Question

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.

Answer 1

(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}
\[
\begin{aligned}
&\sigma=\sqrt{\frac{\sum f x^{2}}{\sum f}-\left(\frac{\sum f x}{\sum f}\right)^{2}} \\
&=\sqrt{\frac{244}{5}-\left(\frac{30}{5}\right)^{2}} \\
&=\sqrt{48.8-36} \\
&=\sqrt{12.8}=3.58
\end{aligned}
\]