Question

compute the mean, range, and standard deviation for the data items in each of the three samples. then describe one way in which the samples are alike and one way in which they are different. sample a: 30, 36, 42, 48, 54, 60, 66 sample b: 30, 33, 38, 48, 60, 63, 66 sample c: 30, 30, 30, 48, 66, 66, 66 mean range standard deviation sample a 48 36 12.96 (round to two decimal places as needed) mean range standard deviation sample b 48 36 15.20 (round to two decimal places as needed) mean range standard deviation sample c 48 36 18 (round to two decimal places as needed.) in which of the following ways are the samples alike? standard deviation mean and range mean range

Explanation:

Step1: Analyze Mean

For Sample A: \(\frac{30 + 36+42 + 48+54 + 60+66}{7}=\frac{336}{7} = 48\)
For Sample B: \(\frac{30+33 + 38+48+60+63+66}{7}=\frac{338}{7}\approx48\)
For Sample C: \(\frac{30+30 + 30+48+66+66+66}{7}=\frac{336}{7}=48\)

Step2: Analyze Range

Range = Maximum - Minimum
For Sample A: \(66 - 30=36\)
For Sample B: \(66 - 30 = 36\)
For Sample C: \(66 - 30=36\)

Step3: Analyze Standard Deviation

Standard deviation formula \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
For Sample A: \(\sum_{i=1}^{7}(x_{i}-48)^{2}=(30 - 48)^{2}+(36 - 48)^{2}+(42 - 48)^{2}+(48 - 48)^{2}+(54 - 48)^{2}+(60 - 48)^{2}+(66 - 48)^{2}=(- 18)^{2}+(-12)^{2}+(-6)^{2}+0^{2}+6^{2}+12^{2}+18^{2}=324 + 144+36+0 + 36+144+324 = 1008\), \(s=\sqrt{\frac{1008}{6}}\approx12.96\)
For Sample B: \(\sum_{i = 1}^{7}(x_{i}-48)^{2}=(30 - 48)^{2}+(33 - 48)^{2}+(38 - 48)^{2}+(48 - 48)^{2}+(60 - 48)^{2}+(63 - 48)^{2}+(66 - 48)^{2}=(-18)^{2}+(-15)^{2}+(-10)^{2}+0^{2}+12^{2}+15^{2}+18^{2}=324+225 + 100+0+144+225+324 = 1342\), \(s=\sqrt{\frac{1342}{6}}\approx15.20\)
For Sample C: \(\sum_{i = 1}^{7}(x_{i}-48)^{2}=(30 - 48)^{2}+(30 - 48)^{2}+(30 - 48)^{2}+(48 - 48)^{2}+(66 - 48)^{2}+(66 - 48)^{2}+(66 - 48)^{2}=3\times(-18)^{2}+0^{2}+3\times(18)^{2}=3\times324+0 + 3\times324=1944\), \(s=\sqrt{\frac{1944}{6}} = 18\)

Answer:

B. Mean and Range