Question

the three data sets have the same mean and range, but is the variation the same? prove your answer by computing the standard deviation. assume the data were obtained from samples.
(a) 3, 3, 9, 11
(b) 3, 4, 8, 11
(c) 4, 4, 6, 12
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select, select has the smallest variation; select has the largest variation.

Explanation:

To determine the variation, we calculate the sample standard deviation for each data set. The formula for the sample standard deviation \( s \) is:

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

where \( \bar{x} \) is the sample mean, \( x_i \) are the data points, and \( n \) is the sample size.

Step 1: Calculate the mean (\( \bar{x} \)) for each data set

All three data sets have \( n = 4 \) data points. Let's verify the mean:

For (a): \( \bar{x}_a = \frac{3 + 3 + 9 + 11}{4} = \frac{26}{4} = 6.5 \)
For (b): \( \bar{x}_b = \frac{3 + 4 + 8 + 11}{4} = \frac{26}{4} = 6.5 \)
For (c): \( \bar{x}_c = \frac{4 + 4 + 6 + 12}{4} = \frac{26}{4} = 6.5 \)

Step 2: Calculate the sum of squared deviations (\( \sum (x_i - \bar{x})^2 \)) for each data set

Data Set (a): \( 3, 3, 9, 11 \)

\[

$$\begin{align*} (3 - 6.5)^2 &= (-3.5)^2 = 12.25 \\ (3 - 6.5)^2 &= (-3.5)^2 = 12.25 \\ (9 - 6.5)^2 &= (2.5)^2 = 6.25 \\ (11 - 6.5)^2 &= (4.5)^2 = 20.25 \\ \sum (x_i - \bar{x})^2 &= 12.25 + 12.25 + 6.25 + 20.25 = 51 \end{align*}$$

\]

Data Set (b): \( 3, 4, 8, 11 \)

\[

$$\begin{align*} (3 - 6.5)^2 &= (-3.5)^2 = 12.25 \\ (4 - 6.5)^2 &= (-2.5)^2 = 6.25 \\ (8 - 6.5)^2 &= (1.5)^2 = 2.25 \\ (11 - 6.5)^2 &= (4.5)^2 = 20.25 \\ \sum (x_i - \bar{x})^2 &= 12.25 + 6.25 + 2.25 + 20.25 = 41 \end{align*}$$

\]

Data Set (c): \( 4, 4, 6, 12 \)

\[

$$\begin{align*} (4 - 6.5)^2 &= (-2.5)^2 = 6.25 \\ (4 - 6.5)^2 &= (-2.5)^2 = 6.25 \\ (6 - 6.5)^2 &= (-0.5)^2 = 0.25 \\ (12 - 6.5)^2 &= (5.5)^2 = 30.25 \\ \sum (x_i - \bar{x})^2 &= 6.25 + 6.25 + 0.25 + 30.25 = 47 \end{align*}$$

\]

Step 3: Calculate the sample standard deviation (\( s \)) for each data set

The sample standard deviation is:

\[
s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}}
\]

For (a):
\[
s_a = \sqrt{\frac{51}{4 - 1}} = \sqrt{\frac{51}{3}} \approx \sqrt{17} \approx 4.123
\]

For (b):
\[
s_b = \sqrt{\frac{41}{4 - 1}} = \sqrt{\frac{41}{3}} \approx \sqrt{13.6667} \approx 3.697
\]

For (c):
\[
s_c = \sqrt{\frac{47}{4 - 1}} = \sqrt{\frac{47}{3}} \approx \sqrt{15.6667} \approx 3.958
\]

Step 4: Compare the standard deviations

• \( s_b \approx 3.697 \) (smallest)
• \( s_c \approx 3.958 \)
• \( s_a \approx 4.123 \) (largest)

Thus, data set (b) has the smallest variation, and data set (a) has the largest variation.

Answer:

(b) \( 3, 4, 8, 11 \) has the smallest variation; (a) \( 3, 3, 9, 11 \) has the largest variation.