Statistics

We first find the mean (x̅ ) of the given data

The respective absolute values of the deviations from mean, i.e., |x_i – x̅| are

6, 3, 2, 1, 0, 2, 3, 7

Therefore

And

Hence, the mean deviation about the mean for the given data is 3.

We first find the mean (x̅) of the data

The respective absolute values of the deviations from mean, i.e., |x_i – x̅| are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

Therefore

And

Hence, the mean deviation about the mean for the given data is 8.4.

Here the number of observations is 12 which is even. Arranging the data into ascending order, we have 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

Now, Median

The absolute values of the respective deviations from the median, i.e., |x_i – M| are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

Therefore

And

Hence, the mean deviation about the median for the given data is 2.33.

Here the number of observations is 10 which is even. Arranging the data into ascending order, we have 36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

Now, Median

The absolute values of the respective deviations from the median, i.e., |x_i – M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

Therefore

And

Hence, the mean deviation about the median for the given data is 7.

We make the following table and add other columns after calculations.

So, we can calculate the absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

Thus, we have

Therefore

Hence, the mean deviation about the mean for the given data is 6.32.

We make the following table and add other columns after calculations.

So, we can calculate the absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

Thus, we have

Therefore

Hence, the mean deviation about the mean for the given data is 16.

The given observations are already in ascending order. We make a table for the given data, as shown below, adding other columns after calculations.

Now, N = 26 which is even.

Median is the mean of 13^th and 14^th observations. Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

Therefore, Median

So, we can calculate the absolute values of the deviations from the median, i.e., |x_i – M|, as shown in the table.

Thus, we have

And

Hence, the mean deviation about the median for the given data is 3.23.

The given observations are already in ascending order. We make a table for the given data, as shown below, adding other columns after calculations.

Now, N = 30 which is even.

Median is the mean of 15^th and 16^th observations. We see that both the 15^th as well as the 16^th observations lie in the cumulative frequency 21, whose corresponding observation is 30.

Therefore, the Median

So, we can calculate the absolute values of the deviations from the median, i.e., |x_i – M|, as shown in the table.

Thus, we have

And

Hence, the mean deviation about the median for the given data is 5.1.

We make the following table and add other columns after calculations.

So, we can calculate the absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

Therefore, we have

And

Hence, the mean deviation about the mean for the given data is 157.92.

We make the following table and add other columns after calculations.

So, we can calculate the absolute values of the deviations from the mean, i.e., |x_i – x̅|, as shown in the table.

Therefore, we have

And

Hence, the mean deviation about the mean for the given data is 11.28.

We make the following table and add other columns after calculations.

The class interval containing or 25^th item is 20-30. Therefore, 20-30 is the median class.

Now, we know that

Median

Here, l = 20, C = 14, f = 14, h = 10 and N = 50

Therefore, Median

So, we can calculate the absolute values of the deviations from the median, i.e., |x_i – Med.|, as shown in the table.

Therefore, we have

And

Hence, the mean deviation about the median for the given data is 10.34.

We first convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval and then form the following table adding the other columns after calculations.

The class interval containing or 50^th item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

Now, we know that

Median

Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

Therefore, Median

So, we can calculate the absolute values of the deviations from the median, i.e., |x_i – Med.|, as shown in the table.

Therefore, we have

And

Hence, the mean deviation about the median for the given data is 7.35.

We know that Mean,

Where n = number of observations and = Sum of total observations

∴

From the given data, we can form the table:

We know that Variance, σ² =

∴ σ² = (1/8) × 74 = 9.25

Ans. Mean = 9 and Variance = 9.25

We know that Mean =

∴ Mean,

We know that Variance, σ² = =

We know that (a-b)² = a² – 2ab + b²

We know that (a + b) (a – b) = a² – b²

Ans. Mean = and Variance =

The 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

We know that Mean,

Where n = number of observations and = Sum of total observations

From the given data, we can form the table:

We know that Variance, σ² =

∴ σ² = (1/10) × 742.5 = 74.25

Ans. Mean = 16.5 and Variance = 74.25

Presenting the data in the tabular form, we get

We know that Mean,

Where N =

∴

We know that Variance, σ² =

∴ σ² = (1/40) × 1736 = 43.4

Ans. Mean = 19 and Variance = 43.4

Presenting the data in the tabular form, we get

We know that Mean,

Where N =

∴

We know that Variance, σ² =

∴ σ² = (1/22) × 640 =

Ans. Mean = 100 and Variance =

Let the assumed mean, A = 64 and h = 1

We obtain the following table from the given data:

We know that Mean,

∴

We know that Variance σ² =

∴ σ² =

We know that Standard Deviation = σ

∴ σ = √2.86 = 1.691

Ans. Mean = 64 and Standard Deviation = 1.691

Presenting the data in the tabular form, we get

We know that Mean,

Where N =

∴

We know that Variance, σ² =

∴ σ² = (1/30) × 68280 =2276

Ans. Mean = 107 and Variance = 2276

Presenting the data in the tabular form, we get

We know that Mean,

Where N =

∴

We know that Variance, σ² =

∴ σ² = (1/50) × 6600 = 132

Ans. Mean = 27 and Variance = 132

Let the assumed mean, A = 92.5 and h = 5

We obtain the following table from the given data:

We know that Mean,

∴

We know that Variance σ² =

∴ σ² =

We know that Standard Deviation = σ

∴ σ = √105.583 = 10.275

Ans. Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Let the assumed mean, A = 42.5 and h = 4

