Measures Of Central Tendency

It is given that,

The height of 5 person are = 140 cm, 150 cm, 152 cm, 158 cm and 161 cm

Therefore, Mean height =

=

=

= 152.2

Given numbers are: 994, 996, 998, 1002 and 1000

Therefore, Mean =

=

=

= 998

Given that the first five natural numbers are: 1, 2, 3, 4, 5

Therefore, Mean =

=

=

= 3

All factors of 10 are: 1, 2, 5, 10

Therefore, Mean =

=

=

= 4.5

Given that the first 10 even natural numbers be 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

Mean =

=

=

= 11

Numbers be: x, x+2, x+4, x+6, x+8

Therefore, Mean =

=

=

= x + 4

First five multiples of 3 are:

3, 6, 9, 12, 15

Therefore, Mean =

=

=

= 9

The weight (in kg) of 10 new born babies = 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6

Therefore, Mean =

=

=

= 4 kg

The percentage marks obtained by students are:

64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1

Therefore, Mean =

=

=

= 39.5

The number of children in 10 families:

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

Therefore, Mean =

=

=

= 3

Let M is the mean of x₁, x₂, x₃, x₄, x₅ and x₆

Then, M =

6M = x₁ + x₂ + x₃ + x₄ + x₅ + x₆

To prove: (x₁-M) + (x₂-M) + (x₃-M) + (x₄-M) + (x₅-M) + (x₆-M) = 0

Proof: L.H.S

= (x₁ – M) + (x₂ – M) + (x₃ + M) + (x₄ – M) + (x₅ – M) + (x₆ – M)

= (x₁ + x₂ + x₃ + x₄ + x₅ + x₆) – (M+ M + M + M + M + M)

= 6M – 6M

= 0

= R.H.S

Duration of sunshine (in hours) for 10 days are:

9.6, 5.2, 3.5, 1.5, 1.6, 2.4, 2.6, 8.4, 10.3, 10.9

(i) Mean,

=

=

=

= 5.6

(ii) L.H.S =

= (x₁ - ) + (x₂ - ) + (x₃ -) + (x₄ - ) + ………. + (x₁₀ - )

= (9.6 – 5.6) + (5.2 – 5.6) + (3.5 – 5.6) + (1.5 – 5.6) + (1.6 – 5.6) + (2.4 – 5.6)

= (4) + (-0.4) + (-2.1) + (-4.1) + (-4) + (-3.2) + (-3) + (2.8) + (4.7) + (5.3)

= 16.8 – 16.8

= 0

Let us say numbers are be

3, 4, 5

Therefore, Mean =

=

=

= 4

(i) = 2 on each term

New numbers are = 5, 6, 7

New mean =

=

= 6

Therefore new mean will be 2 more than the original mean.

(ii) Subtracting constant terms ‘k’ = 2 in each term

New numbers are = 1, 2, 3

Therefore, new mean =

=

= 2

Therefore, new mean will be 2 less than the original mean.

(iii) = 2 in each term

New numbers are = 6, 7, 8

Therefore, new mean =

=

= 8

Therefore, new mean will be 2 times of the original mean.

(iv) Divide by constant term ‘k’ = 2 in each term

New numbers are = 1.5, 2, 2.5

Therefore, new mean =

=

= 2

Therefore, new mean will be half of the original mean.

Mean marks of 100 students = 40

Sum of marks of 100 students = 100 * 40

= 4000

Correct value = 53

Incorrect value = 83

Correct sum = 4000 – 83 + 53

= 3970

Therefore, correct mean =

= 39.7

The speed of 10 Motorist are:

47, 53, 49, 60, 39, 42, 55, 57, 52, 48

Later on it was discovered that the instrument recorded 5 km/h less than in each case

