Statistics

We have, 7, 8, 4, 13, 9, 5, 16, 18

Mean of the given data is

The respective absolute values of the deviations from the mean , i.e. are

3, 2, 6, 3, 1, 5, 6, 8

Thus, the required mean deviation about the mean is

We have, 39, 72, 48, 41, 43, 55, 60, 45, 54, 43

Mean of the given data is

The respective absolute values of the deviations from mean , i.e are

11, 22, 2, 9, 7, 5, 10, 5, 4, 7

Thus, the required mean deviation about the mean is

We have, 17, 20, 12, 13, 15, 16, 12, 18, 15, 19, 12, 11

Mean of the given data is

The respective absolute values of the deviations from the mean , i.e. are

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

Thus, the required mean deviation about the mean is

Here the number of observations is 9 which is odd.

Arranging the data into ascending order, we have 5, 6, 8, 10, 11, 12, 13, 14, 17

Now,

The respective absolute values of the deviations from median , i.e. are

6, 5, 3, 1, 0, 1, 2, 3, 6

Thus, the required mean deviation about the median is

Here the number of observations is 11 which is odd.

Arranging the data into ascending order, we have 4, 6, 7, 8, 9, 11, 13, 15, 19, 21, 25

Now,

The respective absolute values of the deviations from median , i.e. are

7, 5, 4, 3, 2, 0, 2, 4, 8, 10, 14

Thus, the required mean deviation about the median is

Here the number of observations is 10 which is odd.

Arranging the data into ascending order, we have 23, 28, 32, 34, 35, 37, 40, 44, 46, 50

Now,

The respective absolute values of the deviations from median , i.e. are

13, 8, 4, 2, 1, 1, 4, 8, 10, 14

Thus, the required mean deviation about the median is

Here the number of observations is 12 which is odd.

Arranging the data into ascending order, we have 34, 42, 45, 48, 50, 54, 56, 63, 65, 67, 70, 78

Now,

The respective absolute values of the deviations from median , i.e are

21, 13, 10, 7, 5, 1, 1, 8, 10, 12, 15, 23

Thus, the required mean deviation about the median is

We have,

Therefore,

Now,

Thus, the required mean deviation about the mean is

We have,

Therefore,

Now,

Thus, the required mean deviation about the mean is

We have,

Therefore,

Now,

Thus, the required mean deviation about the mean is

The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get,

Now, N=29 which is odd.

Since, 15^th observation lie in the cumulative frequency 21, for which the corresponding observation is 30.

Now, absolute values of the deviations from the median,

We have, and

The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get,

Now, N=50 which is even.

Median is the mean of the 25^th observation and 26^th observation. Both of these observations lie in the cumulative frequency 30, for which the corresponding observation is 13.

Now, absolute values of the deviations from the median,

We have, and

The given observations are in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get,

Now, N=50 which is even.

Median is the mean of the 25^th observation and 26^th observation. Both of these observations lie in the cumulative frequency 29, for which the corresponding observation is 30.

Now, absolute values of the deviations from the median,

We have, and

we make the following table from the given data:

Therefore,

Thus, the required mean deviation about the mean is

we make the following table from the given data:

Therefore,

Thus, the required mean deviation about the mean is

we make the following table from the given data:

Therefore,

Thus, the required mean deviation about the mean is

we make the following table from the given data:

The class interval containing or 25th item is 20-30. Therefore, 20–30 is the median class. We know that

Here, l = 20, C = 13, f = 15, h = 10 and N = 50

Therefore,

Now,

We have, and

we make the following table from the given data:

The class interval containing or 25th item is 20-30. Therefore, 20–30 is the median class. We know that

