Question

Find the median for the following frequency distribution table:Class Intervals110-119 120-129 130-139 140-149 150-159 160-169 Frequency 6 8 15 10 6 5

Answer 1

Answer: Median = 136.83

Solution:

Sum of frequencies

$6+8+15+10+6+5 \\ \\ = 50$
So,

$\frac{n}{2} = 25$
Class-interval _____Frequency _CF

109.5 - 119.5 _________6______6

119.5- 129.5_________8______14

129.5 - 139.5_________15______29

139.5 - 149.5________10______39

149.5 - 159.5 ________6_______45

159.5- 169.5 _________5_______50

so just upper nearby class of n/2 is 129.5-139.5

So, 129.5-139.5 is median class

l= 129.5

cf= 14

f =15

h = 10

put all these values in the formula

$Median= l + ( \frac{ \frac{n}{2} - cf }{f} )h \\ \\ = 129.5 + ( \frac{25 - 14}{15})10 \\ \\ = 129.5+ \frac{22}{3} \\ \\ = 129.5 + 7.33 \\ \\ = 136.83$
Median = 136.83

Hope it helps you.

Answer 2

Answer:

Answer: Median = 136.83

Solution:

Sum of frequencies

So,

Class-interval _____Frequency _CF

109.5 - 119.5 _________6______6

119.5- 129.5_________8______14

129.5 - 139.5_________15______29

139.5 - 149.5________10______39

149.5 - 159.5 ________6_______45

159.5- 169.5 _________5_______50

so just upper nearby class of n/2 is 129.5-139.5

So, 129.5-139.5 is median class

l= 129.5

cf= 14

f =15

h = 10

put all these values in the formula

Median = 136.83

Hope it helps you.

Step-by-step explanation: