Question

question #eight
give the data set 29 38 49 50 53 54 55 60 65 67 68 68. calculate the following:
a) σx²=
b) variance = s²=
c) standard deviation = s=

Explanation:

Step1: Calculate $\Sigma x^{2}$

First, square each data - point and sum them.
\[

$$\begin{align*} 29^{2}&=841\\ 38^{2}& = 1444\\ 49^{2}&=2401\\ 50^{2}&=2500\\ 53^{2}&=2809\\ 54^{2}&=2916\\ 55^{2}&=3025\\ 60^{2}&=3600\\ 65^{2}&=4225\\ 67^{2}&=4489\\ 68^{2}&=4624\\ 68^{2}&=4624 \end{align*}$$

\]
\[
\Sigma x^{2}=841 + 1444+2401+2500+2809+2916+3025+3600+4225+4489+4624+4624=36498
\]

Step2: Calculate the mean $\bar{x}$

The number of data - points $n = 12$.
\[
\bar{x}=\frac{29 + 38+49+50+53+54+55+60+65+67+68+68}{12}=\frac{656}{12}=\frac{164}{3}\approx54.67
\]

Step3: Calculate the variance $s^{2}$

The formula for the variance of a sample is $s^{2}=\frac{\Sigma(x - \bar{x})^{2}}{n - 1}$.
\[

$$\begin{align*} (x_1-\bar{x})^{2}&=(29-\frac{164}{3})^{2}=(\frac{87 - 164}{3})^{2}=(\frac{- 77}{3})^{2}=\frac{5929}{9}\\ (x_2-\bar{x})^{2}&=(38-\frac{164}{3})^{2}=(\frac{114 - 164}{3})^{2}=(\frac{-50}{3})^{2}=\frac{2500}{9}\\ &\cdots\\ (x_{12}-\bar{x})^{2}&=(68-\frac{164}{3})^{2}=(\frac{204 - 164}{3})^{2}=(\frac{40}{3})^{2}=\frac{1600}{9} \end{align*}$$

\]
\[

$$\begin{align*} \Sigma(x - \bar{x})^{2}&=\sum_{i = 1}^{12}(x_i-\bar{x})^{2}\\ s^{2}&=\frac{\Sigma(x - \bar{x})^{2}}{n - 1}=\frac{\sum_{i = 1}^{12}(x_i-\bar{x})^{2}}{11} \end{align*}$$

\]
Another formula for variance is $s^{2}=\frac{\Sigma x^{2}-n\bar{x}^{2}}{n - 1}$.
\[
n\bar{x}^{2}=12\times(\frac{164}{3})^{2}=12\times\frac{26896}{9}=\frac{322752}{9}
\]
\[
s^{2}=\frac{36498- \frac{322752}{9}}{11}=\frac{\frac{328482-322752}{9}}{11}=\frac{\frac{5730}{9}}{11}=\frac{5730}{99}\approx57.88
\]

Step4: Calculate the standard deviation $s$

The standard deviation $s=\sqrt{s^{2}}$.
\[
s=\sqrt{\frac{5730}{99}}\approx\sqrt{57.88}\approx7.61
\]

Answer:

a) $\Sigma x^{2}=36498$
b) $s^{2}\approx57.88$
c) $s\approx7.61$