Let [] be the mean and [] be the standard deviation of the distribution x_i 0 1 2 3 4 5 f_…

Question

Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution

${x_i}$
    0
    1
    2
    3
    4
    5

${f_i}$
    $k + 2$
    $2k$
    ${k^2} - 1$
    ${k^2} - 1$
    ${k^2} + 1$
    $k - 3$

where $$\sum f_{i}=62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\left[\mu^{2}+\sigma^{2}\right]$$ is equal to :

Accepted Answer

We have, $\Sigma f_i=62$

$$
\begin{aligned}
& \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\
& \Rightarrow 3 K^2+4 K-64=0 \\
& \Rightarrow (3 K+16)(K-4)=0 \\
& \Rightarrow K=4 \quad
\end{aligned}
$$

$$
\left(\because k=\frac{-16}{3} \text { is not possible }\right)
$$

$$
\[\begin{gathered} x_i \quad  f_i \quad  f_i x_i \quad  f_i x_i^2 \ 0 \quad  6 \quad  0 \quad  0 \ 1 \quad  8 \quad  8 \quad  8 \ 2 \quad  15 \quad  30 \quad  60 \ 3 \quad  15 \quad  45 \quad  135 \ 4 \quad  17 \quad  68 \quad  272 \ 5 \quad  1 \quad  5 \quad  25 \ \text { Total } \quad  62 \quad  156 \quad  500 \end{gathered}$$
\]

$$
\mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}
$$

$$
\begin{aligned}
\sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\
= & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\
& =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2
\end{aligned}
$$

$$
\begin{aligned}
& \therefore \sigma^2+\mu^2=\frac{500}{62} \\
& \text { Hence, }\left[\sigma^2+\mu^2\right]=8
\end{aligned}
$$

Solutions · 1