What do the lower and upper quartiles split off from a data set?

The lower quartile has a quarter of the data below it, and the upper quartile has three quarters below it. Together with the median they cut the ordered data into four equal parts.

You have 20 values in order and want the quartiles by the halves method. What are the two halves, and what do you do with each?

With an even number of values the first 10 form the lower half and the last 10 the upper half. The lower quartile is then the median of the lower half, and the upper quartile is the median of the upper half.

With an odd number of values, what happens to the median when finding quartiles?

It is left out of both halves. Everything below the median forms the lower half and everything above it forms the upper half, and the median itself belongs to neither.

For 20 values the lower quartile is at position 5.25, and the 5th and 6th values are 36 and 37. What is the lower quartile?

36.25 : divide the gap from 36 to 37 into four equal parts and take the first one. A position ending in .25 means a quarter of the way between the two values on either side of it.

True or False? The two methods for finding quartiles always give the same answer.

False. For the same 20 values the halves method gives a lower quartile of 36.5 while the position formula gives 36.25. Both are accepted as correct answers.

What do percentiles and deciles divide a data set into?

Percentiles divide it into 100 equal parts and deciles into 10. So the 30th percentile has 30 per cent of the data below it, and percentiles need not be whole numbers, such as the 2.5th percentile.

What does a measure of dispersion tell you that an average does not?

How spread out the data is, rather than what a typical value looks like. A small measure of spread means the values cluster closely around the average, and a large one means many of them lie far from it.

Germination times run from 5 to 23 days, giving a range of 18, but the interquartile range is only 2. What does that tell you?

That the 23 is an extreme value sitting far above the rest of the data. The range is stretched by that one value, so on its own it gives a badly misleading picture of how spread out the data really is.

Why is the interquartile range unaffected by extreme values?

Because it uses only the two quartiles, which sit a quarter and three quarters of the way through the data. The largest and smallest values lie outside that middle half altogether, so changing them does not move either quartile.

Define interpercentile range .

An interpercentile range is the difference between two named percentiles of a data set. You have to say which two, such as the 10th to 90th interpercentile range, which is the 90th percentile minus the 10th.

True or False? The standard deviation is measured in the same units as the data.

True. If the data is measured in seconds or in centimetres, the standard deviation comes out in seconds or centimetres too. Every measure of spread on this course shares that property.

True or False? Removing an outlier changes the median as much as it changes the mean.

False. The mean uses every value, so taking away an extreme one moves it noticeably. The median depends only on the middle position, so removing a single value from one end usually shifts it very little or not at all.

When can outliers be identified without calculating any boundaries?

When a value is so obviously out of line with the rest that inspection is enough, such as an age of 3 in a list of GCSE candidates. A boundary calculation is the more reliable method, because it gives a definite cut-off rather than a judgement.

The outlier boundaries for the number of eggs in a nest work out as 14.17 and 74.47. What is the largest number of eggs that is not an outlier?

74 , because a count of eggs has to be a whole number and 74.47 is not attainable. Always check whether the data can only take particular values before quoting a boundary itself as the answer.

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Measures of Dispersion (AQA GCSE Statistics: Higher): Flashcards

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What do the lower and upper quartiles split off from a data set?

