If the median I.Q of the students is 135, find the missing frequencies f1 and f2 in the following frequency distribution for 220 students:
I.Q 100-110 110-120 120-130 130-140 140-150 150-160 160-170 170-180
Number of students 15 35 f1 f2 33 23 19 8

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Solution

CLASS INTERVALS FREQUENCY CUMULATIVE FREQUENCY

100-110 15 15

110-120 35 50

120-130 f₁ 50+f₁

130-140 f₂ 50+f₁+f₂

140-150 33 83+f₁+f₂

150-160 23 106+f₁+f₂

160-170 19 125+f₁+f₂

170-180 8 133+f₁+f₂

N=133+f₁+f₂

$Now, N = 220 \Rightarrow 133 + f_{1} + f_{2} = 220 \Rightarrow f_{1} + f_{2} = 87 . . . . (1) Now, median is 135 . So, median class = 130 - 140 lower limit of median class, l = 130 cumulative frequency of class preceeding median class, F = 50 + f_{1} frequency of median class, f = f_{2} class size, h = 10 Median = l + [\frac{\frac{N}{2} - F}{f}] \times h \Rightarrow 135 = 130 + \frac{110 - 50 - f_{1}}{f_{2}} \times 10 \Rightarrow 2 f_{1} + f_{2} = 120 . . . . . (2) Subtracting (1) from (2), we get f_{1} = 33 Now, from (1), f_{2} = 54$

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154

Classes	$120 - 130$	$130 - 140$	$140 - 150$	$150 - 160$	$160 - 170$	$170 - 180$
Frequency	$2$	$8$	$12$	$f$	$8$	$7$

\begin{matrix} H e i g h t (i n c m) & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 & 150 - 160 & 160 - 170 & 170 - 180 N o . o f g o a l s s c o r e d & 8 & 10 & 15 & 9 & 12 & 14 & 13 & 10 \end{matrix}

f_{1}

f_{2}

145

Class	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$	$180 - 200$	Total
Frequency	$10$	$f_{1}$	$f_{2}$	$15$	$5$	$80$

Class Intervals	$110 - 119$	$120 - 129$	$130 - 139$	$140 - 149$	$150 - 159$	$160 - 169$
Frequency	$6$	$8$	$15$	$10$	$6$	$5$

Which of the given less than cumulative frequency curve (ogive) correctly represents the following distribution?

$\begin{matrix} Height & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 & 150 - 160 Number of students & 7 & 9 & 12 & 10 & 4 \end{matrix}$

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