The table gives the distribution of marks of 60 candidates in a test. Marks 23-25 26-28 29-31 32-34 35-37 38-40 Frequency 3 7 15 21 10 4 (a) Draw a cumulati...

Answer Details

Setting up class boundaries and cumulative frequencies

The marks are recorded in whole numbers, so each class boundary lies halfway between the top of one class and the bottom of the next. The lower boundary of the first class is \(22.5\), and we plot the cumulative frequency against each upper class boundary.

Total frequency \(N = 60\).

(a) The cumulative frequency curve (ogive)

Plot the points \((22.5,\,0)\), \((25.5,\,3)\), \((28.5,\,10)\), \((31.5,\,25)\), \((34.5,\,46)\), \((37.5,\,56)\) and \((40.5,\,60)\), taking the horizontal axis as the marks and the vertical axis as the cumulative frequency. Join the points with a smooth increasing S-shaped curve. The estimates below are read from this curve.

(b) Estimates from the curve

(i) 80th percentile

The position on the vertical axis is

\[ \frac{80}{100}\times 60 = 48. \]

This falls in the class \(35\text{-}37\) (boundaries \(34.5\) to \(37.5\), frequency \(10\), with cumulative frequency \(46\) reached at \(34.5\)). By interpolation,

\[ P_{80} = 34.5 + \left(\frac{48 - 46}{10}\right)\times 3 = 34.5 + 0.6 = 35.1. \]

So the 80th percentile is approximately \(\mathbf{35.1}\) marks.

(ii) Median

The median is at position

\[ \frac{N}{2} = \frac{60}{2} = 30, \]

which lies in the class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) reached at \(31.5\)). Thus

\[ \text{Median} = 31.5 + \left(\frac{30 - 25}{21}\right)\times 3 = 31.5 + \frac{15}{21} \approx 31.5 + 0.71 = 32.2. \]

The median is approximately \(\mathbf{32.2}\) marks.

(iii) Semi-interquartile range

Lower quartile \(Q_1\) is at position \(\dfrac{N}{4} = 15\), lying in class \(29\text{-}31\) (boundaries \(28.5\) to \(31.5\), frequency \(15\), cumulative frequency \(10\) at \(28.5\)):

\[ Q_1 = 28.5 + \left(\frac{15 - 10}{15}\right)\times 3 = 28.5 + 1 = 29.5. \]

Upper quartile \(Q_3\) is at position \(\dfrac{3N}{4} = 45\), lying in class \(32\text{-}34\) (boundaries \(31.5\) to \(34.5\), frequency \(21\), cumulative frequency \(25\) at \(31.5\)):

\[ Q_3 = 31.5 + \left(\frac{45 - 25}{21}\right)\times 3 = 31.5 + \frac{60}{21} \approx 31.5 + 2.86 = 34.4. \]

Therefore the semi-interquartile range is

\[ \frac{Q_3 - Q_1}{2} = \frac{34.4 - 29.5}{2} = \frac{4.9}{2} \approx 2.4. \]

The semi-interquartile range is approximately \(\mathbf{2.4}\) marks.