Q76 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

For a group of 100 observations, the arithmetic mean and standard deviation are 8 and $\sqrt{10.5}$ respectively. The mean and standard deviation of 50 items selected from these 100 observations are 10 and 2 respectively. Then the standard deviation of the remaining 50 observation is

  • A
    2
  • Correct answer
    B
    3
  • C
    3.5
  • D
    4

Solution

Let $\sigma_1$ and $\bar{x}_1$ are the standard deviation and mean of 50 observation and $\sigma_2$ and $\bar{x}_2$ are standard deviation one mean of another 50 observation respectively $$ \begin{gathered} \bar{x}_1=10, \sigma_1=2, n_1=50 \\ \bar{x}=\text { Mean of } 100 \text { observation }=8 \\ \sigma=\text { SD of } 100 \text { observation }=\sqrt{10.5} \\ \bar{x}=\frac{50 \bar{x}_1+50 \bar{x}_2}{50+50} \\ 800=500+50 \bar{x}_2 \\ \bar{x}=6 \\ \sigma^2=\frac{n_1\left(\sigma_1\right)^2+n_2\left(\sigma_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2} \\ 10.5=\frac{50(4)+50\left(\sigma_2\right)^2}{100}+\frac{2500(10-6)^2}{(100)^2} \\ 10.5=\frac{200+50\left(\sigma_2\right)^2}{100}+\frac{25 \times 16}{100} \\ \Rightarrow \quad 1050=600+50\left(\sigma_2\right)^2 \\ \Rightarrow \quad \sigma_2^2=\frac{1050-600}{50}=9 \Rightarrow \sigma_2=3 \end{gathered} $$

Hence, the correct answer is option: 'B'.

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