The problem asks us to find the mean and coefficient of variation for the given data. The data represents sales in different ranges, along with their relative frequencies and actual frequencies.

1. Problem Description

The problem asks us to find the mean and coefficient of variation for the given data. The data represents sales in different ranges, along with their relative frequencies and actual frequencies.

2. Solution Steps

First, we calculate the midpoint ($x$) of each sales range. This is given in the table already.

Second, we multiply each midpoint ($x$) by its frequency ($f$) to get $fx$. This is also given in the table.

Third, we calculate the mean, which is the sum of the $fx$ values divided by the sum of the frequencies.

$\text{Mean} = \frac{\sum fx}{\sum f}$

From the image, $\sum fx = 697.5 + 1732.5 + 3150 + 4347.5 + 6532.5 + 8917.5 + 10525 + 11250 = 47152.5$.

The sum of the frequencies is $9+21+36+47+67+87+98+100 = 465$.

So,

$\text{Mean} = \frac{47152.5}{465} = 101.4032 \approx 101.40$

Next, we need to compute the standard deviation to find the coefficient of variation. To do that we need $\sum fx^2$.

$fx^2$:

$77.5^2 \times 9 = 54165.625$

$82.5^2 \times 21 = 143203.125$

$87.5^2 \times 36 = 275625$

$92.5^2 \times 47 = 402453.125$

$97.5^2 \times 67 = 635859.375$

$102.5^2 \times 87 = 911421.875$

$107.5^2 \times 98 = 1131612.5$

$112.5^2 \times 100 = 1265625$

$\sum fx^2 = 54165.625 + 143203.125 + 275625 + 402453.125 + 635859.375 + 911421.875 + 1131612.5 + 1265625 = 4819965.625$

The variance is given by

$\sigma^2 = \frac{\sum fx^2}{\sum f} - (\frac{\sum fx}{\sum f})^2 = \frac{\sum fx^2}{n} - \mu^2$

$\sigma^2 = \frac{4819965.625}{465} - (101.4032)^2 \approx 10365.517 - 10282.608 = 82.909$

$\sigma = \sqrt{82.909} = 9.105$

The coefficient of variation (CV) is the ratio of the standard deviation to the mean, expressed as a percentage:

$CV = \frac{\sigma}{\mu} \times 100 = \frac{9.105}{101.40} \times 100 = 8.979\%$

3. Final Answer

Mean = 101.40

Coefficient of Variation = 8.98%