Here, l = 20, C = 14, f = 11, h = 10 and N = 50

Therefore,

Now,

We have, and

Given data: 4, 6, 10, 12, 7, 8, 13, 12

To find: MEAN

We know that,

To find: VARIANCE

= 9.25

To find: STANDARD DEVIATION

= 3.04

Odd natural numbers = 1, 3, 5, 7, 9, …

First Six Odd Natural Numbers = 1, 3, 5, 7, 9, 11

To find: MEAN

We know that,

To find: VARIANCE

= 11.67

To find: STANDARD DEVIATION

= 3.41

To find: MEAN

Now,

=14

To find: VARIANCE

= 45.8

To find: STANDARD DEVIATION

= 6.77

To find: MEAN

Now,

=19

To find: VARIANCE

= 43.4

To find: STANDARD DEVIATION

= 6.58

To find: MEAN

Now,

=19.5

To find: VARIANCE

= 19.25

To find: STANDARD DEVIATION

= 4.39

To find: MEAN

Now,

=100

To find: VARIANCE

= 29.09

To find: STANDARD DEVIATION

= 5.39

Here, we apply the step deviation method with A = 25 and h = 10

To find: MEAN

Now,

To find: VARIANCE

= 132

To find: STANDARD DEVIATION

= 11.49

Here, we apply the step deviation method with A = 65 and h = 10

To find: MEAN

Now,

To find: VARIANCE

= 201

To find: STANDARD DEVIATION

= 14.17

Here, we apply the step deviation method with A = 50 and h = 10

To find: MEAN

Now,

To find: VARIANCE

= 130.67

To find: STANDARD DEVIATION

= 11.43

Given: Standard Deviation, σ = 3.5

and Numbers are 2, 3, 2x, 11

We know that,

⇒ 12.25 × 16 = 280 – 64x + 12x²

⇒ 196 = 280 – 64x + 12x²

⇒ 12x² – 64x + 280 – 196 = 0

⇒ 12x² – 64x + 84 = 0

⇒ 3x² – 16x + 21 = 0

⇒ 3x² – 9x – 7x + 21 = 0

⇒ 3x(x – 3) – 7(x – 3) = 0

⇒ (3x – 7)(x – 3) = 0

Putting both the factors equal to 0, we get

3x – 7 = 0 and x – 3 = 0

⇒ 3x = 7 and x = 3

Let the observations are x₁, x₂, x₃, x₄, …, x₁₅

and Let mean =

Given: Variance = 6 and n = 15

We know that,

Putting the given values, we get

or …(i)

It is given that each observation is increased by 8, we get new observations

Let the new observation be y₁, y₂, y₃, …, y₁₅

where y_i = x_i + 8 …(ii)

or x_i = y_i – 8 …(iii)

Now, we find the variance of new observations

Now, we calculate the value of

We know that,

[from eq. (ii)]

…(iv)

Putting the value of eq. (iii) and (iv) in eq. (i), we get

So,

= 6

Let the observations are x₁, x₂, x₃, x₄, …, x₂₀

and Let mean =

Given: Variance = 5 and n = 20

We know that,

Putting the given values, we get

or …(i)

It is given that each observation is multiplied by 2, we get new observations

Let the new observation be y₁, y₂, y₃, …, y₂₀

where y_i = 2(x_i) …(ii)

or …(iii)

Now, we find the variance of new observations

Now, we calculate the value of

We know that,

[from eq. (ii)]

…(iv)

Putting the value of eq. (iii) and (iv) in eq. (i), we get

So,

= 20

Given: Mean of 5 observations = 6

and Variance of 5 observations = 4

Let the other two observations be x and y

∴, our observations are 5, 7, 9, x and y

Now, we know that,

⇒ 6 × 5 = 21 + x + y

⇒ 30 – 21 = x + y

⇒ 9 = x + y

or x + y = 9 …(i)

Also,

Variance = 4

So,

⇒ 20 = 11 + (x² + 36 – 12x) + (y² + 36 – 12y)

⇒ 20 – 11 = x² + 36 – 12x + y² + 36 – 12y

⇒ 9 = x² + y² + 72 – 12(x + y)