Therefore, correct values are = 52, 58, 54, 65, 44, 47, 60, 62, 57, 53

Therefore, correct mean =

=

= 55.2 km/h

The mean of the numbers is 27

The sum of 5 numbers = 5 * 27

= 135

If one number is excluded, then the new mean is 25

Therefore sum of new numbers = 4 * 25

= 100

Therefore, excluded number = 135 – 100

= 35

The mean weight per student in a group of 7 students = 55 kg

Weight of 6 students (in kg) = 52, 54, 55, 53, 56 and 54

Let weight of 7^th student = x kg

Therefore, Mean =

55 =

385 = 324 + x

x = 385 – 324

x = 61 kg

Therefore, weight of 7^th student = 61 kg

We have,

The mean weight of 8 numbers is 15

Then, the sum of 8 numbers = 8 * 15

= 120

If each number is multiplied by 2

Then, New mean = 120 * 2

= 240

Therefore, new mean =

= 30

The mean of 5 numbers is 18

Then, the sum of 5 numbers = 5 * 18

= 90

If the one number is excluded

Then, the mean of 4 numbers = 16

Therefore, sum of 4 numbers = 4 * 16

= 64

Excluded number = 90 – 64

= 26

The mean of 200 items = 50

Then the sum of 200 items = 200 * 50

= 10,000

Correct values = 192 and 88

Incorrect values = 92 and 8

Therefore, correct sum = 10,000 – 92 – 8 + 192 + 88

= 10,180

Therefore, correct mean =

= 50.9

(i) Given, = -10

= (x₁ – 12) + (x₂ – 12) + …. + (x_n + 12) = -10

= (x₁ + x₂ + x₃ + x₄ + ….. + x_n) + (12 + 12 + 12 + …. + 12) = -10

= ∑x – 12n = -10 (I)

And = 62

= (x₁ – 3) + (x₂ – 3) + (x₃ – 3) + …. + (x_n – 3) = 62

= (x₁ + x₂ + x₃ + …. + x_n) – (3 + 3 + 3 + …. + 3) = 62

= ∑x – 3n = 62 (II)

BY subtracting (I) from (II), we get

∑x – 3n - ∑x – 12n = 62 + 10

= 9n = 72

= n =

= 8

Put value of n in equation (I), we get

∑x – 12 * 8 = -10

= ∑x – 96 = -10

= ∑x = -10 + 96 = 86

Therefore, =

=

= 10.75

(ii) Given, = 30

= (x₁ – 10) + (x₂ – 10) + …. + (x_n – 10) = 30

= (x₁ + x₂ + x₃ + …. + x_n) – (10 + 10 + 10 + …. + 10) = 30

= ∑x – 10n = 30 (I)

And = 150

= (x₁ – 6) + (x₂ – 6)+ …. + (x_n – 6) = 150

= (x₁ + x₂ + x₃ + …. + x_n) – (6 + 6 + 6 + …. + 6) = 150

= ∑x – 6n = 150

By subtracting (I) from (II), we get

∑x – 6n - ∑x – 10n = 150 – 30

= ∑x - ∑x + 4n = 120

n =

= 30

Put value of n in (I), we get

∑x – 10 * 30 = 30

∑x – 300 = 30

∑x = 30 + 300

= 330

Therefore, =

=

= 11

Given, = -90

= (x₁ – 15) + (x₂ – 15) + …. + (x_n – 15) = - 90

= (x₁ + x₂ + … + x_n) – (15 + 15 + … + 15) = -90

= ∑x – 15n = - 90 (I)

And, = 54

= (x₁ + 3) + (x₂ + 3) + …. + (x_n + 3) = 54

= (x₁ + x₂ + …. + x_n) + (3 + 3 + …. + 3) = 54

= ∑x + 3n = 54 (II)

Subtracting (I) from (II), we get

∑x + 3n - ∑x + 15n = 54 + 90

18n = 144

n =

= 8

Put value of n in (I), we get

∑x – 15 * 8 = - 90

∑x – 120 = - 90

∑x = 30

Therefore, Mean =

=

=

Values are 3, 4, 6, 7, 8, 14

Therefore, Mean =

=

= 42/6

= 7

Therefore, sum of deviation of values from their mean

= (3 – 7) + (4 – 7) + (6 – 7) + (7 – 7) + (8 – 7) + (14 – 7)

= (-4) + (-3) + (-1) + (0) + (1) + (7)

= - 8 + 8

= 0

We have, =

x₁ + x₂ + x₃ + …. + x₁₀ = 10 (i)

Now, (x₁ -) + (x₂ -) + …. + (x₁₀ - )

= (x₁ + x₂ + …. + x₁₀) – ( + + …. + up to 1o terms)

= 10 - 10 [From (i)]

