Question

compute the mean, range, and standard deviation for the data lons 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. 27, 33, 39, 45, 51, 57, 63
sample b. 27. 30, 33, 45, 57, 60, 63
sample c. 27, 27, 27, 45, 63, 63, 63
sample a
mean: 45
range: 36
standard deviation: 12.96 (round to two decimal places as needed)
sample b
mean
range
standard deviation
(round to two decimal places as needed)

Explanation:

Step1: Calculate the mean of Sample B

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). For Sample B: \(x=\{27,30,33,45,57,60,63\}\), \(n = 7\).
\(\sum_{i=1}^{7}x_{i}=27 + 30+33 + 45+57+60+63=315\)
\(\bar{x}=\frac{315}{7}=45\)

Step2: Calculate the range of Sample B

The formula for the range \(R=\text{Max}-\text{Min}\). In Sample B, \(\text{Max}=63\), \(\text{Min}=27\)
\(R = 63-27=36\)

Step3: Calculate the standard deviation of Sample B

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\)
\((27 - 45)^{2}=(-18)^{2}=324\)
\((30 - 45)^{2}=(-15)^{2}=225\)
\((33 - 45)^{2}=(-12)^{2}=144\)
\((45 - 45)^{2}=0^{2}=0\)
\((57 - 45)^{2}=12^{2}=144\)
\((60 - 45)^{2}=15^{2}=225\)
\((63 - 45)^{2}=18^{2}=324\)
\(\sum_{i = 1}^{7}(x_{i}-\bar{x})^{2}=324+225 + 144+0+144+225+324=1386\)
\(s=\sqrt{\frac{1386}{7-1}}=\sqrt{\frac{1386}{6}}=\sqrt{231}\approx15.20\)

Answer:

Mean: \(45\), Range: \(36\), Standard Deviation: \(15.20\)