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Mean = \(\frac {Sum \,of\,items}{No \,of\,items}\)

30 = \(\frac {20+24+36+26+34+K}{6}\)

⇒ 180 = 140 + K 

⇒ K = 40

Grouped data is convenient when the values f_i and x_i are low.

a) Mode = 6 (most repeated value of the data).

b) Mode = 3,7.

c) Mode = 2, 3, 4, 5, 6.

n = 40 

Median = \(\frac{n+1}{2}= \frac{40+1}{2}=\frac{41}{2}\)

= 20.5^th term = 7.

(i) Find the mean deviation from the median

Let us arrange the data in ascending order,

15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – M|

The number of observations are Even then Median = (40+52.3)/2 = 46.15

Median = 46.15

Number of observations, ‘n’ = 10

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/10 × 167.4

= 16.74

∴ The Mean Deviation is 16.74.

(ii) Find the mean deviation from the mean also.

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – x|

So, let ‘x’ be the mean of the given observation.

x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10

= 458/10

= 45.8

Number of observations, ‘n’ = 10

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/10 × 166.4

= 16.64

∴ The Mean Deviation is 16.64

Correct option is (C) 8

Let us calculate the mean deviation for the first data set.

Since the data is arranged in ascending order,

4000, 4200, 4400, 4600, 4800

Median = 4400

Total observations = 5

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – M|

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/5 × 1200

= 240

Let us calculate the mean deviation for the second data set.

Since the data is arranged in ascending order,

3800, 4000, 4200, 4400, 4600, 4800, 5800

Median = 4400

Total observations = 7

We know that,

MD = 1/n ∑ⁿ_i=1|d_i|

Where, |d_i| = |x_i – M|

MD = 1/n ∑ⁿ_i=1|d_i|

= 1/7 × 3200

= 457.14

∴ The Mean Deviation of set 1 is 240 and set 2 is 457.14

Cumulative frequency of less than type for the class 

30 – 40 = 7 + 3 + 12 + 13 = 35

Correct option is (C) 35

∴ Mean = The sum of all observations/Total number of observations

∴ The sum of all observations = Mean x 

Total number of observations mean of nine numbers is 77 

∴ sum of 9 numbers = 11 x 9 = 693 …(i) 

If one more number is added, then mean increases by 5 mean of 10 numbers 

= 77 + 5 = 82 

∴ sum of the 10 numbers = 82 x 10 = 820 …(ii) 

∴ Number added = sum of the 10 numbers – sum of the 9 numbers 

= 820 – 693 … [From (i) and (ii)] = 127 

∴ The number added in the data is 127.

 Mean = \(\cfrac{The\,sum \,of \,all \,observations}{Total \,number \,of \,observations}\)

∴ The sum of all observations = Mean x Total number of observations mean of nine numbers is 77 

∴ sum of 9 numbers = 11 x 9 = 693 …(i)

If one more number is added, then mean increases by 5 

mean of 10 numbers = 77 + 5 = 82 

∴ sum of the 10 numbers = 82 x 10 = 820 …(ii)

∴ Number added = sum of the 10 numbers – sum of the 9 numbers = 820 – 693 … [From (i) and (ii)] = 127 

∴ The number added in the data is 127

Mean = The sum of all observations/Total number of observations

∴ The sum of all observations = Mean x 

Total number of observations 

The mean salary of 20 workers is Rs 10,250. 

∴ Sum of the salaries of 20 workers = 20 x 10,250 

= Rs 2,05,000 …(i) 

If the superintendent’s salary is added, then mean increases by 750 new mean = 10, 250 + 750 = 11,000 

Total number of people after adding superintendent = 20 + 1 = 21 

∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = Rs 2,31,000 …(ii) 

∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers 

= 2, 31,00 – 2,05,000 …[From (i) and (ii)] 

= 26,000 

∴ The salary of the office superintendent is Rs 26,000.

Mean = \(\cfrac{The\,sum \,of \,all \,observations}{Total \,number \,of \,observations}\)

∴ The sum of all observations = Mean x Total number of observations 

The mean salary of 20 workers is ₹ 10,250.

