Question

SOLVE FOR ME QUESTION NUMBER 9 .
AND TAKE OUT THE CORRECT ANSWER ..

ANSWER IS 136.69​

Answer 1

Answer:

ANSWER IS 136.69​

Explanation:

Answer 2

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

Now,

Median =l+

f

2

N

−F

×h

⇒Median=150+

116

195−151

×10=153.80

PLEASE