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 $\sum x^{2}$

The data - set is $x = \{29,38,49,50,53,54,55,60,65,67,68,68\}$.
\[

$$\begin{align*} \sum x^{2}&=29^{2}+38^{2}+49^{2}+50^{2}+53^{2}+54^{2}+55^{2}+60^{2}+65^{2}+67^{2}+68^{2}+68^{2}\\ &=841 + 1444+2401+2500+2809+2916+3025+3600+4225+4489+4624+4624\\ &=39508 \end{align*}$$

\]

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

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

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

The formula for the sample variance is $s^{2}=\frac{\sum x^{2}-\frac{(\sum x)^{2}}{n}}{n - 1}$.
We know that $\sum x^{2}=39508$, $\sum x = 656$, and $n = 12$.
\[

$$\begin{align*} s^{2}&=\frac{39508-\frac{656^{2}}{12}}{12 - 1}\\ &=\frac{39508-\frac{430336}{12}}{11}\\ &=\frac{39508 - \frac{107584}{3}}{11}\\ &=\frac{\frac{118524-107584}{3}}{11}\\ &=\frac{\frac{10940}{3}}{11}\\ &=\frac{10940}{33}\approx331.52 \end{align*}$$

\]

Step4: Calculate the standard deviation $s$

The standard deviation $s=\sqrt{s^{2}}$.
Since $s^{2}=\frac{10940}{33}$, then $s=\sqrt{\frac{10940}{33}}\approx\sqrt{331.52}\approx18.21$.

Answer:

a) $\sum x^{2}=39508$
b) $s^{2}=\frac{10940}{33}\approx331.52$
c) $s=\sqrt{\frac{10940}{33}}\approx18.21$