given: Median = 5000 & N = 60

Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x is the unknown frequency.

Given,
Median = 5000 (as already mentioned in the question)

5000 lies between 4500 - 5500 ⇒ Median class = 4500 - 5500

∴ l = 4500, h = 1000, f = y, N/2 = 60/2=30
and C_f = 5 + x

Median is given by,

⇒

⇒

⇒ 5000 – 4500 = (25000 – 1000x)/y

⇒ 500y = 25000 – 1000x

⇒ 2x + y = 50 …(i)

And given that N = 60

⇒ 25 + x + y = 60

⇒ x + y = 35 …(ii)

Solving equations (i) & (ii), we get

(2x + y) – (x + y) = 50 – 35

⇒ x = 15

Substituting x = 15 in eq.(ii),

15 + y = 35

⇒ y = 20

Thus, the unknown frequencies are x = 15 and y = 20.

Given: Median = 32.5 & N = 40

Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x is the unknown frequency.

Median = 32.5 (as already mentioned in the question)

32.5 lies between 30 - 40 ⇒ Median class = 30 - 40

∴ l = 30, h = 10, f = 12, N/2 = (31 + f₁ + f₂)/2 = 40/2 and C_f = 14 + f₁

Median is given by,

⇒

⇒

⇒ 32.5 – 30 = (60 – 10f₁)/12

⇒ (2.5)(12) = 60 – 10f₁

⇒ 30 = 60 – 10f₁

⇒ f₁ = 3 …(i)

And given that N = 40

⇒ 31 + f₁ + f₂ = 40

⇒ f₁ + f₂ = 9 …(ii)

Substituting f₁ = 3 in eq.(ii),

3 + f₂ = 9

⇒ f₂ = 6

Thus, the unknown frequencies are f₁ = 3 and f₂ = 6.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table and convert it into exclusive - type by adjusting from both ends of a class.

So, N = 340

⇒ N/2 = 340/2 = 170

The cumulative frequency just greater than (N/2 = ) 170 is 199, so the corresponding median class is 32.5 - 39.5 and accordingly we get C_f = 131(cumulative frequency before the median class).

Now, since median class is 32.5 - 39.5.

∴ l = 32.5, h = 7, f = 68, N/2 = 170 and C_f = 131

Median is given by,

⇒

= 32.5 + 4.014

= 36.51

Thus, median is 36.51 years.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table and convert it into exclusive - type by adjusting from both ends of a class.

So, N = 98

⇒ N/2 = 98/2 = 49

The cumulative frequency just greater than (N/2 = )49 is 70, so the corresponding median class is 90.5 - 100.5 and accordingly we get C_f = 40(cumulative frequency before the median class).

Now, since median class is 90.5 - 100.5.

∴ l = 90.5, h = 10, f = 30, N/2 = 49 and C_f = 40

Median is given by,

⇒

= 90.5 + 3

= 93.5

Thus, median is Rs. 93.5.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table and convert it into exclusive - type by adjusting from both ends of a class.

So, N = 123

⇒ N/2 = 123/2 = 61.5

The cumulative frequency just greater than (N/2 = )61.5 is 65, so the corresponding median class is 15.5 - 20.5 and accordingly we get C_f = 33(cumulative frequency before the median class).

Now, since median class is 15.5 - 20.5.

∴ l = 15.5, h = 5, f = 32, N/2 = 61.5 and C_f = 33

Median is given by,

⇒

= 15.5 + 4.45

= 19.95

Thus, median is 19.95.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table and convert it into exclusive - type.

So, N = 200

⇒ N/2 = 200/2 = 100

The cumulative frequency just greater than (N/2 = )100 is 116, so the corresponding median class is 50 - 60 and accordingly we get C_f = 92(cumulative frequency before the median class).

Now, since median class is 50 - 60.

∴ l = 50, h = 10, f = 24, N/2 = 100 and C_f = 92

Median is given by,

⇒

= 50 + 3.33

= 53.33

Thus, median is 53.33.

Here, the maximum class frequency is 45.

The class corresponding to this frequency is the modal class. ⇒ modal class = 30 - 40

∴ lower limit of the modal class (l) = 30

Modal class size (h) = 10

Frequency of the modal class (f₁) = 45

Frequency of class preceding the modal class (f₀) = 35

Frequency of class succeeding the modal (f₂) = 25

Mode is given by,

⇒

⇒

⇒ Mode = 30 + 3.33 = 33.33

Hence, the mode is 33.33

Here, the maximum class frequency is 28.

The class corresponding to this frequency is the modal class. ⇒ modal class = 40 - 60

∴ lower limit of the modal class (l) = 40

Modal class size (h) = 20

Frequency of the modal class (f₁) = 28

Frequency of class preceding the modal class (f₀) = 16

Frequency of class succeeding the modal (f₂) = 20

Mode is given by,

⇒

⇒

⇒ Mode = 40 + 12 = 52

Hence, the mode is 52.

Here, the maximum class frequency is 20.

