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Ungrouped Frequency Table

The height of...

Question

The height of $30$ boys of a class are given in the following table :
Height in cm
Frequency
120 - 129
130 - 139
140 - 149
150 - 159
150 - 159
2
8
10
7
3
If by joining a boy of height $140$ cm, the median of the heights is changed from $M_{1}$ to $M_{2}$ , then $M_{1} - M_{2}$ , in cm is

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Solution

The correct option is C $0$
Height in
ems
Frequency
Cumulative
frequency
Actual class
limit
120-129
130-139
1401-149
150-159
160-169
n = 30
2
8
10
7
3
2
10
20
27
30
119.5-129.5
129.5-189.5
139.5-149.5
149.5-159.5
159.5-169.5
Here $n = 30$
$∴ \frac{n}{2} 1 = 15 + 1 = 16$
$∴$ 16 is under cumulative frequency 20. So median class be 140-149.
$L_{1} = 189.5, L_{2} = 149.5, f = 10, n = 30, c = 10$ .
$M e d i a n M_{1} =: L_{1} + \frac{L_{2} - L_{1}}{f} (\frac{n}{2} - c)$
$= 139.5 + \frac{10}{10} (15 - 10)$
$= 139.5 + \frac{10}{10} \times 5 = 144.5$
If by joining of a boy of height 140 ems, then
$n = 31, f = 11.$
$∴ M e d i a n M 2 = 139.5 + \frac{149.5 - 139.5}{11} (15.5 - 10)$
$= 139.5 + \frac{10}{11} \times 5.5$
$= 144.5$ cms
then $M_{1} - M_{2} = 144.5 - 144.5 = 0$

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Height in cm	Frequency
120 - 129 130 - 139 140 - 149 150 - 159 150 - 159	2 8 10 7 3

Height in ems	Frequency	Cumulative frequency	Actual class limit
120-129 130-139 1401-149 150-159 160-169 n = 30	2 8 10 7 3	2 10 20 27 30	119.5-129.5 129.5-189.5 139.5-149.5 149.5-159.5 159.5-169.5

The height of 30 boys of a class are given in the following table :Height in cmFrequency120 - 129130 - 139140 - 149150 - 159150 - 159281073If by joining a boy of height 140 cm, the median of the heights is changed from M1 to M2, then M1−M2, in cm is

The height of $30$ boys of a class are given in the following table :
Height in cm
Frequency
120 - 129
130 - 139
140 - 149
150 - 159
150 - 159
2
8
10
7
3
If by joining a boy of height $140$ cm, the median of the heights is changed from $M_{1}$ to $M_{2}$ , then $M_{1} - M_{2}$ , in cm is