Question

1. calculate the range and interquartile range for the following set of scores from a continuous variable: 5, 1, 6, 5, 4, 6, 7, 12. identify the score that corresponds to the 75th percentile and the score that corresponds to the 25th percentile. why is the interquartile range a better description of variability in the data than the range?

Explanation:

Step1: Arrange data in ascending order

$1,4,5,5,6,6,7,12$

Step2: Calculate the range

Range = Maximum - Minimum. Here, Maximum = 12, Minimum = 1. So, Range = $12 - 1=11$.

Step3: Calculate the position of quartiles

The formula for the position of the $p$ - th percentile in a data - set of size $n$ is $L_p=(n + 1)\frac{p}{100}$. Here, $n = 8$.
For the 25th percentile ($p = 25$), $L_{25}=(8 + 1)\frac{25}{100}=2.25$. The 25th percentile is the value at the 2.25 - th position. Using interpolation, the 25th percentile $Q_1=4+(0.25)\times(5 - 4)=4.25$.
For the 75th percentile ($p = 75$), $L_{75}=(8 + 1)\frac{75}{100}=6.75$. The 75th percentile is the value at the 6.75 - th position. Using interpolation, the 75th percentile $Q_3=6+(0.75)\times(7 - 6)=6.75$.

Step4: Calculate the inter - quartile range

Inter - quartile range (IQR)=$Q_3 - Q_1$. So, $IQR = 6.75-4.25 = 2.5$.

Step5: Explain why IQR is better than range

The range is affected by extreme values (outliers). In this data - set, the value 12 is an outlier. The inter - quartile range only considers the middle 50% of the data and is not affected by extreme values. So, it gives a better measure of the spread of the central data values.

Answer:

Range = 11, Inter - quartile range = 2.5, 25th percentile score = 4.25, 75th percentile score = 6.75. The inter - quartile range is a better description of variability in the data than the range because it is not affected by extreme values (outliers).