What are the mean, the median and the mode?

The mean is the total of all the values divided by how many values there are. The median is the middle value once the data has been put in numerical order. The mode is the value that occurs the greatest number of times, and it is the only one of the three that works for non-numerical data.

A median position works out as 24.5. What does that tell you, and what do you do next?

A position ending in .5 means there is an even number of values, so there is no single middle one. Take the halfway point between the 24th and 25th values, which is found by adding those two values and dividing by 2.

True or False? It is possible for a data set to have no mode at all.

True. If no value occurs more often than any other, then nothing occurs 'the most', so there is no mode. A data set can also have two modes, in which case it is described as bi-modal.

How do you find the mean of discrete data given in a frequency table?

Multiply each data value by its frequency, then add up all of those products to get the total of the data. Divide that total by the total frequency, which is the number of values altogether.

A frequency table holds 31 values and the median is the 16th. The first five rows hold 11 values between them, and the sixth row holds 6. Which row contains the median?

The sixth row. The first five rows account for values 1 to 11, so the sixth row holds values 12 to 17, and the 16th value falls inside that range.

A value larger than the mean is added to a data set. What happens to the mean?

The mean increases . The new value contributes more than the current average does, so it pulls the average up, while a value below the mean would pull it down and a value exactly equal to the mean would leave it unchanged.

Why can you only estimate the mean from a grouped frequency table?

Because the table does not contain the individual data values, so their exact total can never be found. Each class is represented by its midpoint instead, which assumes every value in that class sits at the centre of it.

What is the modal class of a grouped frequency table?

The modal class is the class interval with the highest frequency . Grouped data cannot have a single mode, because the individual values are not in the table and there is no way to say which one occurs most often, so a class is given instead.

What assumption does linear interpolation make about the data in a class?

That the values are spread evenly across the class. That is what lets a position part of the way through a class be turned into a value part of the way along it, and it is why the answer can only ever be an estimate.

True or False? A median found by linear interpolation is always one of the actual data values.

False. You are estimating a point along a class interval, not picking out a value from a list, so the answer usually is not a value that appears in the data at all. That is also why the odd-or-even question that arises with a list never comes up here.

Before you can work out how far into a class the median lies, what has to be found first?

The class that contains the median, worked out from the cumulative frequencies. Only once that class is known do its lower boundary, its frequency and its width become the numbers you need for the rest of the calculation.

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.

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Measures of Central Tendency (AQA GCSE Statistics: Foundation): Flashcards

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What are the mean, the median and the mode?

