It’s a little complicated but here’s how it works: Imagine a table with the intervals 0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals) Then we have different rows Class width: 4 , 2 , 1 , 3 , 3 Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3 So now calculate frequency where freq = class width * density Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9 So to find median find cumulative frequency (Add all freq) Cfreq = 8.4 now divide by 2 = 4.2 So find the interval where 4.2 lies. 0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.