Empirical Rule, Skewness and Outliers
A report says the mean price of a house in a Dublin suburb is €420,000. But the median price is only €365,000. Which figure tells you what a typical house costs? The answer depends on the shape of the data. By the end of this page you will be able to explain it.
Skewness: the shape of a distribution
A histogram shows how the data are spread out. Its shape tells you a lot.
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Symmetric: the left and right halves are mirror images. The mean and the median are about equal.
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Skewed to the right (positive skew): a long tail towards the higher values. A few large values pull the mean up, so the mean is greater than the median.
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Skewed to the left (negative skew): a long tail towards the lower values. A few small values pull the mean down, so the mean is less than the median.
The skew is named after the tail, not after where the peak is.
Now return to the Dublin houses. The mean (€420,000) is bigger than the median (€365,000), so the distribution is skewed to the right: a few very expensive houses pull the mean up. We come back to which average to quote in the last section.
State whether each distribution is skewed to the left, skewed to the right or symmetric: a) A histogram with a long tail towards the higher values. b) A histogram with a…
State whether each distribution is skewed to the left, skewed to the right or symmetric:
- a) A histogram with a long tail towards the higher values.
- b) A histogram with a long tail towards the lower values.
- c) A bell-shaped histogram with equal tails.
House prices in a Dublin suburb have a mean of €420,000 and a median of €365,000. a) State what this suggests about the skewness of the distribution. b) Explain which…
House prices in a Dublin suburb have a mean of €420,000 and a median of €365,000.
- a) State what this suggests about the skewness of the distribution.
- b) Explain which measure of average is more suitable to describe a typical house price.
The empirical rule
Many real measurements, such as heights, egg masses or journey times, give a symmetric, bell-shaped histogram. We call this an approximately normal distribution. For these data, the empirical rule tells you how much of the data lies close to the mean $\bar{x}$ (the standard deviation is $\sigma$).
Because the curve is symmetric, you can split the percentages into the tails:
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Within $1\sigma$: $68\%$, so $100 - 68 = 32\%$ is outside, i.e. $16\%$ in each tail.
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Within $2\sigma$: $95\%$, so $5\%$ is outside, i.e. $2.5\%$ in each tail.
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Within $3\sigma$: $99.7\%$, so only $0.3\%$ is outside, i.e. $0.15\%$ in each tail.
The masses of a large sample of eggs are approximately normally distributed with mean $60$ g and standard deviation $4$ g. Using the empirical rule, find the interval…
The masses of a large sample of eggs are approximately normally distributed with mean $60$ g and standard deviation $4$ g.
Using the empirical rule, find the interval containing approximately $95\%$ of the masses.
The heights of adult men in a rugby club are approximately normally distributed with mean $180$ cm and standard deviation $7$ cm. a) Find the percentage of men with…
The heights of adult men in a rugby club are approximately normally distributed with mean $180$ cm and standard deviation $7$ cm.
- a) Find the percentage of men with heights between $173$ cm and $187$ cm.
- b) Find the percentage of men taller than $194$ cm.
- c) In a club of $400$ men, how many would you expect to be shorter than $166$ cm?
The times taken by cyclists to complete a route are approximately normally distributed. Approximately $16\%$ of cyclists take more than $52$ minutes and approximately…
The times taken by cyclists to complete a route are approximately normally distributed. Approximately $16\%$ of cyclists take more than $52$ minutes and approximately $2.5\%$ take less than $28$ minutes.
- a) Use the empirical rule to find the mean and the standard deviation.
- b) Hence find the percentage of cyclists who take between $36$ and $60$ minutes.
- c) Comment on whether the data are consistent with a normal distribution if the longest recorded time is $100$ minutes.
Quartiles, IQR and outliers
The standard deviation measures spread around the mean. The interquartile range measures the spread of the middle half of the data and is not affected by extreme values.
An outlier is a value that is unusually far from the rest of the data. A common test says a value is an outlier if it is:
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below $Q_1 - 1.5 \times \text{IQR}$, or
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above $Q_3 + 1.5 \times \text{IQR}$.
The reason is that the IQR describes the usual spread, so a value more than one and a half times that spread beyond the quartiles is far outside it.
A data set has lower quartile $12$ and upper quartile $20$. Find the interquartile range.
A data set has lower quartile $12$ and upper quartile $20$.
Find the interquartile range.
A data set has $Q_1 = 24$, $Q_3 = 40$. A value is considered an outlier if it is more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$. a) Find the IQR and the…
A data set has $Q_1 = 24$, $Q_3 = 40$. A value is considered an outlier if it is more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$.
- a) Find the IQR and the limits for outliers.
- b) The data set contains the values $5$, $18$, $44$ and $67$. State which are outliers.
- c) Explain the effect on the mean and the median of removing the outliers.
How outliers affect the mean and median
Try the data set below. Change the last value and watch what happens to the mean and median.
The mean uses every value, so one very large or very small value drags it towards that value. The median only depends on the middle position, so it hardly moves.
For the hours-online data in the last section, the mean is $\frac{111}{10} = 11.1$ and the median is $8.5$. Remove the outlier $40$ and the mean falls to $\frac{71}{9} \approx 7.9$, while the median only changes from $8.5$ to $8$. The mean is bigger than the median because of the large outlier, which is a sign of right skew.
Back to the Dublin houses: a few very expensive properties pull the mean up to €420,000. The median of €365,000 is the better description of a typical house.
To compare two data sets, use a measure of centre and a measure of spread together: for example the mean and standard deviation for symmetric data, or the median and IQR for skewed data. A larger standard deviation means the data are more spread out.
The number of text messages sent in a day by ten students is: $4, 6, 7, 8, 8, 9, 10, 11, 12, 45$ a) Find the mean and the median. b) Explain which value is an outlier…
The number of text messages sent in a day by ten students is:
$4, 6, 7, 8, 8, 9, 10, 11, 12, 45$
- a) Find the mean and the median.
- b) Explain which value is an outlier and how this affects the mean.
- c) State, with a reason, whether the distribution is skewed to the left or right.
Two classes sit the same test. Class A has mean $62$, median $62$ and standard deviation $5$. Class B has mean $58$, median $65$ and standard deviation $14$. a) Describe…
Two classes sit the same test. Class A has mean $62$, median $62$ and standard deviation $5$. Class B has mean $58$, median $65$ and standard deviation $14$.
- a) Describe the skewness of each distribution, giving a reason.
- b) Compare the two classes, referring to a measure of centre and a measure of spread.
- c) Class B has a single student with a score of $5$. Explain how removing this score would change the mean and the median.