Which of the following statements are TRUE?

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The first group of two samples has 100 items with mean 15 and Standard deviation 3. If the combined sample of two groups has 250 items with mean 15.6 and standard deviation $\sqrt{13.44},$ then the standard deviation of the second group is 4

$\lim_{n \to \infty} \sum_{k = 0}^{n} \frac{^{n} c_{k}}{n {.2}^{n} + k} = 0$

$2 (\tan^{- 1} π - \tan^{- 1} e) - l n π + 1 < 0$

If the mean deviation of the numbers 1, 1 + d, 1 + 2d …… 1 + 100d from their mean is 255, then d is equal to 10.2

answer is A, B, C.

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Detailed Solution

$A) 2 (\tan^{- 1} π - \tan^{- 1} e) - l n π + 1 < 0 C o n s i d e r f (x) = 2 (\tan^{- 1} x) - \ln x f' (x) = \frac{2}{1 + x^{2}} - \frac{1}{x} = \frac{2 x - 1 - x^{2}}{x (1 + x^{2})} = \frac{- {(x - 1)}^{2}}{x (x^{2} + 1)} < 0 f o r x > 0 e < π \Rightarrow f (e) < f (π) 2 \tan^{- 1} (e) - \ln e > 2 \tan^{- 1} (π) - \ln (π) 0 > 2 (\tan^{- 1} (π) - \tan^{- 1} (e)) - \ln π + 1 H e n c e i t i s t r u e$

$B) \sum_{k = 0}^{n} \frac{^{n} c_{k}}{n (2^{n} + 1)} \leq \sum_{k = 0}^{n} \frac{^{n} c_{k}}{n {.2}^{n} + k} \leq \sum_{k = 0}^{n} \frac{^{n} c_{k}}{n {.2}^{n}} \frac{^{n} c_{k}}{n {.2}^{n} + n} \leq \frac{^{n} c_{k}}{n {.2}^{n} + k} \leq \frac{^{n} c_{k}}{n {.2}^{n} + 0} 0 \leq \sum_{k = 0}^{\infty} \frac{^{n} c_{k}}{2^{n} + k} \leq 0$

$C) U s e σ^{2} = \frac{n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}$

Where $d_{1} = m_{1} - a, d_{2} = m_{2} - a, a$ being the mean of the whole ground, Let $m_{2}$ =mean of the second group

$15.6 = \frac{100 \times 15 + 150 \times m_{2}}{250} \Rightarrow m_{2} = 16, (100 \times 9 + 150 \times σ^{2}) T h u s, 13.44 = \frac{+ 100 \times {(0.6)}^{2} + 150 \times {(0.4)}^{2}}{250} \Rightarrow σ = 4$

$D) \bar{x} = \frac{1}{101} [1 + (1 + d) + (1 + 2 d) + ............ + (1 + 100 d)] = \frac{1}{101} \times \frac{101}{2} [1 + (1 + 100 d)] = 1 + 50 d$

Mean deviation from mean $= \frac{1}{101} [| 1 - (1 + 50 d) | + | 1 + d -] (1 + 50 d) | + ........... + | 1 + 100 d - (1 + 50 d) | 255 = \frac{2 | d |}{101} {1 + 2 + ............. + 50} \Rightarrow 2 \frac{| d |}{101} \frac{50 (51)}{2} = \frac{2550}{101} | d | N o w \frac{2550}{101} | d | = 10.1$