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The class intervals are 9.5 − 19.5, 19.5 − 29.5, 29.5 − 39.5, 39.5 − 49.5, 49.5 – 59.5

∴ Range = Maximum value − minimum value

= 59.5 – 9.5

= 50

We know that empirical relation between mean, median and mode is 

Mode = 3 Median - 2 Mean

Here, the variable are1,2,3,…, n

∴ Mean, \(\bar{x}=\frac{\sum x_i}{n}\)

\(=\frac{1+2+3+...+n}{n}\)

\(=\frac{n(n+1)}{2n}\)

 \(=\frac{n+1}{2}\)

Variance = \(\sum x_i^2-(\bar{x})^2\)

= \(\frac{1^2+2^2+3^2+...+n^2}{n}-(\frac{n+1}{2})^2\)

= \(\frac{n(n+1)(2n+1)}{6n}-(\frac{n+1}{2})\)

= \((n+1)\{\frac{4n+2-3n-3}{12}\}\) 

 = \(\frac{(n+1)(n-1)}{12}\)

= \(\frac{n^2-1}{12}\)

Here,

N = 10,

\(\bar{x}\) = 50

And,

\(\sum_{i=1}^{10} (x_i-50)^2=250\)

⇒ \(\frac{\sum_{i=1}^{10} (x_i-50)^2}{10}=25\)

⇒ σ² = 25

 ⇒ σ = 5

∴ Co-effecient of variation,

C.V = \(\frac{σ}{x}\times100\)

= \(\frac{5}{50}\times100\)

= \(\frac{500}{50}\)

 = 10

∴ 10 is the co-effecient of variation.

Given,

\(\bar{x}=50\),

n = 100,

σ = 5

We have,

\(σ^2=\frac{\sum x_i^2}{n}-(\bar{x})^2\)

⇒ \(\frac{\sum x_i^2}{n}=σ^2+(\bar{x})^2\)

⇒ \(\sum x_i^2=n[σ^2+(\bar{x})^2]\)

= 100 [5² + (50)²]

= 100 (25 + 2500)

= 100 × 2525

= 252500

Hence, the sum of all squares of all the observation in 252500.

Given,

CV₁ = 60, CV₂ = 70

σ₁ = 21, σ₂ = 16

We Know,

CV = \(\frac{σ}{x}\times100\)

 \(\bar{x}=\frac{σ}{CV}\times100\)

\(\bar{x_1}=\frac{σ_1}{CV_1}\times100\)

\(=\frac{21}{60}\times100\)

= 35

and,

\(\bar{x_2}=\frac{σ_2}{CV_2}\times100\)

\(=\frac{16}{70}\times100\)

 = 22.85

∴ Their arithmetic mean are 35 and 22.85

Given,

C.V₁= 50, 

C.V₂ = 60

\(\bar{x_1}=30\),

\(\bar{x_2}=25\) 

We Know:

C.V = \(\frac{σ}{x}\times100\) 

⇒ \(\bar{x}=\)\(\frac{σ}{C.V}\times100\) 

∴ \(\bar{x_1}=\)\(\frac{σ_1}{C.V_1}\times100\) 

⇒ 30 = \(\frac{σ_1}{50}\times100\) 

⇒ \(σ_1\) = 15 

And,

\(\bar{x_2}=\) \(\frac{σ_2}{C.V_2}\times100\) 

⇒ 25 = \(\frac{σ_2}{60}\times100\) 

⇒ \(σ_2\) \(=25\frac{3}{5}\)

⇒ \(σ_2\) \(=15\)

∴ Required difference

= \(σ_1-σ_2\) 

= 15 − 15

= 0

Arranging the given data in ascending order

27, 28, 28, 29, 29, 30, 31, 31, 31, 31, 31, 32, 33, 34, 34, 35, 35, 36, 38, 38. 

Here the number 31 occurs 5 times which is the maximum. 

∴ Mode of this data is 31.

