7. The mean and median of a group of 25 observations are 143.144, and 147
respectively. A set of 6 observations is added to this data with values 132.125
130, 160, 165 and 157. Find mean and median for the combined group of 31
observations.
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Answer:
Here we are given the mid-values. So, should first find the upper and lower limits of the various classes. The difference between two consecutive values is h=125−115=10.
∴ Lower limit of a class = Mid-value- h/2, Upper limit = Mid-value + h/2.
Calculation of Median
Mid-value Class groups Frequency Cumulative Frequency
115 110-120 6 6
125 120-130 25 31
135 130-140 48 79
145 140-150 72 151
155 150-160 116 267
165 160-170 60 327
175 170-180 38 365
185 180-190 22 387
195 190-200 3 390
N=∑f
i
=390
We have,
N=390∴
2
N
=
2
390
=195
The cumulative frequency just greater than N/2 i.e.195 is 267 and the corresponding class is 150−160. So, 150−160 is the median class.
∴l=150,f=116,h=10,F=151
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