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If the variance of the distribution
\begin{array}{cccccccc}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 8 & 11 & 17 & 20 & 24 & 32 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1
\end{array}
is 45.8 , then the variance of the distribution.
\begin{array}{llllllll}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 10 & 18 & 24 & 36 & 42 & 50 & 66 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1 \\
\hline
\end{array}
Options:
\begin{array}{cccccccc}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 4 & 8 & 11 & 17 & 20 & 24 & 32 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1
\end{array}
is 45.8 , then the variance of the distribution.
\begin{array}{llllllll}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 10 & 18 & 24 & 36 & 42 & 50 & 66 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 3 & 5 & 9 & 5 & 4 & 3 & 1 \\
\hline
\end{array}
Solution:
1537 Upvotes
Verified Answer
The correct answer is:
183.2
Mean of given observation $\left(x_i\right)$ is
$$
\begin{array}{r}
\frac{(4 \times 3)+(8 \times 5)+(11 \times 9)+(17 \times 5)+(20 \times 4)}{+(24 \times 3)+(32 \times 1)} \\
=\frac{420}{30}=14
\end{array}
$$
So,
$$
\begin{aligned}
\mathrm{ce} & =k \frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30} \\
& =45.8 \text { (given) }
\end{aligned}
$$
Now, mean of given observation $\left(y_i\right)$ is
$$
\begin{array}{r}
\begin{array}{r}
(10 \times 3)+(18 \times 5)+(24 \times 9)+(36 \times 5)+(42 \times 4) \\
+(50 \times 3)+(66 \times 1)
\end{array} \\
\begin{aligned}
30 & \frac{900}{30}=30
\end{aligned}
\end{array}
$$
So, variance
$$
\begin{aligned}
& =k \frac{20^2+12^2+6^2+6^2+12^2+20^2+36^2}{30} \\
& =4 \times k\left(\frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30}\right) \\
& =4 \times 45.8=183.2
\end{aligned}
$$
$$
\begin{array}{r}
\frac{(4 \times 3)+(8 \times 5)+(11 \times 9)+(17 \times 5)+(20 \times 4)}{+(24 \times 3)+(32 \times 1)} \\
=\frac{420}{30}=14
\end{array}
$$
So,
$$
\begin{aligned}
\mathrm{ce} & =k \frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30} \\
& =45.8 \text { (given) }
\end{aligned}
$$
Now, mean of given observation $\left(y_i\right)$ is
$$
\begin{array}{r}
\begin{array}{r}
(10 \times 3)+(18 \times 5)+(24 \times 9)+(36 \times 5)+(42 \times 4) \\
+(50 \times 3)+(66 \times 1)
\end{array} \\
\begin{aligned}
30 & \frac{900}{30}=30
\end{aligned}
\end{array}
$$
So, variance
$$
\begin{aligned}
& =k \frac{20^2+12^2+6^2+6^2+12^2+20^2+36^2}{30} \\
& =4 \times k\left(\frac{10^2+6^2+3^2+3^2+6^2+10^2+18^2}{30}\right) \\
& =4 \times 45.8=183.2
\end{aligned}
$$
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