Latika Leekha
Last Activity: 9 Years ago
Hello student,
First of all, let us tabulate the data in a systematic form:
Class Interval f cf
161-167 79 79
167-173 92 171
173-179 60 231
179-185 22 253
185-191 5 258
191-197 2 260
Q1 = 161 + 6/79 .(65-0)
= 165.93
Q3 = 161 + 6/79 .(65-0)
= 165.93
D2 is given by 52nd value lying in C.I 161-167 and D9 lies in C.I. 179-185.
So, D2 = 161 + 6/79 .(52-0)
= 164.95
D9 = 179 + 6/22 .(234-231)
= 179.82
Similarly,
P45 = 167 + 6/92 .(117-79-0)
= 169.47
P57 = 167 + 6/92 .(148-79)
= 171.5