Analysing data for SQA National 5 Maths
This page covers the following topics:
2. Inter-quartile range
3. Standard deviation
Quartiles are another measure of spread. The lower quartile is the median of the lower half of the data and it is found by taking the (n+1)/4 value, where n is the number of values in the data. The upper quartile is the median of the upper half of the data and it is found by taking the 3(n+1)/4 value. If there are an even number of values, the quartile is found by taking the average of the two middle values.
The interquartile range (IQR) is the difference between the lower and upper quartile. The IQR is a measure of spread which is not affected by outliers, unlike the range of the set of values.
Standard deviation is a measure of spread which gives a quantitative measure of how much the values of a set differ from their mean value. The difference between each value and the mean is found and is squared. These squared differences are then added together and divided by the number of values in the set. Taking the square root of this gives the standard deviation.
What is the interquartile range of the given set of values?
47, 56, 58, 59, 62, 64, 67
There are 7 values.
For the lower quartile, (7 + 1)/4 = 2, therefore it is the second value.
So the lower quartile is 56.
For the upper quartile, 3(7 + 1)/4 = 6, therefore it is the sixth value.
So the upper quartile is 64.
Therefore, the IQR = 64 − 56 = 8.
What is the upper quartile of the following set of values: 14, 19, 20, 21, 26, 27, 30, 32.
There 8 values. So, upper quartile = 3(8 + 1)/4 = 6.75. To calculate the upper quartile, the average of the 6th and 7th values must thus be calculated. So, upper quartile = (27 + 30)/2 = 28.5.
Can the standard deviation of a set of values be calculated using only the number of values in the set and their mean?
The standard deviation cannot be calculated using only these two values, since the difference must be found between each value and the mean. Thus, the values must be known for the standard deviation to be calculated.
Use the given table to calculate the standard deviation for the set of values.
standard deviation = √(((9 + 0 + 1 + 1 + 9) ÷ 5) = 2
The standard deviation of a set of values is found to be 2.3. Given that the sum of the squared difference between each value and their mean is 42, calculate the number of values in the set.
2.3 = √(42/n), therefore n = 42/2.3² = 8 (rounded to an integer).
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