0
$\begingroup$

In a frequency distribution the upper boundary of each class interval has a constant ratio to the lower boundary. Show that the Geometric Mean $g$ can be expressed as $$\log g=a+\dfrac{k}{n}\sum_{i=1}^r f_i(i-1)$$

where $a$ is the class mark of the first class, $k$ is the logarithm of the ratio between upper and lower boundary and $n=\sum_{i=1}^r i.$

My try:

$\displaystyle g=\left\{\left(\dfrac{l_i+u_i}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}=\left\{\left(\dfrac{l_i+Rl_i}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}=\left\{{l_i}^{f_i}\left(\dfrac{1+R}{2}\right)^{f_i}\right\}^{\dfrac{1}{\displaystyle\sum f_i}}$

Then $\log g={\dfrac{1}{\displaystyle\sum f_i}}\left[f_i\left(\log l_i+\log \left(\dfrac{1+R}{2}\right)\right)\right]$

I can't proceed further.

$\endgroup$
  • 2
    $\begingroup$ I've added the tag self-study. Normally you are expected to show your ideas so far, not just ask us to do your homework for you. $\endgroup$ – Nick Cox Mar 9 '14 at 9:32
  • $\begingroup$ Please see the tag wiki info. What have you tried? $\endgroup$ – Glen_b -Reinstate Monica Mar 9 '14 at 9:40
  • 1
    $\begingroup$ Is your expression for $n$ correct? $\endgroup$ – Glen_b -Reinstate Monica Mar 9 '14 at 9:54
  • 1
    $\begingroup$ Please explain what "$f_i$" means and tell us what $g$ is supposed to be the GM of. (I cannot find any interpretation of the "geometric mean" and of the $f_i$ that makes a formula like this generally true.) $\endgroup$ – whuber Mar 9 '14 at 23:34
  • 1
    $\begingroup$ $f_i$ is the frequency of the $i^{th}$ class, $g$ is the gm of the class marks considering frequencies $\endgroup$ – s.jan Mar 10 '14 at 15:30
2
$\begingroup$

Quite a few errors in the entire post.

First of all, it would have made sense to say $n=\sum_{i=1}^r f_i$, the total frequency being the total number of observations for weighted means.

The derivation is plain algebraic manipulation to simply arrive at a formula.

Based on comments, $g$ is the weighted geometric mean of the class marks and $f_i$ is the frequency of the $i$th class (This should be added to the post itself).

Let $$\frac{\text{Upper class boundary}}{\text{Lower class boundary}}=c\qquad\text{for all classes}$$

$$x_i=\text{class mark of the $i$th class},\quad i=1,2,\cdots,r$$

$$f_i=\text{frequency of the $i$th class},\quad i=1,2,\cdots,r$$

Denote the $i$th class by $[a_i,a_{i+1}],\quad i=1,2,\cdots,r$.

We are given that $$\frac{a_2}{a_1}=\frac{a_3}{a_2}=\cdots=\frac{a_{r+1}}{a_r}=c$$

Now,

\begin{align} \frac{a_3}{a_2}&=\frac{a_2}{a_1} \\\implies \frac{a_3}{a_2}+1&=\frac{a_2}{a_1}+1 \\\implies \frac{(a_2+a_3)/2}{(a_1+a_2)/2}&=\frac{a_2}{a_1} \\\implies \frac{x_2}{x_1}&=\frac{a_2}{a_1} \end{align}

Similar arguments lead to

$$\frac{x_2}{x_1}=\frac{x_3}{x_2}=\cdots=\frac{x_{r+1}}{x_r}=c$$

From this we have $$x_i=x_1c^{\,i-1}\quad,i=1,2,\cdots,r$$

Thus,

\begin{align} g=\left(\prod_{i=1}^rx_i^{\,f_i}\right)^{1/n}&=\left\{\prod_{i=1}^r\left(x_1c^{\,i-1}\right)^{f_i}\right\}^{1/n} \\&=\left\{\prod_{i=1}^r x_1^{\,f_i}\,c^{\,(i-1)f_i}\right\}^{1/n} \end{align}

On simplification we get

\begin{align} \log g&=\frac{1}{n}\log\left[x_1^{\,\sum_{i=1}^rf_i}\,c^{\,\sum_{i=1}^r(i-1)f_i}\right] \\&=\frac{1}{n}\left[\log x_1^n+\log c^{\,\sum_{i=1}^r(i-1)f_i}\right] \\&=\frac{1}{n}\left[n\log x_1+\left\{\sum_{i=1}^r(i-1)f_i\right\}\log c\right] \\&=\log x_1+\frac{\log c}{n}\sum_{i=1}^r(i-1)f_i \end{align}

This is the expression we can arrive at. Thus in the formula asked to be proved in the original post, the $a$ should be the logarithm of the class mark of the first class, and not just the class mark.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.