find median of the following data class interval 130-139 , 140 -149,150-159,160-169,170-179,180-189,190-199 frequency 4,9,18, 28,24,10,7,
Answers
Given:
Class interval and frequency:
Class interval 130-139 140 -149 150-159 160-169 170-179 180-189 190-199
frequency 4 9 18 28 24 10 7
To find:
The median of the data given:
Solution:
Class interval 130-139 140 -149 150-159 160-169 170-179 180-189 190-199
Continuous C.I 129.5-139.5 139.5-149.5 149.5-159.5 159.5-169.5 169.5-179.5 179.5-189.5 189.5-199.5
frequency 4 9 18 28 24 10 7
Cumulative freq. 4 13 31 59 83 93 100
N/2 = 100/50 = 50 lies in cumulative frequency 59, so 160-169 is the median class.
Lower limit of median = 159.5
And the height of the class is 10.
Preceding median of class frequency = 31.
Median = 159.5 + ( (50 - 31)/ 28)*10
= 159.5 + 0.679*10
≈ 166.29
Therefore the median is 166.29.
Answer:
Step-by-step explanation:
Given data
C.I 130−139 140−149 150−159 160−169 170−179 180−189 190−199
Frequency 4 9 18 28 24 10 7
Now we can prepare a table or calculating median
C.I Continous C.I Frequency Cumulative
130−139 129.5−139.5 4 4
140−149 139.5−149.5 9 4+9=13
150−159 149.5−159.5 18 13+18=31cf
160−169 159.5−169.5 28(f) 31+28=59
170−179 169.5−179.5 24 59+24=83
180−189 179.5−189.5 10 83+10=93
190−199 189.5−199.5 7 93+7=100
Here
N=100
⇒N/2= 50
30 median class 159.5−169.5
Because the cf(59) is near to (50)
∴ Lower limit of median
Class(l)=159.5
Class width(h)=10,
cf=preceding cf of median class
f=frequency of median class
∴ Median=l+( N/2−cf/f) ×h
=159.5+( 50−31/28 )×10
=159.5+ (19/28) ×10
=159.5+ 190/28
=159.5+6.7
Median=166.2