find the range and the standard deviation for the three samples below.\nsample a: 41, 43, 45, 47, 49, 51…

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 A	(12)	(4.32)
Sample B	(12)	(5.05)
Sample C	(12)	(6)