= 0

Therefore, (x₁ - ) + (x₂ - ) + …. + (x₁₀ - ) = 0

Hence, proved

Therefore, Mean () =

=

= 7.025

Therefore, Mean () =

=

= 25

It is given that,

Mean = 20.6

= 20.6

= 20.6

= 1030

25P = 500

P = = 20

Therefore, P = 20

Given, mean = 15

= 15

= 15

15p – 10p = 445 – 405

5p = 40

p =

= 8

Given, mean = 16.6

= 16.6

= 16.6

24p = 432

p = = 18

Given, mean = 12.58

= 12.58

= 12.58

7p = 629 – 524

7p = 105

p =

= 15

Given, mean = 7.68

= 7.68

= 7.68

9P – 7.68P = 314.88 – 303

1.32P = 11.88

P =

P = 9

Mean, =

20 =

60 + 20p = 59 + 20p + p²

p² – 1 = 0

p = 1

Therefore, mean () =

=

= 20

Therefore, = 20

Let no. of students appeared from school III = x

Given, Average score of all schools = 66

= 66

= 66

= 66

= 66

10340 – 9768 = 66x – 55x

11x = 572

x =

= 52

Therefore, no. of candidates appeared from school (III) are 52

Therefore, mean number of heads per toss =

=

= 2.47

It is given that, mean = 50

= 50

= 50

3480 + 30f₁ + 70f₂ = 50 (120)

30f₁ + 70f₂ = 6000 – 3480

10 (3f₁ + 7f₂) = 10 (252)

3f₁ + 7f₂ = 252 (i) [Dividing by 10]

And N = 120

17 + f₁ + 32 + f₂ + 19 = 120

68 + f₁ + f₂ = 120

f₁ + f₂ = 52

Multiplying by ‘3’ on both sides, we get

3f₁ + 3f₂ = 156 (ii)

Subtracting (ii) from (i), we get

3f₁ + 7f₂ – 3f₁ – 3f₂ = 252 – 156

4f₂ = 96

f₂ =

= 24

Put value of f₂ in (i), we get

3f₁ + 7 * 24 = 252

3f₁ = 252 – 168

3f₁ = 84

f₁ =

= 28

Given numbers are:

83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Arrange the numbers in ascending order:

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

n = 10 (even)

Therefore, Median =

=

=

=

= = 49.5

Given numbers are:

133, 73, 89, 108, 94, 104, 94, 85, 100, 120

Arrange in ascending order:

78, 85, 89, 94, 94, 100, 104, 108, 120, 133

n = 10 (even)

Therefore, median =

=

=

= = 9.7

Given numbers are:

31, 38, 27, 28, 36, 25, 35, 40

Arranging in increasing order:

25, 27, 28, 31, 35, 36, 38, 40

n = 8 (even)

Therefore, Median =

=

=

=

= = 33

Given, numbers are 15, 6, 16, 8, 22, 21, 9, 18, 25

Arrange in increasing order:

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

n = 9 (odd)

Therefore, median = value

= value

= 5^th value

= 16

Given numbers are:

41, 43, 127, 99, 71, 92, 71, 58, 57

Arrange in increasing order:

41, 43, 57, 58, 71, 71, 92, 99, 127

n = 9 (odd)

Therefore, median = value

= value

= 5^th value = 71

Given numbers are:

25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Arrange in increasing order:

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

n = 10 (even)

Therefore, median =

=

=

=

=

Given, numbers are:

12, 17, 3, 14, 5, 8, 7, 15

Arrange in increasing order:

3, 5, 7, 8, 12, 14, 15, 17

n = 8 (even)

Therefore, median =

=

=

= = 10

Given, numbers are:

92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Arrange in increasing order:

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

n = 10 (even)

Therefore, median =

=

=

= =

= 68

Given, no. of observations, n = 8

Median =

=

= 2x + 1

Given, median = 25

2x + 1 = 25

2x = 24

x = 12

Given, numbers are:

46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33

Arrange in increasing order:

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

n = 11 (odd)

Therefore, median =

=

= = 6^th value

= 58

If 92 is replaced by 99 and 41 by 43, then the new values are:

33, 35, 43, 46, 55, 58, 64, 77, 81, 90, 99

Therefore, n = 11 (odd)

New median = value

=

= 6^th value = 58

Given, numbers are:

41, 43, 127, 99, 61, 92, 71, 58, 57

Arrange in ascending order:

41, 43, 57, 58, 61, 71, 92, 99, 127

n = 9 (odd)

Therefore, median = value

=

= 5^th value = 61

If 58 is replaced by 85

Then the new value be in order:

41, 43, 57, 61, 71, 85, 92, 99, 127

n = 9 (odd)

Therefore, median =

=

= 5^th value = 71

Given, numbers are:

31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30

Arrange in increasing order:

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

n = 15 (odd)

Therefore, median = value

=

= 8^th value = 35 kg

If the weight 44kg is replaced by 46 kg and 27 kg is replaced by 25 kg

Then new values in order be

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46

n = 15 (odd)

Therefore, new median =

=

= 8^th value = 35kg

Total number of observation in the given data is 10 (even number). So median of this data will be mean of i.e., 5^th and + 1 i.e. 6^th observations.

