True or False? Putting data into a frequency table loses the individual data values.

False. An ungrouped frequency table keeps every value: it records how many times each individual value occurred, so exact averages, ranges and summary statistics can still be calculated from it. It is grouping the data that loses the values. Once a table records only that eight items fell between 240 and 260, those eight values are gone, and every statistic calculated from it is an estimate .

For grouped data you give the modal class rather than the mode. Why can you not give a mode?

Grouping the data throws away the individual values, so there is no longer any way to tell which single value occurred most often. What survives is which class the data is most concentrated in, so that is what you report. The same limitation applies to the mean and the median from a grouped table: everything you can calculate is an estimate.

International A Level (IAL)MathsEdexcelStatistics 1FlashcardsData Presentation & InterpretationStatistical Measures

Statistical Measures (Edexcel International A Level (IAL) Maths: Statistics 1): Flashcards

Exam code: YMA01

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FrontFrequency Tables
True or False?
Putting data into a frequency table loses the individual data values.
BackFrequency Tables
False.
An ungrouped frequency table keeps every value: it records how many times each individual value occurred, so exact averages, ranges and summary statistics can still be calculated from it.
It is grouping the data that loses the values. Once a table records only that eight items fell between 240 and 260, those eight values are gone, and every statistic calculated from it is an estimate.
BackFrequency Tables
The median is the $\frac{n + 1}{2}$ th value, where $n$ is the total frequency.
Add the frequencies up in order until the running total first reaches that position, and the median is the data value you are on when it does.
So, for example, with $n = 60$ the median is the 30.5th value, meaning halfway between the 30th and the 31st; grouped data uses a different position, because the individual values are no longer known.

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Cards in this collection (14)