The class corresponding to this frequency is the modal class. ⇒ modal class = 160 - 165

∴ lower limit of the modal class (l) = 160

Modal class size (h) = 5

Frequency of the modal class (f₁) = 20

Frequency of class preceding the modal class (f₀) = 8

Frequency of class succeeding the modal (f₂) = 12

Mode is given by,

⇒

⇒

⇒ Mode = 160 + 3 = 163

Hence, the mode is 163 cm.

Mode represents frequency, hence 163 cm is the height of maximum number of students.

To find the mean, we will solve by using direct method.

We have got

Σf_i = 60 & Σf_ix_i = 9670

∵ mean is given by

⇒

⇒

Thus, mean is 161.17

Mean represents average, thus 161.7 cm is the average height of all the students.

Here, the maximum class frequency is 25.

The class corresponding to this frequency is the modal class. ⇒ modal class = 26 - 30

∴ lower limit of the modal class (l) = 26

Modal class size (h) = 4

Frequency of the modal class (f₁) = 25

Frequency of class preceding the modal class (f₀) = 20

Frequency of class succeeding the modal (f₂) = 22

Mode is given by,

⇒

⇒

⇒ Mode = 26 + 2.5 = 28.5

Hence, the mode is 28.5.

Expenditure done by maximum number of manual workers is estimated by finding mode.

So here, the maximum class frequency is 40.

The class corresponding to this frequency is the modal class. ⇒ modal class = 1500 - 2000

∴ lower limit of the modal class (l) = 1500

Modal class size (h) = 500

Frequency of the modal class (f₁) = 40

Frequency of class preceding the modal class (f₀) = 24

Frequency of class succeeding the modal (f₂) = 31

Mode is given by,

⇒

⇒

⇒ Mode = 1500 + 320 = 1820

Hence, the mode is Rs.1820.

Here, the maximum class frequency is 150.

The class corresponding to this frequency is the modal class. ⇒ modal class = 5000 - 10000

∴ lower limit of the modal class (l) = 5000

Modal class size (h) = 5000

Frequency of the modal class (f₁) = 150

Frequency of class preceding the modal class (f₀) = 90

Frequency of class succeeding the modal (f₂) = 100

Mode is given by,

⇒

⇒

⇒ Mode = 5000 + 2727.27 = 7727.27

Hence, the mode is Rs.7727.27.

Here, the maximum class frequency is 24.

The class corresponding to this frequency is the modal class. ⇒ modal class = 15 - 20

∴ lower limit of the modal class (l) = 15

Modal class size (h) = 5

Frequency of the modal class (f₁) = 24

Frequency of class preceding the modal class (f₀) = 18

Frequency of class succeeding the modal (f₂) = 17

Mode is given by,

⇒

⇒

⇒ Mode = 15 + 2.30 = 17.30

Hence, the mode is 17.30 years.

Here, the maximum class frequency is 32.

The class corresponding to this frequency is the modal class. ⇒ modal class = 85 - 95

∴ lower limit of the modal class (l) = 85

Modal class size (h) = 10

Frequency of the modal class (f₁) = 32

Frequency of class preceding the modal class (f₀) = 30

Frequency of class succeeding the modal (f₂) = 6

Mode is given by,

⇒

⇒

⇒ Mode = 85 + 0.71 = 85.71

Hence, the mode is 85.71.

Since, the given data is in inclusive series, it needs to get converted in exclusive series.

Here, the maximum class frequency is 28.

The class corresponding to this frequency is the modal class. ⇒ modal class = 15.5 - 20.5

∴ lower limit of the modal class (l) = 15.5

Modal class size (h) = 5

Frequency of the modal class (f₁) = 28

Frequency of class preceding the modal class (f₀) = 18

Frequency of class succeeding the modal (f₂) = 20

Mode is given by,

⇒

⇒

⇒ Mode = 15.5 + 2.78 = 23.28

Hence, the mode is 23.28.

To find frequencies, we have Sum of frequencies that is, 181.

Using Sum of frequencies = 181,

x + 15 + 18 + 30 + 50 + 48 + x = 181

⇒ 2x + 161 = 181

⇒ 2x = 181 – 161 = 20

⇒ x = 10

Thus we have,

Here, the maximum class frequency is 50.

The class corresponding to this frequency is the modal class. ⇒ modal class = 13 - 15

∴ lower limit of the modal class (l) = 13

Modal class size (h) = 2

Frequency of the modal class (f₁) = 50

Frequency of class preceding the modal class (f₀) = 30

Frequency of class succeeding the modal (f₂) = 48

Mode is given by,

⇒

⇒

⇒ Mode = 13 + 1.82 = 14.82

Hence, the mode is 14.82.

To find mean, we will solve by direct method:

We have got

Σf_i = 50 & Σf_ix_i = 1910

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 37, so the corresponding median class is 40 - 50 and accordingly we get C_f = 25(cumulative frequency before the median class).

Now, since median class is 40 - 50.

