When is the median a better average than the mean?

When the data contains one or two unusually high or low values. Those extremes pull the mean away from what is typical, but the median is simply the middle value and is barely affected by them.

Which average would you use for people's favourite ice cream flavour, and why?

The mode , because it is the only one of the three averages that works for qualitative data. Flavours are names rather than numbers, so they cannot be added up or placed in a meaningful numerical order.

True or False? A data set can have two modes, or none at all.

True. If two values occur most often and equally often, the data set has two modes and is described as bi-modal . If no value occurs more often than any other, there is no mode at all.

A value larger than the mean is added to a data set. What happens to the mean, and what happens instead if such a value is removed?

Adding a value larger than the mean increases it, because the new value pulls the average upwards. Removing a value larger than the mean decreases it, for the same reason in reverse.

The median of a frequency table is the 16th value. How do you find which data value that actually is?

Add the frequencies row by row to build a running total. The first row whose running total reaches 16 is the row containing the 16th value, so that row's data value is the median.

Why can you only ever estimate the mean of grouped data?

Because the original data values are not in the table, only the class that each one fell into. Without the actual values there is no way to find their exact total, so the mean has to be estimated.

Grouped data is presented as a histogram and you need to estimate the mean. What should you do first?

Rewrite the histogram as a grouped frequency table. Read the class intervals off the horizontal axis and work out each class's frequency from its bar, then carry on with midpoints as usual.

True or False? An estimated mean from grouped data can be a value that never appears in the data.

True. An average is a calculated figure rather than one of the data values, so it need not be a value that actually occurred. With grouped data that is doubly so, because the calculation uses class midpoints rather than the real values at all.

What assumption does linear interpolation make about the data?

That the values are spread evenly across the class containing the median. That is almost certainly not exactly true of real data, which is why the result can only ever be an estimate.

Why does transforming data make averages easier to calculate by hand?

Because it replaces awkward numbers with much simpler ones. Values such as 1001.2 and 1002.4 become 1.2 and 2.4 by subtracting 1000, and then 12 and 24 by multiplying by 10, which are far quicker to add up.

You subtracted 1000 from each value and then multiplied by 10. How do you get back to the original mean?

Undo the transformations in reverse order : divide the mean by 10 first, and only then add 1000. Doing them the other way round gives the wrong answer, because the two operations were applied in a particular order.

Define weighted mean .

A weighted mean is an average in which some of the values count for more than others. It is used when the values differ in statistical importance, such as exam papers that carry different proportions of a final mark.

A company's profits rise by 1%, 7%, 3% and 12% over four years. Which numbers go into the geometric mean?

The percentage change multipliers , not the percentages themselves, so 1.01, 1.07, 1.03 and 1.12. A 2% rise would be 1.02 and a 2% fall would be 0.98, because the geometric mean is for numbers that get multiplied together.

The fourth root of four annual growth multipliers comes to 1.056671. What does that number tell you?

That a rise of about 5.67% every year would produce exactly the same total growth over the four years. The geometric mean is the single percentage change that could stand in for all of them.

True or False? The geometric mean should be used whenever you are averaging percentages.

False. The geometric mean is for values that get multiplied together, such as year-on-year growth multipliers. For percentages that are simply being averaged, such as three test scores, the ordinary or weighted mean is the right one.

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Measures of Central Tendency (Edexcel GCSE Statistics: Higher): Flashcards

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FrontMode, Median & Arithmetic Mean
When is the median a better average than the mean?

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Cards in this collection (36)

