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: \(n = 7\), \(\sum_{i=1}^{7}x_i=41 + 43+45+47+49+51+53=329\), \(\bar{x}_A=\frac{329}{7}=47\)
For Sample B: \(n = 7\), \(\sum_{i=1}^{7}x_i=41+42 + 43+47+51+52+53=329\), \(\bar{x}_B=\frac{329}{7}=47\)
For Sample C: \(n = 7\), \(\sum_{i=1}^{7}x_i=41\times3 + 47+53\times3=123+47 + 159=329\), \(\bar{x}_C=\frac{329}{7}=47\)

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:
\(\sum_{i=1}^{7}(x_i - 47)^2=(41 - 47)^2+(42 - 47)^2+(43 - 47)^2+(47 - 47)^2+(51 - 47)^2+(52 - 47)^2+(53 - 47)^2\)
\(=(-6)^2+(-5)^2+(-4)^2+0^2+4^2+5^2+6^2=36+25 + 16+0+16+25+36=154\)
\(s_B=\sqrt{\frac{154}{6}}\approx5.05\)
For Sample C:
\(\sum_{i=1}^{7}(x_i - 47)^2=3\times(41 - 47)^2+(47 - 47)^2+3\times(53 - 47)^2\)
\(=3\times(-6)^2+0^2+3\times6^2=3\times36+0 + 3\times36=216\)
\(s_C=\sqrt{\frac{216}{6}}=6\)

Answer:

	Range	Standard Deviation
Sample B	\(12\)	\(5.05\)
Sample C	\(12\)	\(6\)