We show how to calculate the relative percentage frequency for one range of values for  $\var{a[r]} \le X \lt \var{a[r+1]}$    - you can then check the rest.

Note that there were $\var{daysopen}$ days  in the year when sales took place. 

There were $\var{norm1[r]}$ days out of the  $\var{daysopen}$ when there were between $\var{a[r]}$ and $\var{a[r+1]}$ thousand pounds worth of sales (including  $\var{a[r]}$ thousand but not $\var{a[r+1]}$ thousand) .

Hence the relative frequency percentage for such sales is given by \[100 \times \frac{\var{norm1[r]}}{\var{daysopen}}\%=\var{rel[r]}\%\] to one decimal place.

As we have to find the median and the interquartile range it is a good idea to order the data and also to total up the data (for the mean) and find the total of the squares of the data (for the variance).

Note that from the above table:

$n=\var{m*n}$.

$\displaystyle  \sum x_i = \var{sum(r)}$ and 

$\displaystyle  \sum x^2_i = \var{sum(map(x^2,x,r))}$ .

The sample mean is $\bar{x}=\displaystyle \frac{ \sum x_i}{n}=\frac{\var{sum(r)}}{\var{m*n}}=\var{mean(r)}=\var{av}$ to 2 decimal places.

The sample deviation is the square root of the sample variance.

Sample variance:\[\begin{eqnarray*}\frac{1}{ n -1}\left(\sum x_i ^ 2 - n \bar{x} ^ 2\right)&=& \frac{1}{\var{m*n-1}}\left(\var{sum(map(x^2,x,r))}-\var{m*n}\times\var{mean(r)^2}\right)\\&=&\var{variance(r,true)}\end{eqnarray*}\] Note that we used the more accurate value $(\var{mean(r)})^2$ for $\bar{x}^2$. 

So the sample standard deviation = $\sqrt{\var{variance(r,true)}}=\var{std}$ to 2 decimal places.

The median is $\var{median(r)} $.

The lower quartile is : $\var{lquartile(r)}$.

The upper quartile is : $\var{uquartile(r)}$.

The interquartile range is the difference between these quartiles =$\var{uquartile(r)}-\var{lquartile(r)}=\var{uquartile(r)-lquartile(r)}$