The mean of the numbers 1, 4, k, (k + 4) and 11 is (k + 1). Calculate the :
(a) value of k ;
(b) standard deviation.
(a) \(\frac{1 + 4 + k + (k + 4) + 11}{5} = k + 1\)
\(\frac{2k + 20}{5} = k + 1\)
\(2k + 20 = 5(k + 1) \implies 2k + 20 = 5k + 5\)
\(20 - 5 = 5k - 2k \implies 15 = 3k\)
\(k = 5\)
(b) Mean \(\bar{x} = 6\)
|
Mark (x) |
freq (f) | fx | \(x^{2}\) | \(fx^{2}\) |
| 1 | 1 | 1 | 1 | 1 |
| 4 | 1 | 4 | 16 | 16 |
| 5 | 1 | 5 | 25 | 25 |
| 9 | 1 | 9 | 81 | 81 |
| 11 | 1 | 11 | 121 | 121 |
| 5 | 30 | 244 |
\(\sigma = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\)
= \(\sqrt{\frac{244}{5} - (\frac{30}{5})^{2}}\)
= \(\sqrt{48.8 - 36}\)
= \(\sqrt{12.8} = 3.58\)
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