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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 :
begin{tabular}{|c|c|}
hline Length in (mm) & Number of leaves
hline 100 - 120 & 8
hline 120 - 140 & 9
hline 140 - 160 & 12
hline 160 - 180 & 5
hline 180 - 200 & 6
hline
end{tabular}
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.
begin{tabular}{|c|c|c|c|c|c|c|}
hline Number of days & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 & 20 - 24
hline Number of Absent students & 1 & 8 & x & 6 & 5 & y
hline
end{tabular}
If the mean number of days a student was absent is $12$, find the values of $x$ and $y$.
begin{tabular}{|c|c|}
hline Length in (mm) & Number of leaves
hline 100 - 120 & 8
hline 120 - 140 & 9
hline 140 - 160 & 12
hline 160 - 180 & 5
hline 180 - 200 & 6
hline
end{tabular}
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.
begin{tabular}{|c|c|c|c|c|c|c|}
hline Number of days & 0 - 4 & 4 - 8 & 8 - 12 & 12 - 16 & 16 - 20 & 20 - 24
hline Number of Absent students & 1 & 8 & x & 6 & 5 & y
hline
end{tabular}
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)
begin{tabular}{|c|c|c|}
hline C.I. & f & CF
hline 100 - 120 & 8 & 8
hline 120 - 140 & 9 & 17
hline 140 - 160 & 12 & 29
hline 160 - 180 & 5 & 34
hline 180 - 200 & 6 & 40
hline & 40 &
hline
end{tabular}
Correct table: [2 marks]
median class: $140 - 160$
$\text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h$
$= 140 + \frac{20 - 17}{12} \times 20$ [2 marks]
$= 145$ [1 mark]
$\therefore \text{The median length of the leaves is } 145$ mm
OR
(b)
begin{tabular}{|c|c|c|c|}
hline C.I. & $f_i$ & $x_i$ & $f_i x_i$
hline 0 - 4 & 1 & 2 & 2
hline 4 - 8 & 8 & 6 & 48
hline 8 - 12 & x & 10 & 10x
hline 12 - 16 & 6 & 14 & 84
hline 16 - 20 & 5 & 18 & 90
hline 20 - 24 & y & 22 & 22y
hline & $20 + x + y$ & & $224 + 10x + 22y$
hline
end{tabular}
Correct table: [2 marks]
$x + y + 20 = 30 \Rightarrow x + y = 10$ ---(i) [1 mark]
$12 = \frac{10x + 22y + 224}{30} \Rightarrow 5x + 11y = 68$ ---(ii) [1 mark]
Solving (i) and (ii) we get
$x = 7$ [1/2 mark]
$y = 3$ [1/2 mark]
begin{tabular}{|c|c|c|}
hline C.I. & f & CF
hline 100 - 120 & 8 & 8
hline 120 - 140 & 9 & 17
hline 140 - 160 & 12 & 29
hline 160 - 180 & 5 & 34
hline 180 - 200 & 6 & 40
hline & 40 &
hline
end{tabular}
Correct table: [2 marks]
median class: $140 - 160$
$\text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h$
$= 140 + \frac{20 - 17}{12} \times 20$ [2 marks]
$= 145$ [1 mark]
$\therefore \text{The median length of the leaves is } 145$ mm
OR
(b)
begin{tabular}{|c|c|c|c|}
hline C.I. & $f_i$ & $x_i$ & $f_i x_i$
hline 0 - 4 & 1 & 2 & 2
hline 4 - 8 & 8 & 6 & 48
hline 8 - 12 & x & 10 & 10x
hline 12 - 16 & 6 & 14 & 84
hline 16 - 20 & 5 & 18 & 90
hline 20 - 24 & y & 22 & 22y
hline & $20 + x + y$ & & $224 + 10x + 22y$
hline
end{tabular}
Correct table: [2 marks]
$x + y + 20 = 30 \Rightarrow x + y = 10$ ---(i) [1 mark]
$12 = \frac{10x + 22y + 224}{30} \Rightarrow 5x + 11y = 68$ ---(ii) [1 mark]
Solving (i) and (ii) we get
$x = 7$ [1/2 mark]
$y = 3$ [1/2 mark]