True or False?
Putting data into a frequency table loses the individual data values.
False.
An ungrouped frequency table keeps every value: it records how many times each individual value occurred, so exact averages, ranges and summary statistics can still be calculated from it.
It is grouping the data that loses the values. Once a table records only that eight items fell between 240 and 260, those eight values are gone, and every statistic calculated from it is an estimate.
For discrete data in an ungrouped frequency table, which position gives the median, and how do you find the value there?
The median is the $\frac{n + 1}{2}$ th value, where $n$ is the total frequency.
Add the frequencies up in order until the running total first reaches that position, and the median is the data value you are on when it does.
So, for example, with $n = 60$ the median is the 30.5th value, meaning halfway between the 30th and the 31st; grouped data uses a different position, because the individual values are no longer known.
Complete the formula for the mean of data given in a frequency table, where $x$ is a data value and $f$ is its frequency:
$\bar{x} = \frac{Σ______}{Σ______}$
The completed formula is:
$\bar{x} = \frac{Σ x f}{Σ f}$
Multiply each value by its frequency and add the results, then divide by the total frequency.
$Σ x f$ is the sum of all the data values, exactly as $Σ x$ would be if they were listed out one by one, and $Σ f$ is how many values there are.
A grouped frequency table for continuous data has the classes $10 \leq x \leq 19$ and $20 \leq x \leq 29$ . What is wrong with them, and what do you do about it?
There is a gap: nothing between 19 and 20 has a class to go in, and a continuous quantity can take those values.
Close the gap before doing any calculation: for data that has been rounded, the boundaries become
$9.5 \leq x < 19.5 and 19.5 \leq x < 29.5$
For data that has been truncated, such as an age counted in whole years, they become $10 \leq x < 20$ and $20 \leq x < 30$ instead, since someone is 19 right up until their 20th birthday.
For grouped data you give the modal class rather than the mode. Why can you not give a mode?
Grouping the data throws away the individual values, so there is no longer any way to tell which single value occurred most often.
What survives is which class the data is most concentrated in, so that is what you report.
The same limitation applies to the mean and the median from a grouped table: everything you can calculate is an estimate.
What do you use in place of the actual data values when estimating the mean from a grouped frequency table?
The midpoint of each class, which stands in for every value in that class; the midpoint is the mean of the class's lower and upper boundaries, so $150 \leq h < 155$ has midpoint 152.5.
This is why the answer is only an estimate: it assumes the values in each class average out at its midpoint, which they will not do exactly.
Watch for a class of a different width from the others, since its midpoint still has to be worked out from its own boundaries.
Complete the assumption that linear interpolation makes about grouped data:
The data values are assumed to be $______$ spread throughout each class.
The completed assumption is:
The data values are assumed to be evenly spread throughout each class.
That assumption is what makes the method work. If the values are evenly spread, then a position one third of the way through a class's frequency corresponds to a data value one third of the way through that class's range, so proportion can be used.
It is also why an interpolated median or quartile is an estimate: real data is rarely spread evenly.
$Q_{1}$ is the 6.25th value, and it lies in the class $155 \leq h < 160$ , which runs from a cumulative frequency of 3 to a cumulative frequency of 8. How does linear interpolation give $Q_{1}$ ?
Set the fraction of the way through the cumulative frequency equal to the fraction of the way through the class boundaries:
$\frac{6.25 - 3}{8 - 3} = \frac{Q_{1} - 155}{160 - 155}$
$\frac{3.25}{5} = \frac{Q_{1} - 155}{5}$
so $Q_{1} = 155 + 3.25 = 158.25$ .
The part that is easiest to forget is adding the lower class boundary back on at the end. The proportion gives you how far into the class the value sits, not the value itself.
Define coding a set of data.
Coding is applying the same formula to every value in a data set, to simplify the numbers you have to work with.
It is most useful when the data consists of very large or very small numbers, so coding such as $x = \frac{h - 250}{20}$ turns awkward values into small, convenient ones.
The coding must be applied to every value, or the coded data no longer represents the original.
Data $x$ is coded using $y = a x + b$ . What happens to the mean?
The mean is coded in exactly the same way as the data:
$\bar{y} = a \bar{x} + b$
The reason is that the mean is a measure of location.
Changing every value changes where the data sits, and the mean follows it: multiply everything by $a$ and the centre is multiplied by $a$ ; add $b$ to everything and the centre moves up by $b$ .
Data $x$ is coded using $y = a x + b$ . What happens to the standard deviation?
Only the multiplier affects it:
$σ_{y} = |a| σ_{x}$
Adding or subtracting $b$ makes no difference at all, because the standard deviation is a measure of spread: sliding every value along by the same amount leaves every gap between values exactly as it was.
Multiplying by $a$ does stretch those gaps, so it stretches the standard deviation with them.
Data $x$ was coded using $y = a x + b$ , and the mean of the coded data is $\bar{y}$ . Complete the way the original mean is recovered:
$\bar{x} = \frac{\bar{y} -______}{______}$
The completed formula is:
$\bar{x} = \frac{\bar{y} - b}{a}$
This is just $\bar{y} = a \bar{x} + b$ rearranged, which is worth doing rather than memorising: whatever the coding formula was, write it down with the means in it and solve for the one you want.
Undo the operations in reverse order, so subtract before dividing.
In $σ_{y} = |a| σ_{x}$ , why is the modulus of $a$ taken?
Because a standard deviation can never be negative: it measures how spread out data is, and a spread of less than nothing means nothing.
Coding with a negative multiplier, such as $y = - 3 x$ , reflects the data as well as stretching it, but reflecting does not change how spread out it is.
So the standard deviation is multiplied by the positive value 3, not by $- 3$ .
Heights $h$ , in cm, are coded using $x = \frac{h - 250}{20}$ , and the coded data has standard deviation 1.3102. What is the standard deviation of the heights?
Only the division by 20 has to be reversed:
$σ_{h} = 20 \times 1.3102 = 26.2 cm$
Subtracting 250 slid every height down the scale without changing how spread out they were, so it has no part in the calculation.
Note that the answer carries the units of the original data, cm, not of the coded values, which have no units at all.

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