Ask & Answer Question

Ncert Solutions Maths class 11th Maths Class 11th Statistics

3. Find the mean deviation about the median for the data in Exercises 3 and 4.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

2 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

3. Arranging the data in ascending order we get,

10,11,1112,13,13,14,16,16,17,17,18

As n=12, even

So, median is the mean of ${(\frac{M}{2})th}^{}$ and ${(\frac{M}{2} + 1)th}^{}$ observation.

$\Rightarrow = \frac{{6thobservation}^{} + {7thobservation}^{}}{2}$

$M= \frac{13 + 14}{2} = \frac{27}{2} = 13.5 .$

So, deviation of respective observation about the median. $M, | x_{i} - M |$ are

x_i	10	11	11	12	13	13	14	16	16	17	17	18
\|x_i- M\|	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 the mean deviation about the mean is

$M.D. (M) = \frac{1}{n} \times \sum_{i = 1}^{n} | x_{i} - M |$

$= \frac{1}{12} \times (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5)$

$= \frac{28}{12} = 2.33 .$

0 Upvotes 0 Downvotes

Similar Questions for you

Given data 60, 60, 44, 58, 68, α, β, 56 has mean 58, variance = 66.2 then find α² + β²

alok kumar singh

Variance = $\frac{\sum x^{2}}{n} - {(\bar{x})}^{2}$

$\frac{6 0^{2} + 6 0^{2} + 4 4^{2} + 5 8^{2} + 6 8^{2} + α^{2} + β^{2} + 5 6^{2}}{8} = {(58)}^{2} = 66.2$

$\frac{7200 + 1936 + 3364 + 4624 + 3136 + α^{2} + β^{2}}{8} = 3364 = 66.2$

$2532.5 + \frac{α^{2} + β^{2}}{8} - 3364 = 66.2$

α² + β² = 897.7 × 8

= 7181.6

For the following data table

x_i

f_i

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

Find the value of 20 M (where M is median of the data)

alok kumar singh

x_i

f_i

c.f.

0 – 4

4 – 8

8 – 12

12 – 16

16 – 20

$N = \sum_{}^{} f = 27$

$(\frac{N}{2}) = \frac{27}{2} = 13.5$

So, we have median lies in the class 12 – 16

I₁ = 12, f = 8, h = 4, c.f. = 13

So, here we apply formula

$M = I_{1} + \frac{\frac{N}{2} - c . f .}{f} \times h = 12 + \frac{13.5 - 13}{8} \times 4$

$= 12 + \frac{5}{2}$

$M = \frac{24.5}{2} = 12.25$

20 M = 20 × 12.25

= 245

Let mean and variance of 6 observations a, b, 68, 44, 40, 60 be 55 and 194. If a > b then find a + 3b

alok kumar singh

$\frac{a + b + 68 + 44 + 40 + 60}{6} = 55$

212 + a + b = 330

⇒ a + b = 118

$\frac{\sum_{}^{} x_{i}^{2}}{n} - {(\bar{x})}^{2} = 194$

$\frac{a^{2} + b^{2} + {(68)}^{2} + {(44)}^{2} + {(40)}^{2} + {(60)}^{2}}{6} = {(55)}^{2} = 194$

= 3219

11760 + a² + b² = 19314

⇒ a² + b² = 19314 – 11760

= 7554

(a + b)² –2ab = 7554

From here b = 41.795

a + b = 118

⇒ a + b + 2b = 118 + 83.59

= 201.59

Let X₁, X₂,…..X₁₈ be eighteen observations such that $\sum_{i = 1}^{18} (X_{i} - α) = 36 a n d \sum_{i = 1}^{18} {(X_{i} - β)}^{2} = 90,$ where a and b are distinct real numbers. If the standard deviation of these observations is 1, then the value of $| α - β | i s . . . . . . . .$

alok kumar singh

Kindly go throuigh the solution

Given $\sum_{i = 1}^{18} (x_{i} - α) = 36 i . e \sum_{i = 1}^{18} x_{i} - 18 α = 36 . . . . . . . . . . (i)$

& $\sum_{i = 1}^{18} {(x_{i} - β)}^{2} = 90 i . e \sum_{i = 1}^{18} {x_{i}}^{2} - 2 β \sum_{}^{} x_{i} + 18 β^{2} = 90 . . . . . . . . . . . . . (i i)$

(i) & (ii) $\sum_{i = 1}^{18} {x_{i}}^{2} = 90 - 18 β + 36 β (α + 2) . . . . . . . . . . . . . (i i i)$

Now variance $σ^{2} = \frac{\sum {x_{i}}^{2}}{n} - {(\frac{\sum x_{i}}{n})}^{2}$ = 1 given

->(a - b) (a - b + 4) = 0

Since $α \neq β s o | α - β | = 4$

The number of values of $a \in N$ such that the variance of 3, 7, 12, a, 43 -a is a natural number is:

Vishal Baghel

$M e a n = \frac{3 + 12 + 7 + a + (43 - a)}{5} = 13$

Variance = $\frac{3^{2} + 1 2^{2} + 7^{2} + a^{2} + {(43 - a)}^{2}}{5} - {(13)}^{2}$

$\Rightarrow \frac{2 a^{2} - a + 1}{5} \to N a t u r a l n u m b e r$

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4 λ o r 4 λ + 1 f r o m .$

As each square form is $4 λ o r 4 λ + 1$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

3. Find the mean deviation about the median for the data in Exercises 3 and 4.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

3. Find the mean deviation about the median for the data in Exercises 3 and 4. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

3. Find the mean deviation about the median for the data in Exercises 3 and 4.

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17