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: 28, 32, 36, 40, 44, 48, 52
sample b: 28, 30, 32, 40, 48, 50, 52
sample c: 28, 28, 28, 40, 52, 52, 52
mean
range
standard deviation
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the Mean

The mean ($\bar{x}$) is calculated by summing all the data points and dividing by the number of data points ($n$). For Sample A: \(28, 32, 36, 40, 44, 48, 52\). The sum is \(28 + 32 + 36 + 40 + 44 + 48 + 52 = 280\). There are \(n = 7\) data points. So, \(\bar{x}=\frac{280}{7}=40\).

Step2: Calculate the Range

The range is the difference between the maximum and minimum values. For Sample A, the maximum value is \(52\) and the minimum is \(28\). So, Range \(= 52 - 28 = 24\).

Step3: Calculate the Standard Deviation

First, find the deviation of each data point from the mean, square it, sum these squared deviations, divide by \(n - 1\) (for sample standard deviation), and then take the square root.

• Deviations: \(28 - 40 = -12\), \(32 - 40 = -8\), \(36 - 40 = -4\), \(40 - 40 = 0\), \(44 - 40 = 4\), \(48 - 40 = 8\), \(52 - 40 = 12\)
• Squared deviations: \((-12)^2 = 144\), \((-8)^2 = 64\), \((-4)^2 = 16\), \(0^2 = 0\), \(4^2 = 16\), \(8^2 = 64\), \(12^2 = 144\)
• Sum of squared deviations: \(144 + 64 + 16 + 0 + 16 + 64 + 144 = 448\)
• Variance ($s^2$) for sample: \(\frac{448}{7 - 1}=\frac{448}{6}\approx74.67\)
• Standard deviation ($s$): \(\sqrt{74.67}\approx8.64\)

Answer:

• Mean: \(40\)
• Range: \(24\)
• Standard Deviation: \(\approx 8.64\)