⇒ x² + y² + 72 – 12(9) – 9 = 0 [from (i)]

⇒ x² + y² + 63 – 108 = 0

⇒ x² + y² – 45 = 0

⇒ x² + y² = 45 …(ii)

From eq. (i)

x + y = 9

Squaring both the sides, we get

(x + y)² = (9)²

⇒ x² + y² + 2xy = 81

⇒ 45 + 2xy = 81 [from (ii)]

⇒ 2xy = 81 – 45

⇒ 2xy = 36

⇒ xy = 18

…(iii)

Putting the value of x in eq. (i), we get

x + y = 9

⇒ y² + 18 = 9y

⇒ y² – 9y +18 = 0

⇒ y² – 6y – 3y + 18 = 0

⇒ y(y – 6) – 3(y – 6)= 0

⇒ (y – 3)(y – 6) = 0

⇒ y – 3 = 0 and y – 6 = 0

⇒ y = 3 and y = 6

For y = 3

Hence, x = 6, y = 3 are the remaining two observations

For y = 6

Hence, x = 3, y = 6 are the remaining two observations

Thus, remaining two observations are 3 and 6.

Given: Mean of 5 observations = 4.4

and Variance of 5 observations = 8.24

Let the other two observations be x and y

∴, our observations are 1, 2, 6, x and y

Now, we know that,

⇒ 5 × 4.4 = 9 + x + y

⇒ 22 – 9= x + y

⇒ 13 = x + y

or x + y = 13 …(i)

Also,

Variance = 8.24

So,

⇒ 41.2 = 19.88 + (x² + 19.36 – 8.8x) + (y² + 19.36 – 8.8y)

⇒ 41.2 – 19.88 = x² + 19.36 – 8.8x + y² + 19.36 – 8.8y

⇒ 21.32 = x² + y² + 38.72 – 8.8(x + y)

⇒ x² + y² + 38.72 – 8.8(13) – 21.32 = 0 [from (i)]

⇒ x² + y² + 17.4 – 114.4 = 0

⇒ x² + y² – 97 = 0

⇒ x² + y² = 97 …(ii)

From eq. (i)

x + y = 17.4

Squaring both the sides, we get

(x + y)² = (13)²

⇒ x² + y² + 2xy = 169

⇒ 97 + 2xy = 169 [from (ii)]

⇒ 2xy = 169 – 97

⇒ 2xy = 72

⇒ xy = 36

…(iii)

Putting the value of x in eq. (i), we get

x + y = 13

⇒ y² + 36 = 13y

⇒ y² – 13y + 36 = 0

⇒ y² – 4y – 9y + 36 = 0

⇒ y(y – 4) – 9(y – 4)= 0

⇒ (y – 4)(y – 9) = 0

⇒ y – 4 = 0 and y – 9 = 0

⇒ y = 4 and y = 9

For y = 4

Hence, x = 9, y = 4 are the remaining two observations

For y = 9

Hence, x = 4, y = 9 are the remaining two observations

Thus, remaining two observations are 4 and 9.

Given that number of observations (n) = 18

Incorrect Mean (x̅ ) = 7

and Incorrect Standard deviation, (σ) = 4

We know that,

…(i)

∴ Incorrect sum of observations = 126

Finding correct sum of observations, 12 was misread as 21

So, Correct sum of observations = Incorrect Sum – 21 + 12

= 126 – 21 + 12

= 117

Hence,

= 6.5

Now, Incorrect Standard Deviation (σ)

Squaring both the sides, we get

Since, 12 was misread as 21

So,

= 1170 – 441 + 144

= 873

Now,

Correct Standard Deviation

= 2.5

Hence, Correct Mean = 6.5

and Correct Standard Deviation = 2.5

Given that number of observations (n) = 200

Incorrect Mean (x̅ ) = 40

and Incorrect Standard deviation, (σ) = 15

We know that,

…(i)

