0 like 0 dislike
246 views
The mean of the numbers $1,4, k_{t}(k+4)$ and 11 is $(k+1) .$ Calculate the :

(a) value of $\mathrm{k}$;

(b) standard deviation.
| 246 views

0 like 0 dislike
(a) $\frac{1+4+k+(k+4)+11}{5}=k+1$
\begin{aligned} &\frac{2 k+20}{5}=k+1 \\ &2 k+20=5(k+1) \Longrightarrow 2 k+20=5 k+5 \\ &20-5=5 k-2 k \Longrightarrow 15=3 k \\ &k=5 \end{aligned}
(b) Mean $\bar{x}=6$
\begin{align}
\begin{array}{|l|l|l|l|l|}
\hline \text { Mark } & \text { freq (f) } & \text { fx } & x^{2} & f x^{2} \\
\hline(x) & \text { 1 } & 1 & 1 & 1 \\
\hline 1 & 1 & 4 & 16 & 16 \\
\hline 4 & 1 & 5 & 25 & 25 \\
\hline 5 & 1 & 9 & 81 & 81 \\
\hline 9 & 1 & 11 & 121 & 121 \\
\hline 11 & 1 & 30 & & 244 \\
\hline & 5 & \\
\hline
\end{array}
\end{align}
\begin{aligned} &\sigma=\sqrt{\frac{\sum f x^{2}}{\sum f}-\left(\frac{\sum f x}{\sum f}\right)^{2}} \\ &=\sqrt{\frac{244}{5}-\left(\frac{30}{5}\right)^{2}} \\ &=\sqrt{48.8-36} \\ &=\sqrt{12.8}=3.58 \end{aligned}
by Diamond (88,926 points)

1 like 0 dislike
3 like 0 dislike
0 like 0 dislike
3 like 0 dislike
1 like 0 dislike
0 like 0 dislike
1 like 0 dislike
2 like 0 dislike
2 like 0 dislike
3 like 0 dislike
0 like 0 dislike
2 like 0 dislike
0 like 0 dislike
1 like 0 dislike