 ∴ Sum of the salaries of 20 workers 

= 20 x 10,250 

= ₹ 2,05,000 …(i) 

If the superintendent’s salary is added, then mean increases by 750

new mean = 10, 250 + 750 = 11,000 

Total number of people after adding superintendent 

= 20 + 1 = 21

∴ Sum of the salaries including the superintendent’s salary = 21 x 11,000 = ₹ 2,31,000 …(ii) 

∴ Superintendent salary = sum of the salaries including superintendent’s salary – sum of salaries of 20 workers 

= 2, 31,00 – 2,05,000 …[From (i) and (ii)]

= 26,000

∴ The salary of the office superintendent is ₹ 26,000.

Given mean is 20

Sum 20 × 10 = 200.

We have to find 10 different numbers whose sum is 200 (for this find 5 pairs of sum 40)

(15, 25) (16, 24) (17, 23) (18, 22) (19, 21)

The numbers are 15, 16, 17, 18, 19, 21, 22, 23, 24, 25

Given, Numbers of observations are given in two groups. 

To Find: Calculate the Mean Deviation from their Median. 

Formula Used: Mean Deviation = \(\frac{\Sigma d_i}{n}\)

For Group 1: Since, Median is the middle number of all the observation, 

So, To Find the Median, Arrange the Income of Group 1 in Ascending order, we get 4000, 4200, 4400, 4600, 4800 

Therefore, The Median = 4400

Deviation |d| = |x-Median| 

Now, The Mean Deviation is

Mean Deviation = \(\frac{\Sigma d_i}{n}\)

Mean Deviation Of Group 1 = \(\frac{1200}{5}\) = 240

For Group 2: Since, Median is the middle number of all the observation, 

So, To Find the Median, Arrange the Income of Group 2 in Ascending order, we get 3800,4000,4200,4400,4600,4800,5800 

Therefore, The Median = 4400 

Deviation |d| = |x-Median| 

Now, The Mean Deviation is

Mean Deviation = \(\frac{\Sigma d_i}{n}\)

Mean Deviation Of Group 2 = \(\frac{3200}{7}\) = 457.14

Hence, The Mean Deviation of Group 1 is 240 and Group 2 is 457.14

Correct option is (A) Both triangles and quadrilaterals

\(\because\) 22 is greater than 12.

\(\therefore\) Triangles and quadrilaterals both occur most frequently.

\(\therefore\) Both triangles and quadrilaterals are mode.

(A) Both triangles and quadrilaterals

Correct option is (A) 9: 11

Mean of first 10 whole numbers \(=\frac{0+1+2+....+9}{10}\)

\(=\frac{9\times10}{20}=\frac92\)

Mean of first 10 natural numbers \(=\frac{1+2+3+....+10}{10}\)

\(=\frac{10\times11}{20}=\frac{11}2\)

\(\therefore\) Their ratio \(=\cfrac{\frac92}{\frac{11}2}\)

\(=\frac9{11}\) = 9 : 11

Correct option is   A) 9: 11

Correct option is (C) average of and n/2 + n/2 + 1 observations

When n is even then median of n observations arranged in ascending order is average of \((\frac{n}{2})^{th}\) and \((\frac{n}{2} + 1)^{th}\) observations.

C) average of and n/2 + n/2 + 1 observations 

Let the weight of 11^th student be x

Given that mean weight of 21 students = 52kg

∴ Sum of 21 students weight = 21 × 52 = 1092kg

Mean weight of the first 11 students = 50kg

Sum of first 11 students weight = 11 × 50 = 550kg

Mean weight of the last 11 students = 54kg

Sum of the last 11 students weight = 11 × 54 = 594kg

Therefore,

Sum of first 11 students weight + sum of last 11 students weight – weight of the 11^th student = sum of 21 students weight

550+594 - x = 1092

⇒ 1144 – x = 1092

⇒ x = 52

Hence, weight of 11^th student is 52kg

The correct option is: (B) 31 

Explanation:

36 x 9 = x₁ + x₂ + x₃ +...+x₉      ...(i) 

