Question

determine the standard deviation (σ) by filling in the table as part of your calculation. consider the following data 6, 6, 10, 8, 10, 8
\begin{tabular}{|c|c|c|c|} hline ( x ) & ( \bar{x} ) & ( x - \bar{x} ) & ( (x - \bar{x})^2 ) \\ hline & & & \\ hline & & & \\ hline & & & \\ hline & & & \\ hline & & & \\ hline & & & \\ hline & & & \\ hline end{tabular}
a. 1.63
b. 0.47
c. 0.94
d. 1.15

Explanation:

Step1: Calculate the mean (\(\bar{x}\))

The data set is \(6, 6, 10, 8, 10, 8\). The number of data points \(n = 6\).
The mean \(\bar{x}=\frac{6 + 6 + 10 + 8 + 10 + 8}{6}=\frac{48}{6}=8\).

Step2: Fill in the table

For each \(x\) value:

• When \(x = 6\): \(x-\bar{x}=6 - 8=-2\), \((x - \bar{x})^2=(-2)^2 = 4\)
• When \(x = 6\): \(x-\bar{x}=6 - 8=-2\), \((x - \bar{x})^2=(-2)^2 = 4\)
• When \(x = 10\): \(x-\bar{x}=10 - 8 = 2\), \((x - \bar{x})^2=(2)^2 = 4\)
• When \(x = 8\): \(x-\bar{x}=8 - 8 = 0\), \((x - \bar{x})^2=(0)^2 = 0\)
• When \(x = 10\): \(x-\bar{x}=10 - 8 = 2\), \((x - \bar{x})^2=(2)^2 = 4\)
• When \(x = 8\): \(x-\bar{x}=8 - 8 = 0\), \((x - \bar{x})^2=(0)^2 = 0\)

The table becomes:

\(x\)	\(\bar{x}\)	\(x-\bar{x}\)	\((x - \bar{x})^2\)
6	8	-2	4
10	8	2	4
8	8	0	0
10	8	2	4
8	8	0	0

Step3: Calculate the sum of \((x - \bar{x})^2\)

Sum \(=\ 4+4 + 4+0 + 4+0=16\)

Step4: Calculate the variance (\(\sigma^2\))

Variance \(\sigma^2=\frac{\sum(x - \bar{x})^2}{n}=\frac{16}{6}\approx2.67\) (Wait, no, wait. Wait, actually, for population standard deviation, we use \(n\), but let's recalculate. Wait, the sum of \((x - \bar{x})^2\) is \(4 + 4+4 + 0+4 + 0=16\)? Wait, no: 6-8=-2, squared is 4; another 6-8=-2, squared 4; 10-8=2, squared 4; 8-8=0, squared 0; 10-8=2, squared 4; 8-8=0, squared 0. So 4+4=8, +4=12, +0=12, +4=16, +0=16. Then variance \(\sigma^2=\frac{16}{6}\approx2.6667\)? Wait, no, that can't be. Wait, maybe I made a mistake. Wait, the data is 6,6,10,8,10,8. Let's recalculate the mean: 6+6=12, 10+8=18, 10+8=18; total 12+18+18=48. 48/6=8. Correct. Then deviations: 6-8=-2, 6-8=-2, 10-8=2, 8-8=0, 10-8=2, 8-8=0. Squared deviations: 4,4,4,0,4,0. Sum is 4+4+4+0+4+0=16. Then variance (population) is 16/6 ≈2.6667. Then standard deviation \(\sigma=\sqrt{\frac{16}{6}}=\sqrt{\frac{8}{3}}\approx\sqrt{2.6667}\approx1.63\). Wait, but let's check the options. Option a is 1.63. Wait, but maybe I messed up. Wait, wait, the formula for population standard deviation is \(\sigma=\sqrt{\frac{\sum(x - \bar{x})^2}{N}}\), where \(N\) is the population size. Here \(N = 6\). So \(\sum(x - \bar{x})^2 = 16\), so \(\sigma=\sqrt{\frac{16}{6}}=\sqrt{\frac{8}{3}}\approx1.63\). So the answer should be a. 1.63. Wait, but let's recheck the sum of squared deviations. 6-8=-2, squared 4; two 6s: 24=8. 10-8=2, squared 4; two 10s: 24=8. 8-8=0, squared 0; two 8s: 2*0=0. Total 8+8+0=16. Correct. Then variance 16/6≈2.6667, standard deviation≈1.63. So the answer is a. 1.63.

Wait, but maybe the question is about sample standard deviation? But the symbol is \(\sigma\), which is population standard deviation. If it were sample, we would use \(s\) and divide by \(n - 1\). Let's check: if we divide by \(n - 1 = 5\), then variance is 16/5 = 3.2, standard deviation is \(\sqrt{3.2}\approx1.788\), which is not an option. So it's population standard deviation, so divide by \(n = 6\). So \(\sigma=\sqrt{\frac{16}{6}}\approx1.63\), which is option a.

Answer:

a. 1.63