∴ Incorrect sum of observations = 8000

Finding correct sum of observations, 43 was misread as 34

So, Correct sum of observations = Incorrect Sum – 34 + 43

= 8000 – 34 + 43

= 8009

Hence,

= 40.045

Now, Incorrect Standard Deviation (σ)

Squaring both the sides, we get

Since, 43 was misread as 34

So,

= 365000 – 1156 + 1849

= 125000 + 693

= 365693

Now,

Correct Standard Deviation

= 14.995

Hence, Correct Mean = 40.045

and Correct Standard Deviation = 14.995

Given that number of observations (n) = 100

Incorrect Mean (x̅ ) = 20

and Incorrect Standard deviation, (σ) = 3

We know that,

…(i)

∴ Incorrect sum of observations = 2000

Finding correct sum of observations, incorrect observations 21, 12 and 18 are removed

So, Correct sum of observations = Incorrect Sum – 21 – 12 – 18

= 2000 – 51

= 1949

Hence,

= 20.09

Now, Incorrect Standard Deviation (σ)

Squaring both the sides, we get

Since, 21, 12 and 18 are removed

So,

= 40900 – 441 – 144 – 324

= 40900 – 909

= 39991

Now,

Correct Standard Deviation

= 2.94

Hence, Correct Mean = 20.09

and Correct Standard Deviation = 2.94

Mean wages of both the factories are the same, i.e., Rs. 5300.

To compare variation, we need to find out the coefficient of variation (CV).

We know, CV = , where SD is the standard deviation.

The variance of factory A is 100 and the variance of factory B is 81.

Now, SD of factory A =

And, SD of factory B =

Therefore,

The CV of factory A =

The CV of factory B =

Here, the CV of factory A is greater than the CV of factory B.

Hence, factory A has more variation.

Given: Coefficient of variation of two distributions are 60% and 80% respectively, and their standard deviations are 21 and 16 respectively.

Need to find: Arithmetic means of the distributions.

For the first distribution,

Coefficient of variation (CV) is 60%, and the standard deviation (SD) is 21.

We know that,

⇒

⇒

⇒

⇒

For the first distribution,

Coefficient of variation (CV) is 80%, and the standard deviation (SD) is 16.

We know that,

⇒

⇒

⇒

⇒

Therefore, the arithmetic mean of 1^st distribution is 35 and the arithmetic mean of 2^nd distribution is 20.

In case of heights,

Mean = 63.2 inches and SD = 11.5 inches.

So, the coefficient of variation,

⇒

In case of weights,

Mean = 63.2 inches and SD = 5.6 inches.

So, the coefficient of variation,

⇒

CV of heights > CV of weights

So, heights show more variability.

(i) Both the factories pay the same mean monthly wages.

For factory A there are 560 workers. And for factory B there are 650 workers.

So, factory A totally pays as monthly wage = (5460 x 560) Rs.

= 3057600 Rs.

Factory B totally pays as monthly wage = (5460 x 650) Rs.

= 3549000 Rs.

That means, factory B pays a larger amount as monthly wages.

(ii) Mean wages of both the factories are the same, i.e., Rs. 5460.

To compare variation, we need to find out the coefficient of variation (CV).

We know, CV = , where SD is the standard deviation.

The variance of factory A is 100 and the variance of factory B is 121.

Now, SD of factory A =

And, SD of factory B =

Therefore,

The CV of factory A =

The CV of factory B =

Here,the CV of factory B is greater than the CV of factory A.

Hence, factory B shows greater variability.

To find the more variable, we again need to compare the coefficients of variation (CV).

Here the number of products are n = 50 for length and weight both.

For length,

Mean =

Variance =

=

=

=

So, standard deviation, SD =

Therefore, the coefficient of variation of length,

CV_L =

For weight,

Mean =

Variance =

=

=

=

So, standard deviation, SD =

Therefore, the coefficient of variation of length,

CV_W =

Now, CV_W > CV_L

Therefore, the weight is more variable than height.