∴ l = 40, h = 10, f = 16, N/2 = 25 and C_f = 25

Median is given by,

⇒

= 40 + 0

= 40

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(40) – 2(38.2)

= 120 – 76.4

= 43.6

Hence, mean is 38.2, median is 40 and mode is 43.6.

To find mean, we will solve by direct method:

We have got

Σf_i = 50 & Σf_ix_i = 3120

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 36, so the corresponding median class is 60 - 80 and accordingly we get C_f = 24(cumulative frequency before the median class).

Now, since median class is 60 - 80.

∴ l = 60, h = 20, f = 12, N/2 = 25 and C_f = 24

Median is given by,

⇒

= 60 + 1.67

= 61.67

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(61.67) – 2(62.4)

= 185.01 – 124.8

= 60.21

Hence, mean is 62.4, median is 61.67 and mode is 60.21.

To find mean, we will solve by direct method:

We have got

Σf_i = 25 & Σf_ix_i = 4171

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 25

⇒ N/2 = 25/2 = 12.5

The cumulative frequency just greater than (N/2 = ) 12.5 is 16, so the corresponding median class is 150 - 200 and accordingly we get C_f = 10(cumulative frequency before the median class).

Now, since median class is 150 - 200.

∴ l = 150, h = 50, f = 6, N/2 = 12.5 and C_f = 10

Median is given by,

⇒

= 150 + 20.83

= 170.83

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(170.83) – 2(169)

= 512.49 – 338

= 174.49

Hence, mean is 169, median is 170.83 and mode is 174.49.

To find mean, we will solve by direct method:

We have got

Σf_i = 100 & Σf_ix_i = 4970

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 100

⇒ N/2 = 100/2 = 50

The cumulative frequency just greater than (N/2 = ) 50 is 71, so the corresponding median class is 45 - 55 and accordingly we get C_f = 38(cumulative frequency before the median class).

Now, since median class is 45 - 55.

∴ l = 45, h = 10, f = 33, N/2 = 50 and C_f = 38

Median is given by,

⇒

= 45 + 3.64

= 48.64

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(48.64) – 2(49.7)

= 145.92 – 99.4

= 46.52

Hence, mean is 49.7, median is 48.64 and mode is 46.52.

To find mean, we will solve by direct method:

We have got

Σf_i = 50 & Σf_ix_i = 7490

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 42, so the corresponding median class is 150 - 160 and accordingly we get C_f = 22(cumulative frequency before the median class).

Now, since median class is 150 - 160.

∴ l = 150, h = 10, f = 20, N/2 = 25 and C_f = 22

Median is given by,

⇒

= 150 + 1.5

= 151.5

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(151.5) – 2(149.8)

= 454.5 – 299.6

= 154.9

Hence, mean is 149.8, median is 151.5 and mode is 154.9.

To find mean, we will solve by direct method:

We have got

Σf_i = 50 & Σf_ix_i = 7260

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 26, so the corresponding median class is 120 - 140 and accordingly we get C_f = 12(cumulative frequency before the median class).

Now, since median class is 120 - 140.

∴ l = 120, h = 20, f = 14, N/2 = 25 and C_f = 12

Median is given by,

⇒

= 120 + 18.57

= 138.57

And we know that,

Mode = 3(Median) – 2(Mean)

= 3(138.57) – 2(145.2)

= 415.71 – 290.4

= 125.31

Hence, mean is 145.2, median is 138.57 and mode is 125.31.

To find mean, we will solve by direct method:

We have got

Σf_i = 30 & Σf_ix_i = 6150

∵ mean is given by

⇒

⇒

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 30

⇒ N/2 = 30/2 = 15

The cumulative frequency just greater than (N/2 = ) 15 is 25, so the corresponding median class is 200 - 250 and accordingly we get C_f = 13(cumulative frequency before the median class).

Now, since median class is 200 - 250.

∴ l = 200, h = 50, f = 12, N/2 = 15 and C_f = 13

Median is given by,

⇒

= 200 + 8.33

= 208.33

Hence, mean is 205 and median is 208.33

The frequency distribution table for ‘less than’ type is:

Lets plot a graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

As we have N = 53 by the frequency table.

N/2 = 53/2 = 26.5

Mark 26.5 on y - axis and the corresponding point on x - axis would be the median.

The corresponding point on x - axis is 66.4.

Hence, median is 66.4

Lets plot a graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

As we have N = 80 by the frequency table.

N/2 = 80/2 = 40

Mark 40 on y - axis and the corresponding point on x - axis would be the median.

The corresponding point on x - axis is 76.

Hence, median is 76.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

As we have N = 102 by the frequency table.

N/2 = 102/2 = 51

Mark 51 on y - axis and the corresponding point on x - axis would be the median.

The corresponding point on x - axis is 70.5.

Hence, median is 70.5.

The frequency distribution table for ‘less than’ type is:

Lets plot the graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

As we have N = 230 by the frequency table.

N/2 = 230/2 = 115

Mark 115 on y - axis and the corresponding point on x - axis would be the median.

