The data shows the ordered marks scored by students in a test: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). Given that the median is 13\(\frac{1}{2}\) and y is greater than x by 1, find:
(a) the values of x and y
(b) correct to three significant figures, the standard deviation of the distribution.
Given: 11, 12, (2x + y), (x + 2y), 14, and ((y\(^2\) - 2x). median = 13\(\frac{1}{2}\) = \(\frac{27}{2}\)
Median = \(\frac{(2x + y) + (x + 2y)}{2}\) = \(\frac{27}{2}\)
\(\frac{3x + 3y}{2}\) = \(\frac{27}{2}\)
3x + 3y = 27
x + y = 9 - - - - - - - -(i)
also, y = x + 1 - - - - -(ii)
put y = x + 1 into eqn (ii)
x + x + 1 = 9
2x = 9 - 1 = 8
x = \(\frac{8}{2}\) = 4
y = x + 1 = 4 + 1 = 5
(b) Since x = 4 and y = 5
11, 12, (2(3) + 5), (4 + (2(5)), 14, (5\(^2\) - 2(4))
11, 12, 13, 14, 14, 17
Mean(\(\overline{x}\)) = \(\frac{\sum{x}}{n}\) = \(\frac{11 + 12 + 13 + 14 + 14 + 17}{6}\) = 13.5
| x | f | fx | (x - \(\overline{x}\)) | (x - \(\overline{x}\))\(^2\) | f(x - \(\overline{x}\))\(^2\) |
| 11 | 1 | 11 | - 2.5 | 6.25 | 6.25 |
| 12 | 1 | 12 | - 1.5 | 2.25 | 2.25 |
| 13 | 1 | 13 | - 0.5 | 0.25 | 0.25 |
| 14 | 2 | 28 | 0.5 | 0.25 | 0.50 |
| 17 | 1 | 17 | 3.5 | 12.5 | 12.25 |
\(\sum f(x - \overline{x})^2\) = 21.50
S.D = \(\sqrt{\frac{\sum f(x - \overline{x})^2}{\sum {f}}}\) = \(\sqrt{\frac{21.50}{6}}\) = 1.89
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