When is the median a better average than the mean?
When the data contains one or two unusually high or low values.
Those extremes pull the mean away from what is typical, but the median is simply the middle value and is barely affected by them.
Which average would you use for people's favourite ice cream flavour, and why?
The mode, because it is the only one of the three averages that works for qualitative data.
Flavours are names rather than numbers, so they cannot be added up or placed in a meaningful numerical order.
Complete the rule for the position of the median in a list of $n$ values written in order.
$the median is in position \frac{n +______}{______}$
The completed rule is:
$the median is in position \frac{n + 1}{2}$
With 35 values that gives position 18, and with an even number of values it lands halfway between two of them.
True or False?
A data set can have two modes, or none at all.
True.
If two values occur most often and equally often, the data set has two modes and is described as bi-modal.
If no value occurs more often than any other, there is no mode at all.
A value larger than the mean is added to a data set. What happens to the mean, and what happens instead if such a value is removed?
Adding a value larger than the mean increases it, because the new value pulls the average upwards.
Removing a value larger than the mean decreases it, for the same reason in reverse.
A set of five numbers has a mean of 7, and four of them are 3, 6, 8 and 10. What is the fifth number?
The five numbers must total $7 \times 5 = 35$ altogether, and the four known ones come to 27.
So the missing number is $35 - 27 = 8$ exactly.
A class of 24 has a mean height of 1.56 m. Two more students of equal height join and the mean rises to 1.58 m. Find their height.
Turn both means back into totals: $1.56 \times 24 = 37.44$ metres before, and $1.58 \times 26 = 41.08$ metres after.
The two students account for $41.08 - 37.44 = 3.64$ metres between them, so each of them is 1.82 m tall.
How do you find the mean from a table of data values and their frequencies?
Multiply each data value by its frequency and add the results, which totals up all the data values at once.
Then divide by the total of the frequency column, since that is how many values there are: $\bar{x} = \frac{\sum f x}{\sum f}$ for the whole table.
The median of a frequency table is the 16th value. How do you find which data value that actually is?
Add the frequencies row by row to build a running total.
The first row whose running total reaches 16 is the row containing the 16th value, so that row's data value is the median.
Why can you only ever estimate the mean of grouped data?
Because the original data values are not in the table, only the class that each one fell into.
Without the actual values there is no way to find their exact total, so the mean has to be estimated.
Which value stands in for the data in each class when estimating a mean?
The midpoint of the class interval, found by adding the two endpoints and halving.
For the class $150 \leq x < 160$ every value in it is treated as 155 in the calculation.
Complete the sentence about grouped data.
For grouped data you cannot find the mode, so you give the modal $______$ instead, which is the class with the highest $______$ in the table.
The completed sentence is:
For grouped data you cannot find the mode, so you give the modal class instead, which is the class with the highest frequency in the table.
The highest frequency in a grouped table is 34, belonging to the class $40 \leq x < 50$ . What is the modal class?
The modal class is $40 \leq x < 50$ itself, the interval rather than the frequency.
Writing 34 is a common error: 34 is how many values fall in the modal class, not the class.
A grouped table gives the total of the midpoint times frequency column as 87, with 20 values altogether. What is the estimated mean?
The estimated mean is $\frac{87}{20} = 4.35$ for that data.
That column total stands in for the sum of all the data values, so dividing by the total frequency gives the mean in the ordinary way.
Grouped data is presented as a histogram and you need to estimate the mean. What should you do first?
Rewrite the histogram as a grouped frequency table.
Read the class intervals off the horizontal axis and work out each class's frequency from its bar, then carry on with midpoints as usual.
True or False?
An estimated mean from grouped data can be a value that never appears in the data.
True.
An average is a calculated figure rather than one of the data values, so it need not be a value that actually occurred.
With grouped data that is doubly so, because the calculation uses class midpoints rather than the real values at all.
What assumption does linear interpolation make about the data?
That the values are spread evenly across the class containing the median.
That is almost certainly not exactly true of real data, which is why the result can only ever be an estimate.
Which data position do you find first when interpolating a median?
The value in position $\frac{n}{2}$ counting from the start, where $n$ is the total frequency.
Build up the cumulative frequencies row by row until you find which class interval that position falls in.
The class containing the median holds the 23rd to 34th values, and you want the 25th. How far into the class does it lie?