The corresponding point on x - axis is 590.

Hence, median is 590.

(i) The frequency distribution table for ‘less than’ type is:

(ii) The frequency distribution table for ‘more than’ type is:

Plotting points for ‘less - than ogive’ and ‘more - than ogive’ on the graph,

In this type of graph where ‘less than ogive’ and more than ogive’ are plotted in the same graph, median is found on x - axis by the intersection of these two ogives.

Here, median = 29.5

(i) The frequency distribution table for ‘less than’ type is:

(ii) The frequency distribution table for ‘more than’ type is:

Plotting points for ‘less - than ogive’ and ‘more - than ogive’ on the graph,

In this type of graph where ‘less than ogive’ and more than ogive’ are plotted in the same graph, median is found on x - axis by the intersection of these two ogives.

Here, median = 166

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 26, so the corresponding median class is 30 - 40.

Hence, median class = 30 - 40

Here, the maximum class frequency is 27.

The class corresponding to this frequency is the modal class. ⇒ modal class = 40 - 50

∴ lower limit of the modal class (l) = 40

Here, the maximum class frequency is 30.

The class corresponding to this frequency is the modal class. ⇒ modal class = 150 - 200

∴ lower limit of the modal class (l) = 150

The class mark is found by,

∴ Class mark is 175.

Since we have 25 observations, that is odd number of observations, median is found at position.

So since, n = 25

⇒ Median will be found at position. ⇒ Median = 13^th observation

Given: mode = 1000 and median = 1250

The empirical relationship between mean, median and mode is,

Mode = 3(Median) – 2(Mean)

⇒ 2(Mean) = 3(Median) – Mode

⇒ Mean = [3(Median) – Mode]/2

⇒ Mean = [3(1250) – 1000]/2

⇒ Mean = [3750 – 1000]/2 = 2750/2 = 1375

∴ mean = 1375

Here, the maximum class frequency is 25.

The class corresponding to this frequency is the modal class. ⇒ modal class = 40 - 60

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 35, so the corresponding median class is 40 - 60.

∴ modal class = 40 - 60 and median class = 40 - 60

Class mark is given by

Class mark of class 10 - 25 =

Class mark of class 35 - 55 =

∴ Class mark of class 10 - 25 is 17.5 and 35 - 55 is 45.

We have got

A = 25, Σf_i = 50 & Σf_id_i = 110

∵ By Assumed - mean method, mean is given by

⇒

⇒

Thus, mean is 27.2

According to the question,

⇒ X = 36 × 4 = 144 and Y = 64 × 3 = 192

We have, X = 144 and Y = 192

Mean of distribution (X + Y = 144 + 192 = ) 336 is,

Mean = 336/(36 + 64) = 336/100 = 3.36

Hence, mean = 3.36

Given: number of classes = 12,

Class width = 2.5, and

Lowest class boundary = 8.1

Upper class is given by,

Upper class boundary = Lower class boundary + (width × number of classes)

Substituting values,

⇒ Upper class boundary = 8.1 + (2.5 × 12)

⇒ Upper class boundary = 8.1 + 30 = 38.1

Hence, upper class boundary is 38.1

Since there are 10 observations, that is, even number of observations, median is found by taking average of and observations.

So, median is found at average of and observations.

(5)^th observation = x and (6)^th observation = x + 2

Taking average,

Median = (x + x + 2)/2

⇒ 63 = (2x + 2)/2 [∵ given is median = 63]

⇒ 126 = 2x + 2

⇒ 2x = 126 – 2

⇒ 2x = 124

⇒ x = 124/2 = 62

∴ x = 62

As median is the “middle” number of the sorted list of numbers, and given is median of 19 observations observed to be 30.

⇒ 30 is the middle most value amongst 19 observations.

If two more observations (8 and 32) are added, where 8 is less than 30 and 32 is more than 30. 30 is still the middlemost value as the two values are added on either side of 30.

Hence, median of 21 observations are 30.

Arranging the values x/5, x/4, x/2, x and x/3 in ascending order, we get

x/5, x/4, x/3, x/2 and x

Here, median is x/3 as it is the middle value amongst all values.

Given: median = 8

⇒ x/3 = 8

⇒ x = 24

Hence, x = 24

Here, the maximum class frequency is 23.

The class corresponding to this frequency is the modal class. ⇒ modal class = 12 - 15

Lets form a table.

Since, modal class = 12 - 15, the corresponding cumulative frequency is 53.

Here, the maximum class frequency is 18.

The class corresponding to this frequency is the modal class. ⇒ modal class = 40 - 60

∴ lower limit of the modal class (l) = 40

Modal class size (h) = 20

Frequency of the modal class (f₁) = 18

Frequency of class preceding the modal class (f₀) = 6

Frequency of class succeeding the modal (f₂) = 10

Mode is given by,

⇒

⇒

⇒ Mode = 40 + 12 = 52

Hence, the mode is 52.

In a ‘less than type’ cumulative frequency distribution, upper limit of the classes are considered.

It is given by,

To find p and q, solve by finding cumulative frequency,

⇒ p = 11 + 12 = 23

And 46 = 33 + q ⇒ q = 46 – 33 = 13

∴ p = 23 and q = 13

Lets form the table again,

For modal class,

Here, the maximum class frequency is 20.

The class corresponding to this frequency is the modal class. ⇒ modal class = 500 - 600

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency

and C_f = cumulative frequency

So, N = 80

⇒ N/2 = 80/2 = 40

The cumulative frequency just greater than (N/2 = ) 40 is 46, so the corresponding median class is 400 - 500.

∴ modal class = 500 - 600 and median class = 400 - 500

In a ‘less than type’ cumulative frequency distribution, lower limits of the classes are considered.

It is given by,

The frequency distribution table is:

(a) To convert the given frequency distribution into continuous form, adjust the end - limits of each class.

(b) To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency

and C_f = cumulative frequency

So, N = 2300

⇒ N/2 = 2300/2 = 1150

The cumulative frequency just greater than (N/2 = ) 1150 is 1825, so the corresponding median class is 50.5 - 60.5.

∴ median class = 50.5 - 60.5

The class mark of 50.5 - 60.5 is

(c) For modal class,

Here, the maximum class frequency is 529.

The class corresponding to this frequency is the modal class. ⇒ modal class = 40.5 - 50.5

The cumulative frequency corresponding to the modal class is 1330

We have got

Σf_i = 43 + p and Σf_ix_i = 1245 + 15p

∵ mean is given by

⇒ (∵ given: mean of pocket allowance is 27)

⇒ 1161 + 27p = 1245 + 15p

⇒ 27p – 15p = 1245 – 1161

⇒ 12p = 84

⇒ p = 84/12

⇒ p = 7

Thus, p = 7

Given: Median = 24

Let the unknown frequency be x.

Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x is the unknown frequency.

Median = 24 (as already mentioned in the question)

24 lies between 20 - 30 ⇒ Median class = 20 - 30

∴ l = 20, h = 10, f = x, N/2 = (55 + x)/2 and C_f = 30

Median is given by,

⇒

⇒

⇒ 24 – 20 = (5x – 25)/x

⇒ 4x = 5x – 25

⇒ 5x – 4x = 25

⇒ x = 25

Mean, median and mode are a measure of central tendency but range of a set of data is the difference between the largest and smallest values.

Mean is just the average of some observations. It cannot be determined graphically as the values cannot be Summed up.

Mean is influenced by extreme values in class intervals, while median and mode is not influenced by extreme values as median is the mid value among observations and mode is the value that is come often in a set of data values and they are independent of extreme values.

A histogram shows frequencies of value and mode of frequency distribution can be obtained from a histogram.

An ogive is a type of frequency polygon that shows cumulative frequencies and median is the mid - value among given values. Graphically, median can be found by ogive as corresponding to one axis, we get the value on the other axis in an ogive.

Cumulative frequency table is useful in determining the median in the case of class intervals.

The abscissa of the point of intersection of the ‘less than type’ and ‘more than type’ cumulative frequency curves of the grouped data gives its median as it gives accurate mid - point among all values.

If mean = 7, x_i’s = midpoints of the class intervals and f_i = corresponding frequencies

Mean is given by,

⇒

Or

Or

Or

Since, d_i = x_i – A,

where d_i = deviation and A = Assumed mean

And u_i = d_i/h = (x_i – A)/h,

where h = class width

For finding the mean of the grouped data, d_i’s are deviations from A(Assumed mean) of the midpoints of the classes. It is necessary to find midpoints of the class intervals to find mean of the grouped data.

Class marks are the aggregates value of the classes and is given by:

Class mark = (lower limit + upper limit)/2

And the frequencies are Assumed to be centered at the class marks of the classes.

This relationship between mean, median and mode is also called empirical relationship and is given by,

mode = (3 × median) – (2 × mean)

If ‘less than type’ ogive and ‘more than type’ ogive intersect each other at (20.5,15.5), then median of the given data is 20.5 as median in this kind of graph is found on x - axis, which represents class intervals (upper limit/lower limit).

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency

and C_f = cumulative frequency

So, N = 60

⇒ N/2 = 60/2 = 30

The cumulative frequency just greater than (N/2 = ) 30 is 37, so the corresponding median class is 160 - 165.

∴ upper limit of median class = 165

For modal class,

Here, the maximum class frequency is 16.

The class corresponding to this frequency is the modal class. ⇒ modal class = 150 - 155

∴ lower limit of the modal class = 150

Hence, Sum of lower limit of the modal class and upper limit of the median class = 165 + 150 = 315

For modal class,

Here, the maximum class frequency is 30.

