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a month ago

Standard Deviation Class 11th

Consider three observations a, b and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true?

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Raj Pandey

Contributor-Level 9

Variance of a, b, c & a+2, b+2, c+2, are same.
Given: b = a + c (i)
d² = (1/3) (a² + b² + c²) - [ (a+b+c)/3]²
as a + c = b
d² = (1/3) (a² + b² + c²) - (2b/3)²
9d² = 3 (a² + b² + c²) - 4b²
⇒ b² = 3 (a² + c²) - 9d²

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New answer posted

a month ago

Standard Deviation Class 11th

The mean and the standard deviation (s.d.) of 10 observations are 20 and 2 respectively. Each of these 10 observations is multiplied by $p$ and then reduced by $q$ , where $p = 0$ and $q \neq 0$ . If the new mean and new s.d. become half of their original values, then $q$ is equal to:

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Raj Pandey

Contributor-Level 9

If each observation is multiplied with p and then q is subtracted
New mean x? = px? - q ⇒ 10 = p(20)-q
and new standard deviations σ? = |p|σ? ⇒ 1 = |p|(2) ⇒ |p|=1/2 ⇒ p=±1/2
If p=1/2, then q=0. If p=-1/2, q=-20.

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New answer posted

2 months ago

Standard Deviation Class 11th

The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

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Payal Gupta

Contributor-Level 10

$\bar{x} = \frac{\sum_{i = 1}^{10} x_{i}}{10} = 15 \frac{\sum_{i = 1}^{10} {x_{i}}^{2}}{10} - {(\bar{x})}^{2} = 15$

$\Rightarrow Σ x_{i} = 150 Σ {x_{i}}^{2} = 2400$

Actual mean $\bar{x} = \frac{Σ x_{i} + 15 - 25}{10} = \frac{140}{10} = 14$

Actual variance = $\frac{Σ {x_{i}}^{2} + 1 5^{2} - 2 5^{2}}{10} - {(14)}^{2}$

$= \frac{2400 - 400}{10} - 196$

$\begin{array}{l} σ^{2} = 4 \\ σ = 2 \end{array}$

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