The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and 94 respectively. Then x4+y2 is equal to.

=4+5+6+6+7+8+x+y8=6 ∴x+y=12 …. (i) And variance =22+12+02+02+12+22+ (x−6)2+ (y−6)28 =94 ∴ (x−6)2+ (y−6)2=8 ……. (ii) From (i) and (ii) x = 4 and y = 8 ∴x4+y2=320

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.

Option 1 - 162

Option 2 - 320

Option 3 - 674

Option 4 - 420

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Answered by

Payal Gupta | Contributor-Level 10

8 months ago

Correct Option - 2

Detailed Solution:

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

$∴ x + y = 12$ …. (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:

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

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

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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………

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

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

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Since, variance is independent of origin.

So, we subtract 10 from each observation.

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

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

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