The class corresponding to this frequency is the modal class. ⇒ modal class = 30 - 40

∴ modal class = 30 - 40

Mode is given by,

where,

x_k = lower limit of the modal class,

h = class width,

f_k = frequency of the modal class,

f_{k - 1} = frequency of class preceding the modal class

and f_{k + 1} = frequency of class succeeding the modal class

Medium is given by,

Where

N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Given: mean = 8.9 and median = 9

By empirical formula,

mode = (3 × median) – (2 × mean)

⇒ mode = (3 × 9) – (2 × 8.9)

⇒ mode = 27 – 17.8 = 9.2

To find median,

Assume Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 40, so the corresponding median class is 55 - 65 and accordingly we get C_f = 20(cumulative frequency before the median class).

Now, since median class is 55 - 65.

∴ l = 55, h = 10, f = 20, N/2 = 25 and C_f = 20

Median is given by,

⇒

= 55 + 2.5

= 57.5

Thus, median age is 57.5.

Here, the maximum class frequency is 25.

The class corresponding to this frequency is the modal class. ⇒ modal class = 22 - 26

∴ lower limit of the modal class (l) = 22

Modal class size (h) = 4

Frequency of the modal class (f₁) = 25

Frequency of class preceding the modal class (f₀) = 16

Frequency of class succeeding the modal (f₂) = 19

Mode is given by,

⇒

⇒

⇒ Mode = 22 + 2.4 = 24.4

Hence, the mode is 24.4.

Given: mean = 28 and mode = 16

By empirical formula,

mode = (3 × median) – (2 × mean)

⇒ 3 × median = mode + (2 × mean)

⇒ 3 × median = 16 + (2 × 28)

⇒ 3 × median = 16 + 56

⇒ 3 × median = 72

⇒ median = 72/3 = 24

Given: median = 26 and mode = 29

By empirical formula,

mode = (3 × median) – (2 × mean)

⇒ 2 × mean = (3 × median) – mode

⇒ 2 × mean = (3 × 26) – 29

⇒ 2 × mean = 78 – 29

⇒ 2 × mean = 49

⇒ mean = 49/2 = 24.5

As in a symmetrical frequency distribution, the left and right hand side of the distribution is roughly equally balanced around the mean of the distribution.

From the above table, number of families having income range 20000 - 25000 is 13. (Observe the frequency values corresponding to the monthly income)

Listing out all first 8 prime numbers, we have

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

Since, the median will be at the aggregate of the (8/2 =) 4^th position and (8/2 + 1 = ) 5^th position.

We have, (7 + 11)/2 = 18/2 = 9

Thus, median is 9.

It’s given that mean of 20 numbers is 0, which implies that average of 20 numbers is 0.

This means that Sum of 20 numbers is 0.

If Sum of 19 numbers out of 20 is x(say), then the 20^th number will be absolutely 0 for the average to be 0.

⇒ At the most, 19 numbers will be positive.

Since there are 6 number of observation in all, which is an even number of observation.

Median will be found at the aggregate of (6/2 = ) 3^rd and (6/2 + 1 = ) 4^th position.

3^rd value = x – 1 and 4^th value = x – 3

Taking their aggregate, we get

Median = (x – 1 + x – 3)/2

⇒ 13 = (2x – 4)/2 [∵ median = 13]

⇒ 26 = 2x – 4

⇒ 2x = 26 + 4

⇒ 2x = 30

⇒ x = 15

Thus, median is 15.

Given: mean of 2, 7, 6 and x is 15

Mean =

⇒ _{15 =}

⇒ 60 = 15 + x

⇒ x = 60 – 15 = 45 …(i)

Also, given that mean of 18, 1, 6, x and y is 10

_{10 =} [∵ mean = 10]

⇒ 50 = 25 + 45 + y = 70 + y [from equation (i)]

⇒ y = 50 – 70 = - 20

Thus, y = - 20

The correct answer is:

According to Reason(R),

Relationship between mean, median and mode is,

Mode = 3(Median) – 2(Mean)

If we substitute the values given in the above relationship,

Median = 150 and Mean = 148 (given)

Mode = 3(150) – 2(148)

⇒ Mode = 450 – 296 = 154

We get mode = 154, which satisfies the assertion.

Thus, Assertion(A) and Reason(R) are true and Reason(R) is the correct explanation of the Assertion(A).

Here, the maximum class frequency is 23.

The class corresponding to this frequency is the modal class. ⇒ modal class = 12 - 15

∴ lower limit of the modal class (l) = 12

Modal class size (h) = 3

Frequency of the modal class (f₁) = 23

Frequency of class preceding the modal class (f₀) = 21

Frequency of class succeeding the modal (f₂) = 10

Mode is given by,

⇒

⇒

⇒ Mode = 12 + 0.4 = 12.4

∴ Assertion (A) is true and Reason (R) is true obviously.

But Reason (R) is not the correct explanation of Assertion (A).

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

Mean and mode does not require construction of cumulative frequency, but median necessarily requires construction of cumulative frequency, unless it is raw data (in which median is the (n/2)^th value, when there are n number of observations; and the average of (n/2)^th and (n/2 + 1)^th values, when there are n observations).

Given: mean = 27 and median = 33

We have to find the value of mode.

