Question

use the set of data to calculate the measures that follow. 0,2,3,3,3,3,3,4,4,4,4,5,5,6 choose each correct measure. mean = median = range = interquartile range = 1 2 3.5 3

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 15$, and $\sum_{i=1}^{15}x_{i}=0 + 2+3\times5 + 4\times4+5\times2 + 6=0 + 2+15+16 + 10+6=49$. So $\bar{x}=\frac{49}{15}\approx3.27$.

Step2: Calculate the median

Arrange the data in ascending - order: $0,2,3,3,3,3,3,4,4,4,4,5,5,6$. Since $n = 15$ (odd), the median is the $(\frac{n + 1}{2})$ - th value. $\frac{15+1}{2}=8$ - th value, so the median is $4$.

Step3: Calculate the range

The range is the difference between the maximum and minimum values. The maximum value is $6$ and the minimum value is $0$. So the range is $6-0 = 6$.

Step4: Calculate the inter - quartile range

First, find the lower half and upper half of the data. The lower half of the data (excluding the median since $n$ is odd) is $0,2,3,3,3,3,3$. The median of the lower half ($Q_1$) is the $(\frac{7 + 1}{2})$ - th value, which is $3$. The upper half of the data is $4,4,4,4,5,5,6$. The median of the upper half ($Q_3$) is the $(\frac{7+1}{2})$ - th value, which is $4$. The inter - quartile range $IQR=Q_3 - Q_1=4 - 3=1$.

Answer:

Mean: $3.27$
Median: $4$
Range: $6$
Interquartile range: $1$