5 x 32 = x₁ + x₂ + x₃ + x₄ + x₅    ...(ii)

5 x 39 = x₅ + x₆ + x₇ + x₈ + x₉     ...(iii)

Adding (ii) and (iii), we get 

355 = x₁ + x₂ + x₃ + x₄ + x₅ + ... + x₉ + x₅

Using (i), we get 355 - 36 x 9 = x₅ => x₅ = 31

Let the weight of 13^th student be x

Given that mean weight of 25 students = 60kg

∴ sum of 25 students weight = 25 × 60 = 1500kg

Mean weight of the first 13 students = 57kg

Sum of first 13 students weight = 13 × 57 = 741kg

Mean weight of the last 13 students = 63kg

Sum of the last 13 students weight = 13 × 63 = 819kg

Therefore,

Sum of first 13 students weight + sum of last 13 students weight – weight of the 13^th student = sum of 25 students weight

741 + 819 - x = 1500

⇒ 1560 – x = 1500

⇒ x = 60

Hence, weight of 13^th student is 60kg.

CV=(SD/Mean)*100

Let the 13^th observation be x

Given that mean of 25 observations = 36

∴ sum of 25 observations = 25 × 36 = 900

Mean of the first 13 observations = 32

Sum of first 13 observations = 13 × 32 = 416

Mean of the last 13 observations = 39

Sum of the last 13 observations = 13 × 39 = 507

Therefore,

Sum of first 13 observations + sum of last 13 observations – 13^th observation = sum of 25 observations

416 + 507 - x = 900

⇒ 923 – x = 900

⇒ x = 23

Let the 12^th observation be x

Given that mean of 23 observations = 34

∴ sum of 23 observations = 23 × 34 = 782

Mean of the first 12 observations = 32

Sum of first 12 observations = 12 × 32 = 384

Mean of the last 12 observations = 38

Sum of the last 12 observations = 12 × 38 = 456

Therefore,

Sum of first 12 observations + sum of last 12 observations – 12^th observation = sum of 23 observations

384 + 456 - x = 782

⇒ 840 – x = 782

⇒ x = 58

Let the 6^th number be x

Given that mean of 11 results = 35

∴ sum of 11 numbers = 11 × 35 = 385

Mean of the first 6 results = 32

Sum of first 6 numbers = 6 × 32 = 192

Mean of the last 6 results = 37

Sum of the last 6 results = 6 × 37 = 222

Therefore,

Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers

192 + 222 - x = 385

⇒ 414 – x = 385

⇒ x = 29

11

Explanation: 

We know the square root of variance is standard deviation, i.e.,

σ² =121,

taking square root on both sides, we get

σ=√121=11

So, if the variance of a data is 121, then the standard deviation of the data is 11.

Independent, dependent.

Explanation: 

Change of origin means some value has been added or subtracted in the observation.

And we know the standard deviation doesn’t change if any value is added or subtracted from the observations, so standard deviation is independent of change in origin.

But standard deviation is only affected by a change in scale, i.e., some value is multiplied or divided to observations.

Hence the standard deviation of any data is independent of any change in origin but is dependent of any change of scale.

Least

Explanation:

Mean deviation is sum of all deviations of a set of data about the data's mean.

And it is widely believed that the median is” usually” between the mean and the the mode.

So the mean deviation of the data is least when measured from the median.

Let the 6^th number be x

Given that mean of 11 results = 30

∴ sum of 11 numbers = 11 × 30 = 330

Mean of the first 6 results = 28

Sum of first 6 numbers = 6 × 28 = 168

Mean of the last 6 results = 32

Sum of the last 6 results = 6 × 32 = 192

Therefore,

Sum of first 6 numbers + sum of last 6 numbers – 6^th number = sum of 11 numbers

168 + 192 - x = 330

⇒ 360 – x = 330

⇒ x = 30

The standard deviation of a data is independent of any change in origin but is dependent of charge of scale.

The mean deviation of the data is least when measured from the median.

The standard deviation is greater than or equal to the mean deviation taken from the arithmetic mean

The sum of the squares of the deviations of the values of the variable is minimum when taken about their arithmetic mean.