Question

find the range and the standard deviation for the three samples below.
sample a: 41, 43, 45, 47, 49, 51, 53
sample b: 41, 42, 43, 47, 51, 52, 53
sample c: 41, 41, 41, 47, 53, 53, 53

Explanation:

Step1: Calculate the range

The range is calculated as \( \text{Range} = \text{Max}-\text{Min} \).
For Sample A: \( \text{Max} = 53 \), \( \text{Min}=41 \), so \( \text{Range}=53 - 41=12 \).
For Sample B: \( \text{Max} = 53 \), \( \text{Min}=41 \), so \( \text{Range}=53 - 41=12 \).
For Sample C: \( \text{Max} = 53 \), \( \text{Min}=41 \), so \( \text{Range}=53 - 41=12 \).

Step2: Calculate the mean (\(\bar{x}\))

The formula for the mean of a sample \(x_1,x_2,\cdots,x_n\) is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\).
For Sample A: \(\bar{x}_A=\frac{41 + 43+45+47+49+51+53}{7}=\frac{329}{7} = 47\).
For Sample B: \(\bar{x}_B=\frac{41+42 + 43+47+51+52+53}{7}=\frac{329}{7}=47\).
For Sample C: \(\bar{x}_C=\frac{41+41 + 41+47+53+53+53}{7}=\frac{329}{7}=47\).

Step3: Calculate the standard deviation (\(s\))

The formula for the sample - standard deviation is \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}\).
For Sample A:
\(\sum_{i=1}^{7}(x_i - 47)^2=(41 - 47)^2+(43 - 47)^2+(45 - 47)^2+(47 - 47)^2+(49 - 47)^2+(51 - 47)^2+(53 - 47)^2\)
\(=(-6)^2+(-4)^2+(-2)^2+0^2+2^2+4^2+6^2\)
\(=36 + 16+4+0+4+16+36=112\)
\(s_A=\sqrt{\frac{112}{6}}\approx4.32\).
For Sample B:
\(\sum_{i = 1}^{7}(x_i - 47)^2=(41 - 47)^2+(42 - 47)^2+(43 - 47)^2+(47 - 47)^2+(51 - 47)^2+(52 - 47)^2+(53 - 47)^2\)
\(=(-6)^2+(-5)^2+(-4)^2+0^2+4^2+5^2+6^2\)
\(=36+25 + 16+0+16+25+36=154\)
\(s_B=\sqrt{\frac{154}{6}}\approx5.07\).
For Sample C:
\(\sum_{i = 1}^{7}(x_i - 47)^2=3\times(41 - 47)^2+3\times(53 - 47)^2+(47 - 47)^2\)
\(=3\times(-6)^2+3\times6^2+0^2\)
\(=3\times36+3\times36+0=216\)
\(s_C=\sqrt{\frac{216}{6}} = 6\).

Answer:

Sample	Range	Standard Deviation
Sample B	12	5.07
Sample C	12	6