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What are the mean, the median and the mode?
The mean is the total of all the values divided by how many values there are.
The median is the middle value once the data has been put in numerical order.
The mode is the value that occurs the greatest number of times, and it is the only one of the three that works for non-numerical data.
Complete the rule for the position of the median in an ordered list of $n$ values.
$position of the median = \frac{n +______}{______}$
The completed rule is:
$position of the median = \frac{n + 1}{2}$
The data must be put in numerical order before this position means anything.
A median position works out as 24.5. What does that tell you, and what do you do next?
A position ending in .5 means there is an even number of values, so there is no single middle one.
Take the halfway point between the 24th and 25th values, which is found by adding those two values and dividing by 2.
True or False?
It is possible for a data set to have no mode at all.
True.
If no value occurs more often than any other, then nothing occurs 'the most', so there is no mode.
A data set can also have two modes, in which case it is described as bi-modal.
How do you find the mean of discrete data given in a frequency table?
Multiply each data value by its frequency, then add up all of those products to get the total of the data.
Divide that total by the total frequency, which is the number of values altogether.
A frequency table of shoe sizes gives a total of value times frequency of 278.5, and a total frequency of 31. What is the mean shoe size?
Divide the total by the total frequency, giving $\frac{278.5}{31} = 8.98$ to 3 significant figures.
Notice that a mean does not have to be a value that could actually occur, since there is no such thing as a shoe size of 8.98.
A frequency table holds 31 values and the median is the 16th. The first five rows hold 11 values between them, and the sixth row holds 6. Which row contains the median?
The sixth row.
The first five rows account for values 1 to 11, so the sixth row holds values 12 to 17, and the 16th value falls inside that range.
The mean of 30 test scores is 62. What is the total of all the scores?
Multiply the mean by the number of values, giving $62 \times 30 = 1860$ marks in total.
The mean formula links three quantities, so knowing any two of them gives you the third.
A class of 24 students has a mean height of 1.56 metres. Two more students of equal height join, and the mean rises to 1.58 metres. How tall are they?
The original total is $1.56 \times 24 = 37.44$ metres and the new total must be $1.58 \times 26 = 41.08$ metres.
The two new students account for the difference, so they contribute $41.08 - 37.44 = 3.64$ metres between them, giving 1.82 metres each.
A value larger than the mean is added to a data set. What happens to the mean?
The mean increases.
The new value contributes more than the current average does, so it pulls the average up, while a value below the mean would pull it down and a value exactly equal to the mean would leave it unchanged.
Why can you only estimate the mean from a grouped frequency table?
Because the table does not contain the individual data values, so their exact total can never be found.
Each class is represented by its midpoint instead, which assumes every value in that class sits at the centre of it.
How do you find the midpoint of the class $3.5 \leq w < 4$ ?
Add the two endpoints and divide by 2:
$\frac{3.5 + 4}{2} = \frac{7.5}{2} = 3.75$
A midpoint does not have to be a whole number, or even a value that could actually occur.
Complete the formula for an estimate of the mean from grouped data.
$estimated mean = \frac{total of (midpoint \times______)}{total______}$
The completed formula is:
$estimated mean = \frac{total of (midpoint \times frequency)}{total frequency}$
Adding a midpoint column and a midpoint times frequency column to the table is the quickest way to organise the working.
A grouped table of 20 puppy weights gives a total of midpoint times frequency of 87. What is the estimated mean weight?
Divide the total by the total frequency, giving $\frac{87}{20} = 4.35$ kilograms.
This is an estimate rather than the true mean, so it should never be presented as the exact mean weight of the puppies.
What is the modal class of a grouped frequency table?
The modal class is the class interval with the highest frequency.
Grouped data cannot have a single mode, because the individual values are not in the table and there is no way to say which one occurs most often, so a class is given instead.
True or False?
The highest frequency in a grouped table is 34, for the class $40 \leq x < 50$ , so the modal class is 34.
False.
The modal class is $40 \leq x < 50$ , which is the class interval itself.
34 is that class's frequency, and quoting the frequency instead of the interval is the commonest mistake made on this question.
What assumption does linear interpolation make about the data in a class?
That the values are spread evenly across the class.
That is what lets a position part of the way through a class be turned into a value part of the way along it, and it is why the answer can only ever be an estimate.
Complete the rule for locating the median of grouped data.
The median lies in the class holding the value in position $\frac{n}{______}$ , and that class is found by running down the $______$ frequencies.
The completed rule is:
The median lies in the class holding the value in position $\frac{n}{2}$ , and that class is found by running down the cumulative frequencies.
True or False?
A median found by linear interpolation is always one of the actual data values.
False.
You are estimating a point along a class interval, not picking out a value from a list, so the answer usually is not a value that appears in the data at all.
That is also why the odd-or-even question that arises with a list never comes up here.
Before you can work out how far into a class the median lies, what has to be found first?
The class that contains the median, worked out from the cumulative frequencies.
Only once that class is known do its lower boundary, its frequency and its width become the numbers you need for the rest of the calculation.
A class contains the 23rd to the 34th data values, and the median is the 25th. How far into the class does the median lie?
The 25th value is the 3rd one inside the class, counting 23, 24 and 25, and the class holds 12 values altogether.
So the median sits $\frac{3}{12} = \frac{1}{4}$ of the way into the class.
The median lies a quarter of the way into the class $20 < x \leq 30$ hours. What is the estimated median?
The class width is $30 - 20 = 10$ , and a quarter of that is $10 \times \frac{1}{4} = 2.5$ .
Add that on to the lower boundary of the class, giving an estimated median of $20 + 2.5 = 22.5$ hours.
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.

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