We obtain the following table from the given data:

We know that Mean,

∴

We know that Variance σ² =

∴ σ² =

We know that Standard Deviation = σ

∴ σ = √30.84 = 5.553

Ans. Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

The group having higher coefficient of variation will be more variable

So, we will calculate

Where σ is standard deviation

is mean

For group A

Mean

Where A is assumed mean = 45

h = class size= 20- 10 = 10

Mean,

Variance

=

=

Standard deviation =

C.V._A=

For group B

Variance

=

=

Standard deviation =

Since the coefficient of variance of group B coefficient of Group A

∴ Group B is more variable

Here

For X

Mean where n is number of terms

=

=

=

Standard deviation = √variance= √35=5.91

Coefficient of variation

For Y

where n is number of terms

=

=

=

=

Standard variance = √variance= √4=2

Coefficient of variation

Covariance of x covariance of Y

⇒ X is more variable and Y is more stable than X

Here

Mean monthly wages of firm A = 5253

No. of wage earners = 586

Total amount paid = 586 × 5253 = 3078258

Mean monthly wages of firm B = 5253

No. of wage earners = 648

Total amount paid = 648 × 5253 = 340 3944

(i) Hence the firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

⇒ standard deviation (σ)= √100=10

Variance of firm B = 121

⇒ Standard deviation (σ)=√(121 )=11

Since the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

Mean =

=

Standard deviation σ = √variance = √1.2 = 1.09

For team B

Given

Standard deviation σ=1.25

Since C.V. of firm B is greater

∴ Team A is more consistent.

For length x:

Mean,

=

Standard deviation σ = √variance= √0.0784=0.28

For Weight Y

Mean,

Variance

Standard deviation (

Since C.V._y C.V. _x

⇒ weight is more varying.

Let us assume the remaining two observations to be x and y respectively such that,

Observations: 6, 7, 10, 12, 12, 13, x, y.

∴ Mean,

60 + x + y = 72

x + y = 12 (i)

By using (i)

So, from equation (i) we have:

Thus, from (ii) and (iii), we have

2xy = 64 (iv)

Now by subtracting (iv) from (ii), we get:

x² + y² – 2xy = 80 – 64

x – y = 4 (v)

Hence, from equation (i) and (v) we have:

When x – y = 4 then x = 8 and y = 4

And, when x – y = - 4 then x = 4 and y = 8

∴ The remaining observations are 4 and 8

Let us assume the remaining two observations be x and y

The given observations in the question are 2, 4, 10, 12, 14, x, y

x + y = 14 (i)

It is also given in the question that,

Variance = 16

We know that,

x² + y² = 112 - 12

x² + y² = 100 (ii)

Thus, by using (i) we have:

x² + y² + 2xy = 196 (iii)

Now, from equation (ii) and (iii) we have:

2xy = 196 – 100

2xy = 96 (iv)

Now subtracting equation (iv) from (ii), we get:

x² + y² – 2xy = 100 – 96

(x – y)² = 4

x – y = 2 (v)

Hence, from equation (i) and (v) we have:

When x – y = 2 then x = 8 and y = 6

And, when x – y = - 2 then x = 6 and y = 8

∴ The remaining observations are 6 and 8

Let us assume the observations be x₁, x₂, x₃, x₄, x₅ and x₆

It is given in the question that,

Mean = 8 and Standard deviation = 4

Now, according to question if each observation is multiplied by 3 and the resulting observations are y_i then, we have:

y_i = 3x_i

= 3 × 8

= 24

We know that,

Hence, from (i) and (ii) we have:

Now, by substituting the values of x_i and in (ii) we have:

Therefore, standard deviation of new observation = = 12

The given n observations in the question are x₁, x₂,…..x_n

Also, mean =

And, variance = σ²

We know that,

According to the condition given in the question, if each of the observation is being multiplied by a and the new observation are y_i the, we have:

y_i = ax_i

Hence, mean of the observations ax₁, ax₂,…..ax_n is a

Now, by substituting the values of x_i and in (i), we get:

Hence, the variance of the given observations ax₁, ax₂,….ax_n is a²σ²

It is given in the question that,

Total number of observations (n) = 20

Also, incorrect mean = 20

And, incorrect standard deviation = 2

Thus, incorrect sum of observations = 200

Hence, correct sum of observations = 200 – 8

= 192

= 10.1

Thus,

= 2080 – 64

= 2016

= 2.02

It is given in the question that,

Total number of incorrect sum of observations = 200

Also, correct sum of observations = 200 – 8 + 12

= 204

= 10.2

Now, we know that:

Thus,

= 2080 – 64 + 144

= 2160

= 1.98

It is given in the question that,

Standard deviation of mathematics = 12

Also, standard deviation of physics = 15

And, standard deviation of chemistry = 20

We know that,

Hence, subject with highest variability in marks is chemistry as subject with the greater C.V is more variable than others

And, Subject with lowest variability in marks is mathematics

It is given in the question that,

Total number of observations (n) = 100

Incorrect mean, () = 20

And, Incorrect standard deviation () = 3

Thus, incorrect sum of observations is 2000

Now, correct sum of observations = 2000 – 21 – 21 – 18

= 2000 – 60

= 1940

= 20

= 40900 – 441 – 441 – 324

= 40900 – 1206

= 39694

= 3.036