Question

Compute the A.M., G.M., & H.M. from the following: Class 15-19 20-24 25-29 30-34 35-39 40-44 Frequency 13 32 4 42 58 51

Answer 1

Answer:

$refer \: to \: attachment$

Compute the A.M., G.M., & H.M. from the following: Class 15-19 20-24 25-29 30-34 35-39 40-44 Frequency 13 32 4 42 58 51

$\color{blue}{ Answered \ by \ Srijan}$

Answer 2

Given is a frequency distribution, Find its A.M, G.M, and H.M.

Explanation:  

• A.M stands for the arithmetic mean of the data is given by, $A.M=\frac{\sum f_ix_i}{N}$
• G.M stands for the geometric mean of the data is given by, $G.M=antilog(\frac{\sum f_ilog(x_i)}{N} )$  
• H.M stands for the harmonic mean of the data is given by, $H.M=\frac{N}{\sum \frac{f_i}{x_i} }$  
• Here, $f_i,x_i$ are the frequency and midpoint of each class and $N$ is the total frequencies for the data.
• Table for the frequency distribution is,
• $class\ \ \ \ \ \ \ x_i \ \ \ \ \ f_i\ \ \ \ f_ilog(x_i) \ \ \ \ \ \frac{f_i}{x_i} \\15-19\ \ \ \ 17\ \ \ \ 13\ \ \ \ \ \ 15.9\ \ \ \ \ \ \ \ \ 0.765 \\20-24\ \ \ \ 22\ \ \ \ 32\ \ \ \ \ \ 42.9\ \ \ \ \ \ \ \ \ 1.455\\ 25-29\ \ \ \ 27\ \ \ \ 4\ \ \ \ \ \ \ \ 5.7\ \ \ \ \ \ \ \ \ \ 0.148 \\30-34\ \ \ \ 32\ \ \ \ 42\ \ \ \ \ \ 63.2\ \ \ \ \ \ \ \ \ 1.313 \\ 35-39\ \ \ \ 37\ \ \ \ 58\ \ \ \ \ \ 90.9\ \ \ \ \ \ \ \ \ 1.568\\40-44\ \ \ \ 42\ \ \ \ 51\ \ \ \ \ \ 82.8\ \ \ \ \ \ \ \ \ 1.214$  
• hence, from the above table we get, $N=\sum f_i=200,\ \ \ \sum f_ilog(x_i)=301.4,\ \ \ \sum\frac{f_i}{x_i}=6.463$    
• Now from the above evaluations we get the required AM, GM, and HM as,    $->AM=\frac{\sum f_ix_i}{N}= \frac{6665}{200} =33.325 \\->GM=antilog(\frac{\sum f_ilog(x_i)}{N} )= 10^{1.507}=32.137\\->HM=\frac{N}{\sum \frac{f_i}{x_i} }=\frac{200}{6.463}=30.945$    
• The A.M, G.M, and H.M of the given frequency distribution are $33.325,\ 32.137, \ and\ 30.945$ respectively.