The mean deviation of the numbers 3, 4, 5, 6, 7

Mean \(\bar X = \frac{3+4+5+6+7}{5}\) 

\(\bar X = \frac{25}{5}\) 

\(\bar X = 5\)

Mean deviation = \(\frac{\Sigma d_i}{\Sigma x_i}\) 

Mean deviation = \(\frac{0}{25}\) 

Hence, Mean deviation is 0

Mean \((X)\)= \(\frac{38+70+48+34+42+55+63+46+54+44}{10}\) 

\((\bar X)\) = \(\frac{494}{10}\) 

Here, N= 10, Σd = 0

Mean deviation = \(\Big(\frac{\Sigma d_i}{N}\Big)\) 

M.D. = \(\Big(\frac{0}{10}\Big)\) 

Hence, MD is 0

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation. 

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

Explanation: Here, Observations 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 are Given. Deviation |d| = |x-Mean|

Mean = Σ \(\frac{|x_i|}{n}\)

of the Given Observations = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\)  = \(\frac{500}{10}\)

And, The number of observations is 10. 

Now, The Mean Deviation is

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

Mean Deviation of the given Observations = \(\frac{84}{10}\) = 8.4

Hence, The Mean Deviation is 8.4

Let the assumed mean (A) = 15

We have, 

A = 15

h = 6

Mean = A + h x \(\frac{\sum f_iu_i}{N}\)

= 15 + 6 x \(\frac{5}{40}\)

= 15 + 0.75 = 15.75

Given, Numbers of observations are given. 

To Find: Calculate the Mean Deviation. 

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

Explanation: Here, Observations 38, 70, 48, 34, 42, 55, 63, 46, 54, 44 are Given. 

Since, Median is the middle number of all the observation, 

So, To Find the Median, Arrange the numbers in Ascending order, we get 34, 38,42,44,46,48,54,55,63,70 

Here the Number of observations are Even then Median = Mean Deviation = Hence, The Mean Deviation is 8.6

Therefore, The Median = 47 

Deviation |d| = |x-Median| 

And, The number of observations is 10. 

Now, The Mean Deviation is

Mean Deviation = \(\frac{86}{10}\)

Hence, The Mean Deviation is 8.6 

Let the assumed mean (A) = 20

We have, 

A = 20

h = 8

Mean = A + h x \(\frac{\sum f_iu_i}{N}\)

\(=20 + 8 \times\frac{5}{40}\)

\(=20+1.4 = 21.4\)

Let the assumed mean (A) = 25

We have, 

A = 25

h = 10

Mean = A + h x \(\frac{\sum f_iu_i}{N}\)

\(=25+10\times\frac{8}{60}\)

\(=25+\frac{8}{6}\)

\(=25+\frac{4}{3}=26.333\)

Let the assumed mean (A) = 20

We have,

 A = 20

h = 8

Mean = A + h x \(\frac{\sum f_iu_i}{N}\)

= 20 + 8 x \(\frac{7}{40}\)

\(=20 + 1.4 = 21.4\)

1. False
2. True
3. False
4. True
5. True

Let the assumed mean (A) = 50

We have, 

A = 50

h = 10

Mean = A + h \(\times\frac{\sum f_iu_i}{N }\)

\(=50+10\times\frac{-2}{40}\)

\(=50-0.5\)

\(=49.5\)

By direct method

Mean \(=\frac{\sum f_ix_i}{N}+A\) 

\(=\frac{848}{64} = 13.25\)

By assumed mean method

Let, 

the assumed mean (A) = 6.5

Mean = A + \(\frac{\sum f_iu_i}{N}\)

\(=6.5+\frac{432}{64}\)

\(=6.5+6.75\)

\(=13.25\) 

Correct option is (A) Inclusive classes

The given classes 30–39, 40–49, 50–59 includes end-points.

\(\therefore\) These classes are called inclusive classes.

A) Inclusive classes

(b) discontinuous

Let the assumed mean (A) = 42

We have, 

A = 42

Mean = A + h \(\times\frac{\sum f_iu_i}{N}\)

\(=42+5\times\frac{-79}{10}\)

\(=42-\frac{79}{14}\)

\(=\frac{588-79}{14}\)

\(=36.357\) 

10, 18, 22, 28, 33, 34, 41, 44, 49, 57, 59, 66 (Ascending order) 

N= 10 

∴ Median = \(\frac{34 - 41}{2}\) = \(\frac{75}{2}\) = 37.5