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  • Let m be the mean and be the standard deviation of the distribution <

Let m be the mean and be the standard deviation of the distribution

where \sum f_i=62 . If [x] denotes the greatest integer \leq x , then \left[\mu 2+\sigma^2\right]  is equal to

Option: 1

8


Option: 2

7


Option: 3

6


Option: 4

9


Answers (1)

best_answer

\begin{aligned} & \sum f_i=62 \\ & 3 k^2+16 k-12 k-64=0 \\ & k=4 \text { or }-\frac{16}{3}(\text { rejected }) \end{aligned}

\begin{aligned} & \mu=\frac{\sum f_i x_i}{\sum f_i} \\ & \mu=\frac{8+2(15)+3(15)+4(17)+5}{62}=\frac{156}{62} \\ & \sigma^2=\sum f_i x_i^2-\left(\sum f_i x_i\right)^2 \\ & =\frac{8 \times 1^2+15 \times 13+17 \times 16+25}{62}-\left(\frac{156}{62}\right)^2 \\ & \sigma^2=\frac{500}{62}-\left(\frac{156}{62}\right)^2 \\ & \sigma^2+\mu^2=\frac{500}{62} \\ & {\left[\sigma^2+\mu^2\right]=8} \end{aligned}

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Ritika Jonwal

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