Question

use the following set of data to calculate the measures.
0, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6
choose each correct measure.
mean =
median =
range =
interquartile range =

Explanation:

Step1: Calculate the mean

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here, $n = 14$, and $\sum_{i=1}^{n}x_{i}=0 + 2+3\times5 + 4\times4+5\times2+6=0 + 2+15+16+10+6 = 49$. So, $\bar{x}=\frac{49}{14}=3.5$.

Step2: Calculate the median

Since $n = 14$ (even), the median is the average of the $\frac{n}{2}$-th and $(\frac{n}{2}+1)$-th values. The $7$-th value is $3$ and the $8$-th value is $4$. So, median $=\frac{3 + 4}{2}=3.5$.

Step3: Calculate the range

The range $R=\text{Max}-\text{Min}$. Here, $\text{Max}=6$ and $\text{Min}=0$. So, $R=6 - 0=6$.

Step4: Calculate the inter - quartile range

First, find the first quartile $Q_{1}$. The lower half of the data is $0,2,3,3,3,3,3$. Since $n_{1}=7$ (odd), $Q_{1}$ is the $4$-th value of the lower half, so $Q_{1}=3$.
The upper half of the data is $4,4,4,4,5,5,6$. Since $n_{2}=7$ (odd), $Q_{3}$ is the $4$-th value of the upper half, so $Q_{3}=4$.
The inter - quartile range $IQR = Q_{3}-Q_{1}=4 - 3=1$.

Answer:

Mean $=3.5$, Median $=3.5$, Range $=6$, Interquartile range $=1$