The mean of the numbers 1, 4, k, (k + 4) and 11 is (k + 1). Calculate the :
(b) standard deviation.
(a) Value of k. The mean of \(1, 4, k, (k+4), 11\) is \((k+1)\):
\[ \frac{1 + 4 + k + (k+4) + 11}{5} = k + 1 \Rightarrow \frac{20 + 2k}{5} = k + 1. \]
\[ 20 + 2k = 5k + 5 \Rightarrow 15 = 3k \Rightarrow k = 5. \]
(b) Standard deviation. The numbers become \(1, 4, 5, 9, 11\) with mean \(k+1 = 6\).
Deviations from the mean: \(-5, -2, -1, 3, 5\); their squares: \(25, 4, 1, 9, 25\), summing to \(64\).
\[ \text{Variance} = \frac{64}{5} = 12.8, \qquad \text{S.D.} = \sqrt{12.8} \approx 3.58. \]