Three values in, counting 23, 24 and 25, out of the 12 values that the class holds altogether.
So the median sits $\frac{3}{12} = \frac{1}{4}$ of the way through that class interval.
The median lies a quarter of the way into the class $20 < x \leq 30$ . What is the estimated median?
The class is 10 wide, so a quarter of the way in is $10 \times \frac{1}{4} = 2.5$ along it.
Add that to the lower boundary of the class, giving an estimated median of $20 + 2.5 = 22.5$ in total.
Complete the formula for an estimated median from grouped data.
$estimated median = L + \frac{\frac{n}{2} -______}{______} \times w$
The completed formula is:
$estimated median = L + \frac{\frac{n}{2} - C}{f} \times w$
Here $C$ is the cumulative frequency of everything before the median's class, and $f$ is that class's own frequency.
On a histogram the first two classes hold 4 and 16 dolphins, and the median is the 30th of 60. Which class holds it, and how far in?
The first two classes account for 20, so the median lies in the third class, 10 values into it.
That class has a frequency density of 9.5, so the width needed is $\frac{10}{9.5} = 1.05$ to two decimal places.
Complete the sentence about transforming data.
If every data value has the same amount added to it, the mean, median and mode each $______$ by that same amount, and if every value is multiplied by the same number they are each $______$ by it.
The completed sentence is:
If every data value has the same amount added to it, the mean, median and mode each increase by that same amount, and if every value is multiplied by the same number they are each multiplied by it.
The mean of a data set is 14, and 20 is then added to every value. What is the new mean?
The new mean is $14 + 20 = 34$ exactly.
Adding the same amount to every value shifts the whole data set along the scale, so every average shifts with it by that amount.
Why does transforming data make averages easier to calculate by hand?
Because it replaces awkward numbers with much simpler ones.
Values such as 1001.2 and 1002.4 become 1.2 and 2.4 by subtracting 1000, and then 12 and 24 by multiplying by 10, which are far quicker to add up.
You subtracted 1000 from each value and then multiplied by 10. How do you get back to the original mean?
Undo the transformations in reverse order: divide the mean by 10 first, and only then add 1000.
Doing them the other way round gives the wrong answer, because the two operations were applied in a particular order.
A hen's eggs have a mean weight of 66.21 g, and a new feed increases every egg's weight by 3%. What is the new mean?
Multiply by the decimal multiplier 1.03, giving $66.21 \times 1.03 = 68.1963$ grams.
A percentage increase applied to every value is just a multiplication, so the mean is multiplied by the same amount, giving 68.20 g to two decimal places.
Define weighted mean.
A weighted mean is an average in which some of the values count for more than others.
It is used when the values differ in statistical importance, such as exam papers that carry different proportions of a final mark.
Complete the formula for a weighted mean.
$weighted mean = \frac{\sum (value \times______)}{\sum______}$
The completed formula is:
$weighted mean = \frac{\sum (value \times weight)}{\sum weights}$
Myfanwy scores 64%, 60% and 75% on three papers weighted 25, 35 and 40. What is her final mark?
Multiply each score by its weight, add the results, and divide by the total weight of 100.
$\frac{(64 \times 25) + (60 \times 35) + (75 \times 40)}{100} = 67$
Three classes of 28, 25 and 27 students have mean scores of 64%, 68% and 65%. Why is the overall mean not just the mean of those three figures?
Because the classes are different sizes, so each class mean should count in proportion to the number of students behind it.
The class sizes act as the weights, giving $\frac{5247}{80} = 65.6$ percent to one decimal place.
Half the workers earn £460, 35% earn £600 and 15% earn £820. What plays the part of the weights here?
The percentages themselves are the weights, and they conveniently total 100.
That gives a mean weekly wage of $\frac{56300}{100} = 563$ pounds.
How do you find the geometric mean of three numbers?
Multiply the three numbers together and take the cube root of the result.
For a set of $n$ values you take the root of order $n$ , so for 2, 9 and 12 the geometric mean is $\sqrt[3]{216} = 6$ exactly.
A company's profits rise by 1%, 7%, 3% and 12% over four years. Which numbers go into the geometric mean?
The percentage change multipliers, not the percentages themselves, so 1.01, 1.07, 1.03 and 1.12.
A 2% rise would be 1.02 and a 2% fall would be 0.98, because the geometric mean is for numbers that get multiplied together.
The fourth root of four annual growth multipliers comes to 1.056671. What does that number tell you?
That a rise of about 5.67% every year would produce exactly the same total growth over the four years.
The geometric mean is the single percentage change that could stand in for all of them.
True or False?
The geometric mean should be used whenever you are averaging percentages.
False.
The geometric mean is for values that get multiplied together, such as year-on-year growth multipliers.
For percentages that are simply being averaged, such as three test scores, the ordinary or weighted mean is the right one.

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