Question

which set of population data is the least dispersed from its mean?
2, 3, 2, 9
4, 0, 4, 0
6, 2, 2, 2
9, 3, 5, 3

Explanation:

Step1: Calculate mean and variance for each set

Variance measures dispersion from the mean. For a set of data \(x_1,x_2,\cdots,x_n\), the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\) and the variance \(\sigma^{2}=\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n}\).

For set \(2,3,2,9\):

• Calculate the mean: \(\bar{x}_1=\frac{2 + 3+2 + 9}{4}=\frac{16}{4}=4\).
• Calculate the variance: \(\sigma_1^{2}=\frac{(2 - 4)^2+(3 - 4)^2+(2 - 4)^2+(9 - 4)^2}{4}=\frac{4 + 1+4 + 25}{4}=\frac{34}{4}=8.5\).

For set \(4,0,4,0\):

• Calculate the mean: \(\bar{x}_2=\frac{4+0 + 4+0}{4}=\frac{8}{4}=2\).
• Calculate the variance: \(\sigma_2^{2}=\frac{(4 - 2)^2+(0 - 2)^2+(4 - 2)^2+(0 - 2)^2}{4}=\frac{4+4 + 4+4}{4}=4\).

For set \(6,2,2,2\):

• Calculate the mean: \(\bar{x}_3=\frac{6+2 + 2+2}{4}=\frac{12}{4}=3\).
• Calculate the variance: \(\sigma_3^{2}=\frac{(6 - 3)^2+(2 - 3)^2+(2 - 3)^2+(2 - 3)^2}{4}=\frac{9+1 + 1+1}{4}=\frac{12}{4}=3\).

For set \(9,3,5,3\):

• Calculate the mean: \(\bar{x}_4=\frac{9+3 + 5+3}{4}=\frac{20}{4}=5\).
• Calculate the variance: \(\sigma_4^{2}=\frac{(9 - 5)^2+(3 - 5)^2+(5 - 5)^2+(3 - 5)^2}{4}=\frac{16+4+0 + 4}{4}=\frac{24}{4}=6\).

Step2: Compare variances

We compare the variances \(\sigma_1^{2}=8.5\), \(\sigma_2^{2}=4\), \(\sigma_3^{2}=3\), \(\sigma_4^{2}=6\). The smallest variance indicates the least - dispersed data set from its mean.

Answer:

\(6,2,2,2\)