Averages and measures of spread — mean, median, mode, range

When estimating the mean of grouped data, multiply each frequency by the class MIDPOINT, not by the lower or upper class limit. For $20<t\le30$ the midpoint is $25$, so use $25$ (not $20$ or $30$) in $\sum fx$. Using a limit instead shifts every product and gives the wrong estimate.

After finding $\sum fx$, divide by $\sum f$ — the TOTAL FREQUENCY — to get the estimated mean. Dividing instead by the number of classes (e.g. $5$) or by the sum of the midpoints gives a meaningless value. The denominator is always how many data items there are, i.e. $\sum f$.

The class WIDTH plays no part in the estimated mean. To build $\sum fx$ you multiply each frequency by the class MIDPOINT, never by the width (e.g. $10$). Width matters for histograms (frequency density), not for the mean — mixing the two up is a common slip.

To combine two groups you cannot just add the two means and halve — that only works when the groups are the same size. Multiply each mean by its own count to get each total, add the totals, then divide by the overall count: $\dfrac{n_A\bar{x}_A+n_B\bar{x}_B}{n_A+n_B}$.

For a frequency table, the median is NOT the middle of the row of distinct values. Use the total frequency $n$: the median is the value at position $\tfrac{n+1}{2}$, found from a running (cumulative) total. With $20$ students the median is between the $10$th and $11$th items, not the middle column.

The range is a measure of SPREAD, $\text{largest}-\text{smallest}$; the mode is the most common value. They answer different questions, so do not give the mode when the range is asked (or vice versa). For $2,4,7,7,10$ the range is $10-2=8$ and the mode is $7$.

From a list or table the median is the VALUE in the middle position, not the position number itself — and it need not be one of the listed numbers. For $10$ items the median is the mean of the $5$th and $6$th values, so it can be a half-value like $6.5$.