So, median of data =

63 =

63 =

63 = x + 1

x = 62

Since, the maximum frequency corresponds to the value 7,
then mode = 7 marks.

Since, maximum frequency 4 corresponds to 125, then mode = 125

Since, maximum frequency 4 corresponds to value 7.2, then mode = 7.2

(i) Arranging the data in ascending order:

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

Here observation 14 is having the highest frequency.

So, mode = 14

(ii)

Since, maximum frequency 5 corresponds to value 7.

Then, the mode = 7

Since, maximum frequency 39 corresponds to value 39.

Then, mode = 39

Using empirical formula

Mode = 3 median - 2 mean

Mode = 3 * (3x) - 2(2x)

Mode = 9x - 4x = 5x

Mode: Mean = 5x: 2x = 5: 2

Since range is the difference between the lowest and highest values.

Mode = 3 Median – 2 Mean

=

Mode =

= 3 Median - 2 Mean

2 Mean = 3 Median -

2 Mean =

Mean =

Therefore, = 9: 10

Mean = (x₁ + x₂ + ...+ n terms)/N

_{= x1} + k + x₂ + k +...+n terms/N

_{= Mean + k}

_{=+ k}

Given that,

x + 2 =

4x + 8 = 10x + 14

-6x = 6

x= -1

Mean =

=

= k

= k (Mean)

= k

3 Median = Mode + 2 Mean

3 Median = 12 + 2 (24)

3 Median = 12 + 48

3 Median = 60

Median = 20

Let the discarded number be x,

= 81

(468 + x) = 567

x = 567 - 468

= 99

Mode - Median = 24

Mode = Median + 24 (1)

Since, by the formula

Mode = 3 Median - 2 Mean (2)

From (1)

Median + 24 = 3 Median - 2 Mean

2 Median - 2 Mean = 24

Dividing the whole equation by 2, we get

Median - Mean = 12

1, 3, 3, 3, 5

Mean =

=

= 3

Median = 3^rd term

= 3

Mode = 3 (The highest occurring number)

We know,

Median = term

= term

= 3rd term

Therefore, = 6

x = 24

Hence, =

= 4

Mean =

=

= 5

Mode = 2

Hence, the statement ‘Mean>Mode’ is correct.

Mean =

10 =

50 - 34 = x

x = 16

Mean =

5 =

30 = 20 + x+ y

x + y = 10

Since, the value of mode is 4.

Hence, x = 4

First, you need to solve for x:

= 11

x + x + 2 + x + 4 + x + 6 + x + 8 = 55

5x + 20 = 55

5x = 35

x = 7

Now, substitute for x in each of the last three terms:

x + 4 + x + 6 + x + 8

(7 + 4) + (7 + 6) + (7 + 8)

(11) + (13) + (15)

= 13

The mean is 13

Mode is the value that occurs most frequently in a given set of data.

There are 6 observations i.e. even no. of observations

So, median will be:

Arranging in increasing order:

20, 25, 28, x, 33, 34

If we put any other number at 4^th position the n, we will not get the median = 29

So, = 29

28 + x = 58

x = 58 – 28

= 30

The mode of the data is 8 as it occurs the maximum times.

Since, the number of terms are 5

Hence, the Median will be the 6th term which is x

x = 3

Mean =

=

= 3

This is an approximate relation that holds when the distribution is symmetrical or moderately skewed. It does not hold when the distribution is too skewed. When the distribution is symmetric, this relation holds exactly because in that case, mean = median = mode

Given: The mean of a, b, c, d, and e is 28 and the mean of a, c and e are 24.
To find: the mean of b and d.
Solution:

Mean = = 28

⇒ a + b + c + d + e = 28 × 5 = 140

Also,
= 24

⇒ a + c + e = 72

⇒ a + b + c + d + e – a – c – e = 140 - 72

⇒ b + d = 68

Therefore, mean =
=
= 34

Mean = =

x̅ =

x₁ + x₂ + x₃ + …. + x_n = n (i)

Algebraic sum of the deviation from the mean is:

Sum = (x₁ - ) + (x₂ - ) + (x₃ - ) + …. + (x_n - )

= x₁ + x₂ + x₃ + …. + x_n – ( + + + …. n times)

= x₁ + x₂ + x₃ + …. + x_n - n

Using (i), we get

Sum = n - n

= 0

Hence, algebraic sum of the deviations of a set of n values from mean is 0.

A: 2, 3, 7, 1, 3, 2, 3

A: 1, 2, 2, 3, 3, 3, 7

Mode = 3 (occurs maximum number of times)

Median = 3 (the middle term)

Mean =

=

= 3

Hence, option (d) is true.