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What do the lower and upper quartiles split off from a data set?
The lower quartile has a quarter of the data below it, and the upper quartile has three quarters below it.
Together with the median they cut the ordered data into four equal parts.
You have 20 values in order and want the quartiles by the halves method. What are the two halves, and what do you do with each?
With an even number of values the first 10 form the lower half and the last 10 the upper half.
The lower quartile is then the median of the lower half, and the upper quartile is the median of the upper half.
With an odd number of values, what happens to the median when finding quartiles?
It is left out of both halves.
Everything below the median forms the lower half and everything above it forms the upper half, and the median itself belongs to neither.
Complete the second method for the quartile positions in a list of $n$ values.
$lower quartile at \frac{n + 1}{______}, upper quartile at______\times \frac{n + 1}{4}$
The completed positions are:
$lower quartile at \frac{n + 1}{4}, upper quartile at 3 \times \frac{n + 1}{4}$
For 20 values the lower quartile is at position 5.25, and the 5th and 6th values are 36 and 37. What is the lower quartile?
36.25: divide the gap from 36 to 37 into four equal parts and take the first one.
A position ending in .25 means a quarter of the way between the two values on either side of it.
True or False?
The two methods for finding quartiles always give the same answer.
False.
For the same 20 values the halves method gives a lower quartile of 36.5 while the position formula gives 36.25.
Both are accepted as correct answers.
Complete the sentences about equivalent measures.
The 25th percentile is the same as the lower $______$ of the data, the 50th percentile is the same as the $______$ value, and the 1st decile is the same as the $______$ percentile.
The completed sentences are:
The 25th percentile is the same as the lower quartile of the data, the 50th percentile is the same as the median value, and the 1st decile is the same as the 10th percentile.
What do percentiles and deciles divide a data set into?
Percentiles divide it into 100 equal parts and deciles into 10.
So the 30th percentile has 30 per cent of the data below it, and percentiles need not be whole numbers, such as the 2.5th percentile.
A table of 48 times is given and you need the class containing the 65th percentile. Where do you start?
Work out 65 per cent of 48, which comes to $48 \times \frac{65}{100} = 31.2$ for this data.
The 65th percentile therefore lies between the 31st and 32nd values, so build up cumulative frequencies until you reach them.
Complete the sentence about the range.
The range of a data set is the $______$ value take away the $______$ value, and it is measured in the same units as the data.
The completed sentence is:
The range of a data set is the largest value take away the smallest value, and it is measured in the same units as the data.
What does a measure of dispersion tell you that an average does not?
How spread out the data is, rather than what a typical value looks like.
A small measure of spread means the values cluster closely around the average, and a large one means many of them lie far from it.
Germination times run from 5 to 23 days, giving a range of 18, but the interquartile range is only 2. What does that tell you?
That the 23 is an extreme value sitting far above the rest of the data.
The range is stretched by that one value, so on its own it gives a badly misleading picture of how spread out the data really is.
Why is the interquartile range unaffected by extreme values?
Because it uses only the two quartiles, which sit a quarter and three quarters of the way through the data.
The largest and smallest values lie outside that middle half altogether, so changing them does not move either quartile.
Define interpercentile range.
An interpercentile range is the difference between two named percentiles of a data set.
You have to say which two, such as the 10th to 90th interpercentile range, which is the 90th percentile minus the 10th.
The 2.5th percentile is 274.1 g and the 97.5th percentile is 532.4 g. What is the 2.5th to 97.5th interpercentile range?
It is $532.4 - 274.1 = 258.3$ grams, taking the lower percentile from the upper one.
Like the interquartile range it ignores the most extreme values, but it lets you choose how much of the data to leave out.
You want the 1st to 9th interdecile range and you have a table of percentiles. Which two figures do you use?
The 90th percentile and the 10th percentile, subtracting the smaller from the larger.
For the pepper data that gives $488.3 - 319.7 = 168.6$ grams.
What does the standard deviation of a data set measure?
How spread out the values are around the mean, rather than at the extremes or the quartiles.
It is written with the lower case Greek letter $σ$ and a small value means most of the data sits close to the mean.
For the values 6, 9, 2, 11 and 5 the mean is 6.6 and the squared deviations total 49.2. What is the standard deviation?
Divide that total by the number of values and take the square root: $\sqrt{\frac{49.2}{5}} = \sqrt{9.84} = 3.14$ to three significant figures.
The same data gives the same answer through the other formula, which uses $\sum x$ and $\sum x^{2}$ instead.
A question gives you $\sum x = 33$ and $\sum x^{2} = 267$ for 5 values. Which formula should you use, and what is the standard deviation?
Use the version built from those two sums, since the question has handed them to you directly.
$\sqrt{\frac{267}{5} - {(\frac{33}{5})}^{2}} = \sqrt{9.84} = 3.14$
True or False?
The standard deviation is measured in the same units as the data.
True.
If the data is measured in seconds or in centimetres, the standard deviation comes out in seconds or centimetres too.
Every measure of spread on this course shares that property.
Complete the standard deviation formula for data given in a frequency table.
$σ = \sqrt{\frac{\sum______}{\sum f} - {(\frac{\sum______}{\sum f})}^{2}}$
The completed formula is:
$σ = \sqrt{\frac{\sum f x^{2}}{\sum f} - {(\frac{\sum f x}{\sum f})}^{2}}$
A frequency table gives $\sum f = 20$ with $\sum f x = 60$ and $\sum f x^{2} = 208$ altogether. What is the standard deviation?
Substitute the three totals into the frequency version of the formula.
$\sqrt{\frac{208}{20} - {(\frac{60}{20})}^{2}} = \sqrt{1.4} = 1.18$
How do you find a standard deviation for grouped data?
Use the same formulae as for a frequency table, but with the class midpoints as the $x$ values.
The answer can only be an estimate, because the original values are not available and the mean is itself only an estimate.
A grouped table gives $\sum f = 20$ with $\sum f x = 147$ and $\sum f x^{2} = 1215$ from its midpoints. Estimate the standard deviation.
Substituting the three totals gives:
$\sqrt{\frac{1215}{20} - {(\frac{147}{20})}^{2}} = \sqrt{6.7275} = 2.59$
True or False?
Removing an outlier changes the median as much as it changes the mean.
False.
The mean uses every value, so taking away an extreme one moves it noticeably.
The median depends only on the middle position, so removing a single value from one end usually shifts it very little or not at all.
When can outliers be identified without calculating any boundaries?
When a value is so obviously out of line with the rest that inspection is enough, such as an age of 3 in a list of GCSE candidates.
A boundary calculation is the more reliable method, because it gives a definite cut-off rather than a judgement.
Complete the outlier boundaries based on the quartiles.
$small outlier < LQ -______\times IQR, large outlier > UQ +______\times IQR$
The completed boundaries are:
$small outlier < LQ - 1.5 \times IQR, large outlier > UQ + 1.5 \times IQR$
Complete the outlier rule based on the mean and standard deviation.
$an outlier lies outside μ \pm______σ$
The completed rule is:
$an outlier lies outside μ \pm 3 σ$
Use this version when the mean and standard deviation are known but the quartiles cannot be found.
A data set has a lower quartile of 4 and an upper quartile of 7.5. What are its outlier boundaries?
The interquartile range is 3.5, giving boundaries of $4 - 1.5 \times 3.5 = - 1.25$ and $7.5 + 1.5 \times 3.5 = 12.75$ for this data.
A negative lower boundary is perfectly fine: it simply means no value in the set can be small enough to count as an outlier.
A summary gives $\sum x = 1108$ and $\sum x^{2} = 51632$ for 25 nests. Which outlier method must you use, and why?
The mean and standard deviation method, because quartiles cannot be found from those two totals alone.
They give a mean of 44.32 and a standard deviation of 10.05, so the boundaries are $44.32 \pm 3 \times 10.05$ for this data.
The outlier boundaries for the number of eggs in a nest work out as 14.17 and 74.47. What is the largest number of eggs that is not an outlier?
74, because a count of eggs has to be a whole number and 74.47 is not attainable.
Always check whether the data can only take particular values before quoting a boundary itself as the answer.

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