Measures Of Central Tendency

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 51

∴ mean marks = 51

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 1735

∴ mean salary = 1735

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ 32 = 26 + x

⇒ x = 6

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 43.9

∴ mean runs scored = 43.9

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 22

∴ mean marks = 22

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 400

∴ mean number of books = 400

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 52.75

∴ mean weight of 60 students = 52.75

We know that,

Where, Mean, x_i = observations and n = no. of observations.

Let the excluded no. be x and sum of other numbers be S

⇒ S + x = 90……… (1)

Also,

⇒ S = 64

Substituting value of S in (1), we get -

64 + x = 90

⇒ x = 26

∴ excluded number = 26

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

Let the sum of 13 numbers be S.

⇒ S = 312

Now, let new mean be ̅y.

⇒ = 27

⇒ new mean = 27.

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

Let the salary of the retired employee be x and sum of other employees’ salary be S.

⇒ S + x = 15000……… (1)

Also,

⇒ S = 12800

Substituting value of S in (1), we get -

12800 + x = 15000

⇒ x = 2200

∴ salary of retired employee = 26

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 7.07

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 7.55

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 7.07

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 0.55

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 2

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 23.9

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ 307.5 + 7.5P = 303 + 9P

⇒ 1.5P = 4.5

⇒ P = 3

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

Let first missing frequency be ‘y’ and second one be ‘z’.

⇒ 292 - 140 = y + 2z

⇒ y + 2z = 152……… (1)

Also, 46 + y + z + 25 + 10 + 5 = 200

⇒ y + z = 200 - 86

⇒ y + z = 114

⇒ y = 114 - z

Substituting this in (1), we get -

⇒ 114 - z + 2z = 152

⇒ z = 38

∴, y = 114 - 38

⇒ 76

⇒ The missing frequencies are - 76 and 38.

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 26.33

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 15.45

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 145.71

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 49.5

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 68.2

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 1457.14

We’ll first assume a mean A(the one with highest frequency) = 900 and here h = 820 - 800 = 20.

Now we’ll find the u_i,

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 891.2

We’ll first assume a mean A = 63 and here h = 61 - 60 = 1.

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 62.7

Here first we’ll find the class marks -

We’ll first assume a mean A = 225 and here h = 175 - 125 = 50.

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 266.25

Here first we’ll find the class marks -

We’ll first assume a mean A = 43 and here h = 23 - 18 = 5.

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 39.57

Here first we’ll find the class marks -

We’ll first assume a mean A = 25 and here h = 15 - 5 = 10.

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 23.25

Here first we’ll find the class marks -

We’ll first assume a mean A = 39.5 and here h = 19.5 - 9.5 = 10

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 34.875

Here first we’ll find the class marks -

We’ll first assume a mean A = 65 and here h = 55 - 45 = 10.

Now we’ll find the u_i

Now,

Where, = Mean, u_i = calculated values of x and f_i = frequency for that x.

⇒ = 68.56

Firstly, we’ll arrange the numbers in ascending order -

13, 19, 20, 25, 26, 28, 33, 34, 34, 36

Now, since there are 10 no.s and n = 10 is even

⇒ Median = 27

Firstly, we’ll arrange the numbers in ascending order -

19, 25, 30, 31, 32, 35, 48, 51, 59

Now, since there are 9 no.s and n = 9 is odd

⇒ Median = 32

If 25 is replaced by 52 then series will become -

19, 52, 59, 48, 35, 31, 30, 32, 51

: Firstly, we’ll arrange the numbers in ascending order -

19, 30, 31, 32, 35, 48, 51, 52, 59

Now, since there are 9 no.s and n = 9 is odd

⇒ Median = 35

Firstly, we’ll find the cumulative frequencies(c.f) –

Now, the sum of all frequencies = 99

And,

⇒ m = 50

Now we’ll see to the c.f just greater than this m

Which is 52 and then we’ll see the corresponding marks obtained and that will be the median.

⇒ Median = 30

Firstly, we’ll find the cumulative frequencies(c.f) –

Now, the sum of all frequencies = 99

And,

⇒ m = 50

Now we’ll see to the c.f just greater than this n

Which is 72 and then we’ll see the corresponding marks obtained and that will be the median.

⇒ Median = 2

Firstly, we’ll find the cumulative frequencies(c.f) –

Now, the sum of all frequencies = 160

And,

⇒ m = 80

Now we’ll see to the c.f just greater than this n

Which is 105 and then we’ll see the corresponding marks obtained and that will be the median.

⇒ Median = 35

First we’ll find the median class -

And median class is the class corresponding to cumulative frequency which is just greater than m.

Now,

And c.f just greater is 70 corresponding to 40 - 50 (median class)

Where, l = lower limit of median class, N = sum of all frequencies, C = cumulative frequency of class preceding to median class, h = size of median class, f = frequency of median class.

∴ here l = 40, N = 100, C = 26, h = 10 and f = 44.

⇒ Median = 45.45

First we’ll find the median class -

And median class is the class corresponding to cumulative frequency which is just greater than m.

Now,

And c.f just greater is 74 corresponding to 20 - 30 (median class)

Where, l = lower limit of median class, N = sum of all frequencies, C = cumulative frequency of class preceding to median class, h = size of median class, f = frequency of median class.

∴ here l = 40, N = 100, C = 32, h = 10 and f = 42.

⇒ Median = 24.28

First we’ll find the median class -

And median class is the class corresponding to cumulative frequency which is just greater than m.

Now,

And c.f just greater is65 corresponding to 40 - 50 (median class)

Where, l = lower limit of median class, N = sum of all frequencies, C = cumulative frequency of class preceding to median class, h = size of median class, f = frequency of median class.