Empirical relationship is given by,

Mode = 3(Median) – 2(Mean)

⇒ Mode = 3(33) – 2(27)

⇒ Mode = 99 – 54 = 45

We need to find – (1) Median class

(2) Modal class

First we’ll find (1) Median class.

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency of class intervals

and C_f = cumulative frequency

Lets form a table.

So, N = 66

⇒ N/2 = 66/2 = 33

The cumulative frequency just greater than (N/2 = ) 33 is 37, so the corresponding median class is 10 - 15.

∴ median class is 10 - 15.

To find (2) Modal class,

Here, the maximum class frequency is 20.

The class corresponding to this frequency is the modal class. ⇒ modal class = 15 - 20

Lower limit of median = 10 and lower limit of mode = 15

Sum = 10 + 15 = 25

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency of class intervals

and C_f = cumulative frequency

Lets convert this data into exclusive type of data.

So, N = 57

⇒ N/2 = 57/2 = 28.5

The cumulative frequency just greater than (N/2 = ) 28.5 is 38, so the corresponding median class is 11.5 - 17.5.

∴ Upper limit of this median class = 17.5

Given: mean = 53.4 and mode = 55.2

We have to find the median.

By empirical formula,

Mode = 3(Median) – 2(Mean)

⇒ 3(Median) = Mode + 2(Mean)

⇒ Median = [Mode + 2(Mean)]/3

⇒ Median = [55.2 + 2(53.4)]/3

⇒ Median = [55.2 + 106.8]/3

⇒ Median = 162/3 = 54

∴ Median = 54

We need to form a ‘less than type’ table to solve this.

Here, cumulative frequency shows number of athletes taking different time intervals to run a 100 - m - hurdle race.

So by the table, there are 75 athletes who completed the race in less than 14.6 seconds.

To find median class,

Assume Σf_i = N = Sum of frequencies,

f_i = frequency of class intervals

and C_f = cumulative frequency

Lets convert this data into exclusive type of data.

So, N = 57

⇒ N/2 = 57/2 = 28.5

The cumulative frequency just greater than (N/2 = ) 28.5 is 38, so the corresponding median class is 11.5 - 17.5.

∴ Upper limit of this median class = 17.5

To find frequency corresponding to 20 - 25 class, we need to convert ‘more than or equal to’ type data into class intervals.

Observe in the table above, frequency corresponding to the class 20 - 25 is 4.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 76 & Σf_ix_i = 412

∵ mean is given by

⇒

⇒

Thus, mean is 5.421

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 30

⇒ N/2 = 30/2 = 15

The cumulative frequency just greater than (N/2 = ) 15 is 21, so the corresponding median class is 115 - 130 and accordingly we get C_f = 14(cumulative frequency before the median class).

Now, since median class is 115 - 130.

∴ l = 115, h = 15, f = 7, N/2 = 15 and C_f = 14

Median is given by,

⇒

= 115 + 2.14

= 117.14

Thus, median is 117.14 km/hr.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 92 + p and Σf_ix_i = 2580 + 15p

∵ mean is given by

⇒ (∵ given: arithmetic mean is 50)

⇒ 4600 + 50p = 2580 + 15p

⇒ 50p – 15p = 2580 – 4600

⇒ 35p = 2020

⇒ p = 11

Thus, p is 11.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

So, N = 65

⇒ N/2 = 65/2 = 32.5

The cumulative frequency just greater than (N/2 = ) 32.5 is 52, so the corresponding median class is 20 - 30 and accordingly we get C_f = 22(cumulative frequency before the median class).

Now, since median class is 20 - 30.

∴ l = 20, h = 10, f = 30, N/2 = 32.5 and C_f = 22

Median is given by,

⇒

= 20 + 3.5

= 23.5

Thus, median is 23.5.

The frequency distribution table for ‘less than’ type is:

Lets plot a graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table and convert it into exclusive - type by adjusting from both ends of a class.

So, N = 100

⇒ N/2 = 100/2 = 50

The cumulative frequency just greater than (N/2 = ) 50 is 60, so the corresponding median class is 20 - 30 and accordingly we get C_f = 24(cumulative frequency before the median class).

Now, since median class is 20 - 30.

∴ l = 20, h = 10, f = 36, N/2 = 50 and C_f = 24

Median is given by,

⇒

= 20 + 7.22

= 27.22

Thus, median is 27.22.

Lets plot a graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

As we have N = 100 by the frequency table.

N/2 = 100/2 = 50

Mark 50 on y - axis and the corresponding point on x - axis would be the median.

The corresponding point on x - axis is 55.

Hence, median is 55.

Given: Median = 35 & N = 170

Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table, where x and y are the unknown frequencies.

