Question

3.2 measures of spread: standard deviation and variance
for the sample data shown, answer the questions. round to 2 decimal places.
x
2.3
9.5
15.6
18.2
21.7
27.3
find the mean:
find the median:
find the sample standard deviation:
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Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 6$, $x_1=2.3$, $x_2 = 9.5$, $x_3=15.6$, $x_4=18.2$, $x_5=21.7$, $x_6=27.3$.
$\bar{x}=\frac{2.3 + 9.5+15.6+18.2+21.7+27.3}{6}=\frac{94.6}{6}\approx15.77$

Step2: Calculate the median

First, order the data: $2.3,9.5,15.6,18.2,21.7,27.3$. Since $n = 6$ (an even - numbered data set), the median is the average of the two middle - values. The two middle - values are the 3rd and 4th ordered values. Median $=\frac{15.6 + 18.2}{2}=\frac{33.8}{2}=16.90$

Step3: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}}$.
$(x_1-\bar{x})^2=(2.3 - 15.77)^2=(-13.47)^2 = 181.4409$
$(x_2-\bar{x})^2=(9.5 - 15.77)^2=(-6.27)^2 = 39.3129$
$(x_3-\bar{x})^2=(15.6 - 15.77)^2=(-0.17)^2 = 0.0289$
$(x_4-\bar{x})^2=(18.2 - 15.77)^2=(2.43)^2 = 5.9049$
$(x_5-\bar{x})^2=(21.7 - 15.77)^2=(5.93)^2 = 35.1649$
$(x_6-\bar{x})^2=(27.3 - 15.77)^2=(11.53)^2 = 132.9409$
$\sum_{i = 1}^{6}(x_{i}-\bar{x})^2=181.4409+39.3129 + 0.0289+5.9049+35.1649+132.9409=394.7934$
$s=\sqrt{\frac{394.7934}{6 - 1}}=\sqrt{\frac{394.7934}{5}}=\sqrt{78.95868}\approx8.89$

Answer:

Find the mean: $15.77$
Find the median: $16.90$
Find the sample standard deviation: $8.89$