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(a) The lengths of $40$ leaves of a plant are measured, correct to the nearest millimetre and data obtained is represented in the following table :
Find the median length (in mm) of the leaves.
OR
(b) A class teacher has the following absentees record of $30$ students of a class.
If the mean number of days a student was absent is $12$, find the values of $x$ and $y$.
Find the median length (in mm) of the leaves.
OR
(b) A class teacher has the following absentees record of $30$ students of a class.
If the mean number of days a student was absent is $12$, find the values of $x$ and $y$.
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Solution: (a) Correct table:
median class: $140 - 160$
Median $= l + \frac{\frac{N}{2} - cf}{f} \times h = 140 + \frac{20 - 17}{12} \times 20 = 145$
$\therefore$ The median length of the leaves is $145\text{ mm}$
OR
(b) Correct Table:
$x + y + 20 = 30 \Rightarrow x + y = 10$ ---(i)
$12 = \frac{224 + 10x + 22y}{30} \Rightarrow 5x + 11y = 68$ ---(ii)
Solving (i) and (ii) we get $x = 7, y = 3$
median class: $140 - 160$
Median $= l + \frac{\frac{N}{2} - cf}{f} \times h = 140 + \frac{20 - 17}{12} \times 20 = 145$
$\therefore$ The median length of the leaves is $145\text{ mm}$
OR
(b) Correct Table:
$x + y + 20 = 30 \Rightarrow x + y = 10$ ---(i)
$12 = \frac{224 + 10x + 22y}{30} \Rightarrow 5x + 11y = 68$ ---(ii)
Solving (i) and (ii) we get $x = 7, y = 3$