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Class 11 Maths Statistics exercise 15.1 Solution

1. Find the mean deviation about the mean for the data in Exercises 1 and 2.

4, 7, 8, 9, 10, 12, 13, 17

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Payal Gupta | Contributor-Level 10

4 months ago

1. Mean of the given observation is.

$\bar{x} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8}$

$= \frac{80}{8} = 10 .$

Deviation of the respective observation about the mean $\bar{x}$ i.e., $x_{i} - \bar{x}$ are 4–10,7–10,8–10,9–10,10–10,12–10,13–10,17–10

=6, -3, -2, -1,0,2,3,7

The absolute value of the deviation i.e., $| x_{i} - \bar{x} |$ are 6,3,2,1,0,2,3,7.

Therefore, the required mean deviation about the mean is

$M.D = (\bar{x}) = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} | = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}$

$= \frac{24}{8}$

= 3.

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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 $4 λ o r 4 λ + 1$

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Class 11 Maths Statistics exercise 15.1 Solution

1. Find the mean deviation about the mean for the data in Exercises 1 and 2.

4, 7, 8, 9, 10, 12, 13, 17