∴ here l = 40, N = 100, C = 35, h = 10 and f = 30.

⇒ Median = 45

First we’ll find the median class -

And median class is the class corresponding to cumulative frequency which is just greater than m.

Now,

And c.f just greater is 122 corresponding to 16 - 24 (median class)

Where, l = lower limit of median class, N = sum of all frequencies, C = cumulative frequency of class preceding to median class, h = size of median class, f = frequency of median class.

∴ here l = 16, N = 157, C = 72, h = 8 and f = 50.

⇒ Median = 17.04

(i) We know that, the mode is one with highest frequency, i.e., the one repeats the most.

Here, 5 is repeated four times.

⇒ Mode = 5

(ii) We know that, the mode is one with highest frequency, i.e., the one repeats the most.

Here, 6 is repeated three times.

⇒ Mode = 6

(iii) We know that the mode is one with highest frequency, i.e., the one repeats the most.

Here, 2.5 is repeated four times.

⇒ Mode = 2.5

(i) We know that the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 6 is the highest frequency and is for 5.

⇒ Mode = 5

(ii) We know that the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 80 is the highest frequency and is for 1.3.

⇒ Mode = 1.3

We know that the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 10 is the highest frequency and is for 6.

⇒ Mode = 6

We know that, the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 14 is repeated seven times.

⇒ Mode = 14

We know that the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 26 is the highest frequency and is for 40.

⇒ Mode = 40

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 5 here).

And, modal class is the class with highest frequency(here = 20 - 25)

⇒ Mode = 23.21

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 10 here).

And, modal class is the class with highest frequency(here = 20 - 30)

⇒ Mode = 23.33

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 10 here).

And, modal class is the class with highest frequency(here = 40 - 50)

⇒ Mode = 43.88

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 3 here).

And, modal class is the class with highest frequency(here = 58 - 61)

⇒ Mode = 58.75

By definition, we know that mode is the value with highest frequency.

(A)doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

Firstly, we’ll arrange the numbers in ascending order -

20, 35, 50, 80, 190, 340, 520, 800, 1210

Now, since there are 9 no.s and n = 9 is odd

⇒ Median = 190

(A)doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

We know, Average = Mean

Also,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 60

∴ mean marks = 60

(A)doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 85

∴ mean marks = 85

(A)doesn’t match the solution.

(B) doesn’t match the solution.

(D) doesn’t match the solution.

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ x + 21 = 36

⇒ x = 15

(A)doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

Firstly, we’ll arrange the numbers in ascending order -

1, 2, 3, 4, 5, 7

Now, since there are 6 no.s and n = 6 is even

⇒ Median = 3.5

(A)doesn’t match the solution.

(B) doesn’t match the solution.

(C) doesn’t match the solution.

Firstly, we’ll arrange the numbers in ascending order -

1, 2, 3, 5, 9

Now, since there are 5 no.s and n = 5 is odd

⇒ Median = 3

(B)doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

We know that, the mode is one with the highest frequency, i.e., the one repeats the most.

Here, 4 is repeated three times.

⇒ Mode = 4

(A)doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

We know that, the mode is one with highest frequency, i.e., the one repeats the most.

Here, 41 is the highest frequency and is for 12.

⇒ Mode = 12

(A)doesn’t match the solution.

(C) doesn’t match the solution.

(D) doesn’t match the solution.

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 7.025

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 21.25

We know that,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 25

Where = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 2

We know,

Also,

Where, = Mean, x_i = values of x and f_i = frequency for that x.

⇒ = 50.67

Firstly, we’ll find the cumulative frequencies (c.f) –

Now, the sum of all frequencies = 245

And,

⇒ m = 123

Now we’ll see to the c.f just greater than this m

Which is 150 and then we’ll see the corresponding marks obtained and that will be the median.

⇒ Median = 0.4

Firstly, we’ll find the cumulative frequencies(c.f) –

Now, the sum of all frequencies = 369

And,

⇒ m = 185

Now we’ll see to the c.f just greater than this n

Which is 212 and then we’ll see the corresponding marks obtained and that will be the median.

⇒ Median = 8.0

For Mean -

We know that,

Where, = Mean, x_i = observations and n = no. of observations.

⇒ = 49

For Median -

Firstly, we’ll arrange the numbers in ascending order -

0, 17, 26, 26, 26, 41, 57, 57, 91, 115

Now, since there are 11 no.s and n = 11 is odd

⇒ Median = 41

For Mode -

We know that, the mode is one with highest frequency, i.e., the one repeats the most.

Here, 26 is repeated three times.

⇒ Mode = 26

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 10 here).

And, modal class is the class with highest frequency(here = 20 - 30)

⇒ Mode = 26

For ungrouped frequency distribution -

Where, l = lower limit of modal class, f₁ = frequency of modal class, f₀ = frequency of class previous to modal class, f₂ = frequency of class next to modal class, and h = width of modal class(h = 20 here).

And, modal class is the class with highest frequency(here = 40 - 60)

⇒ Mode = 47.2

The mean or average of individual observations is the sum of all the values of all the observations divided by the total no. of observations. i.e.,

Merits:

(i)It is easy to be calculated and can be easily understood.

(ii)It is possible to verify it.

Merits:

(i)It is a suitable method to study qualitative characteristics.

(ii)It is not affected by extreme values.

Formula:

Where, l = lower limit of median class, N = sum of all frequencies, C = cumulative frequency of class preceding to median class, h = size of median class, f = frequency of median class.