Question

If a is the missing frequency such that 1 < a < 8, then the median of the given frequency distribution cannot be

Class interval	Frequency
100 - 110	12
110 - 120	6
120 - 130	10
130 - 140	a
140 - 150	8
150 - 160	12

• A. 126
• B. 128
• C. 129
• D. 127

Answer 1

The correct option is A 126

Class interval	Frequency	Cumulative Frequency
100 -110	12	12
110 - 120	6	18
120 -130	10	28
130 - 140	a	28 + a
140 -150	8	36+ a
150 -160	12	48 + a

Thus the total number of observations (N) is 48 + a
$\therefore \frac{N}{2}=\frac{48+a}{2}=24+\frac{a}{2}$ ........(i)
As, $1<a<8$
$\Rightarrow 0.5<\frac{a}{2}<4$
$\Rightarrow 24+0.5<24+\frac{a}{2}<24+4$
$\Rightarrow 24.5<24+\frac{a}{2}<28$
$\Rightarrow 24.5<\frac{N}{2}<28$ [from (i)]
Thus, the median lies in the class 120 - 130.
Here,
Lower limit, (l) = 120
Cumulative frequency of class preceding the median class, (cf) = 18
Frequency of the median class (f) = 10
Class size (h) = 10
$\therefore$ Median = $l+\frac{\frac{N}{2}-cf}{f}\times h$
$=120+\frac{24+\frac{a}{2}-18}{10}\times 10$
$=120+6+\frac{a}{2}$
$=126+\frac{a}{2}$
$\therefore 126.5<126+\frac{a}{2}<130(\therefore 1<a<8)$

$\Rightarrow 126.5<Median<130$
Thus, the median of the given frequency distribution cannot be 126.
Hence, the correct answer is option a.