Median = 35 (as already mentioned in the question)

35 lies between 30 - 40 ⇒ Median class = 30 - 40

∴ l = 30, h = 10, f = 40, N/2 = (110 + x + y)/2 = 170/2 = 85 and C_f = 30 + x

Median is given by,

⇒

⇒

⇒ 35 – 30 = (55 – x)/4

⇒ 5 × 4 = 55 – x

⇒ 20 = 55 – x

⇒ x = 55 – 20 = 35 …(i)

And given that N = 170

⇒ 110 + x + y = 170

⇒ x + y = 170 – 110

⇒ x + y = 60 …(ii)

Substituting x = 35 in eq.(ii),

35 + y = 60

⇒ y = 60 – 35 = 25

Thus, the unknown frequencies are x = 35 and y = 25.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Mean = 50 and N = 120 (as given in the question)

Σf_i = 68 + f₁ + f₂ and Σf_ix_i = 3480 + 30f₁ + 70f₂

∵ mean is given by

⇒ (∵ given: mean is 50)

⇒ 3400 + 50f₁ + 50f₂ = 3480 + 30f₁ + 70f₂

⇒ 50f₁ – 30f₁ + 50f₂ – 70f₂ = 3480 – 3400

⇒ 20f₁ – 20f₂ = 80

⇒ f₁ – f₂ = 4 …(i)

As given in the question, frequency(Σf_i) = 120

And as calculated by us, frequency (Σf_i) = 68 + f₁ + f₂

Equalizing them, we get

68 + f₁ + f₂ = 120

⇒ f₁ + f₂ = 120 – 68 = 52

⇒ f₁ + f₂ = 52 …(ii)

We will now solve equations (i) and (ii), adding them we get

(f₁ + f₂) + (f₁ – f₂) = 52 + 4

⇒ 2f₁ = 56

⇒ f₁ = 56/2

⇒ f₁ = 28

Substitute f₁ = 28 in equation (ii),

28 + f₂ = 52

⇒ f₂ = 52 – 28

⇒ f₂ = 24

Thus, f₁ = 28 and f₂ = 24.

We will find the mean using step - deviation method, where A = Assumed mean and h = length of class interval.

Here, let A = 99 and h = 6

Since, the class intervals are inclusive type, we’ll first convert it into exclusive type by extending the class interval from both the ends.

We have got

A = 99, h = 6, Σf_i = 120 & Σf_iu_i = 81

∵ mean is given by

⇒

⇒

Thus, mean is 103.05.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 50 and Σf_ix_i = 1500

∵ mean is given by

⇒

⇒

Thus, mean is 30.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

We have got

So, N = 50

⇒ N/2 = 50/2 = 25

The cumulative frequency just greater than (N/2 = ) 25 is 39, so the corresponding median class is 30 - 40 and accordingly we get C_f = 24(cumulative frequency before the median class).

Now, since median class is 30 - 40.

∴ l = 30, h = 10, f = 15, N/2 = 25 and C_f = 24

Median is given by,

⇒

= 30 + 0.67

= 30.67

Thus, median is 30.67.

Since, we have got mean = 30 and median = 30.67

Applying the empirical formula,

Mode = 3(Median) – 2(Mean)

⇒ Mode = 3(30.67) – 2(30)

⇒ Mode = 92.01 – 60 = 32.01

∴ Mean = 30, Median = 30.67 and Mode = 32.01

The frequency distribution table for ‘less than’ type is:

The frequency distribution table for ‘more than’ type is:

Plotting points for ‘less - than ogive’ and ‘more - than ogive’ on the graph,

In this type of graph where ‘less than ogive’ and more than ogive’ are plotted in the same graph, median is found on x - axis by the intersection of these two ogives.

Here, median = 15.5

The frequency distribution table for ‘less than’ type is:

Lets plot the graph of ‘less than ogive’, taking upper limits of the class intervals on x - axis and cumulative frequencies on y - axis.

The frequency distribution table for ‘more than’ type is:

Lets plot a graph of ‘more than’ ogive, taking lower limits of the class intervals on x - axis and cumulative frequencies on y - axis.

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Σf_i = 47 and Σf_ix_i = 1516

∵ mean is given by

⇒

⇒

Thus, mean is 32.26.

To find median, Assume

Σf_i = N = Sum of frequencies,

h = length of median class,

l = lower boundary of the median class,

f = frequency of median class

and C_f = cumulative frequency

Lets form a table.

We have got

So, N = 47

⇒ N/2 = 47/2 = 23.5

The cumulative frequency just greater than (N/2 = ) 23.5 is 29, so the corresponding median class is 30.5 - 35.5 and accordingly we get C_f = 15(cumulative frequency before the median class).

Now, since median class is 30.5 - 35.5.

∴ l = 30.5, h = 5, f = 14, N/2 = 23.5 and C_f = 15

Median is given by,

⇒

= 30 + 3.03

= 33.03

Thus, median is 33.03.

Since, we have got mean = 32.26 and median = 33.03

Applying the empirical formula,

Mode = 3(Median) – 2(Mean)

⇒ Mode = 3(33.03) – 2(32.26)

⇒ Mode = 99.09 – 64.52 = 34.57

∴ Mean = 32.26, Median = 33.03 and Mode = 34.57