Let the mean and the variance of 5 observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} b e \frac{24}{5} a n d \frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then (4a + x₅) is equal to:

Option 1 - 13

Option 2 - 15

Option 3 - 17

Option 4 - 18

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Alok Kumar Singh | Contributor-Level 10

8 months ago

Correct Option - 2

Detailed Solution:

$\frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} = \frac{7}{2}$

$x_{1} + x_{2} + x_{3} + x_{4} = 14$

and $\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5} = \frac{24}{5}$

x₅= 10

Variance $\frac{\sum_{i = 1}^{4} x_{i}^{2}}{4} - {(\frac{Σ x i}{4})}^{2} = a$

$\frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}}{4} - \frac{49}{4} = a$

$x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} = 4 a + 49$

and $\frac{x_{1}^{2} + x_{2}^{2} + x_{5}^{2} + x_{4}^{2} + x_{5}^{2}}{5} - {(\frac{24}{5})}^{2} = \frac{194}{25}$

$\frac{4 a + 49 + x_{5}^{2}}{5} = \frac{576 + 194}{25}$

49 + 49 + $x_{5}^{2} = \frac{770}{5}$

49 $+ x_{5}^{2} + 49 = 154$

4a + 149 = 154

4a = 5

now 4a + x₅= 15

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Similar Questions for you

The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $σ$ is the standard deviation of the data after omitting the two wrong observations from the data, then $38 σ^{2}$ is equal to………

Alok Kumar Singh

$\bar{x} = \frac{\sum x i}{40} = 30 \Rightarrow \sum x i = 1200$ ………. (i)

$α^{2} = \frac{1}{40} \sum x i_{i}^{2} - {(30)}^{2} = 25$

$\Rightarrow \sum x_{i}^{2} = 37000$

after omitting two wrong observations

$\sum y_{i}^{2} = 37000 - 144 - 100 = 36756$

$∴ 38 a^{2} = 36756 - 36158 = 238$

The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations are $5,7, 10,12,14,15$ , then the absolute difference of the remaining two observations is:

Alok Kumar Singh

$\overline{x} = 10$

$\Rightarrow \overset{?}{x} = \frac{63 + a + b}{8} = 10$

$\Rightarrow a + b = 17$

Since, variance is independent of origin.

So, we subtract 10 from each observation.

So, $σ^{2} = 13.5 = \frac{79 + (a - 10)^{2} + (b - 10)^{2}}{8}$

$\Rightarrow a^{2} + b^{2} - 20 (a + b) = - 171$

$\Rightarrow a^{2} + b^{2} = 169$

From (1) and (2) ; $a = 12$ and $b = 5$

Let the mean and the variance of 20 observations x1, x2,…….x20 be 15 and 9, respectively. For $\in R$ , if the mean of ${(x_{1} + α)}^{2}, {(x_{2} + α)}^{2}, . . . ., {(x_{20} + α)}^{2}$ is 178, then the square of the maximum value of is equal to………….

Vishal Baghel

$\bar{x} = 15 \Rightarrow \sum_{i = 1}^{20} x_{i} = 300$

$\frac{\sum_{i = 1}^{20} {(x_{i} - 15)}^{2}}{20} = 9 \Rightarrow \sum_{i = 1}^{20} {(x_{i} - 15)}^{2} = 180$

$\sum_{i = 1}^{20} {(x_{i} + α)}^{2} = 178 \times 20 = 3560$

$4680 + 2 α (300) + 20 α^{2} = 3560$

$α^{2} + 30 α + 234 - 178 = 0$

= 2, 28

$α_{m a x} 2 = {(- 2)}^{2} = 4$

If the variance of the following frequency distribution :

$\begin{array}{l} Class & : & 10 - 20 & 20 - 30 & 30 - 40 \\ Frequency & : & 2 & x & 2 \end{array}$ is 50 , then $x$ is equal to:

Alok Kumar Singh

Since, ${l i m}_{x \to 0} ? \frac{f (x)}{x}$ exist $\Rightarrow f (0) = 0$

Now, $f^{'} (x) = {l i m}_{h \to 0} ? \frac{f (x + h) - f (x)}{h} = {l i m}_{h \to 0} ? \frac{f (h) + x h^{2} + x^{2} h}{h} ($ take $y = h)$

$= {l i m}_{h \to 0} ? \frac{f (h)}{h} + {l i m}_{h \to . 0} ? (x h) + x^{2}$

$\Rightarrow f^{'} (x) = 1 + 0 + x^{2}$

$\Rightarrow f^{'} (3) = 10$

The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $\frac{9}{4}$ respectively. Then $x^{4} + y^{2}$ is equal to.

Payal Gupta

$= \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$

$∴ x + y = 12$ …. (i)

And variance

$= \frac{2^{2} + 1^{2} + 0^{2} + 0^{2} + 1^{2} + 2^{2} + {(x - 6)}^{2} + {(y - 6)}^{2}}{8}$

$= \frac{9}{4}$

$∴ {(x - 6)}^{2} + {(y - 6)}^{2} = 8$ ……. (ii)

From (i) and (ii)

x = 4 and y = 8

$∴ x^{4} + y^{2} = 320$

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Let the mean and the variance of 5 observations $x_{1}, x_{2}, x_{3}, x_{4}, x_{5} b e \frac{24}{5} a n d \frac{194}{25}$ respectively. If the mean and variance of the first 4 observation are $\frac{7}{2}$ and a respectively, then (4a + x₅) is equal to: