chapter 3.2 homework\nscore: 12/102 answered: 8/33\nquestion 9\n3.2 measures of spread: standard deviation…

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 9$, and $x_{i}$ are the data - points. $\sum_{i=1}^{9}x_{i}=6.2 + 6.5+10.9+11.8+14.1+18.7+23.7+26.9+28.5 = 147.3$. So, $\bar{x}=\frac{147.3}{9}\approx16.37$.

Step2: Calculate the median

Arrange the data in ascending order: $6.2,6.5,10.9,11.8,14.1,18.7,23.7,26.9,28.5$. Since $n = 9$ (odd), the median is the $\left(\frac{n + 1}{2}\right)$-th value. $\frac{9+1}{2}=5$-th value, so the median is $14.10$.

Step3: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. First, calculate $(x_{i}-\bar{x})^{2}$ for each $x_{i}$: $(6.2-16.37)^{2}=(-10.17)^{2}=103.4289$ $(6.5 - 16.37)^{2}=(-9.87)^{2}=97.4169$ $(10.9-16.37)^{2}=(-5.47)^{2}=29.9209$ $(11.8-16.37)^{2}=(-4.57)^{2}=20.8849$ $(14.1-16.37)^{2}=(-2.27)^{2}=5.1529$ $(18.7-16.37)^{2}=(2.33)^{2}=5.4289$ $(23.7-16.37)^{2}=(7.33)^{2}=53.7289$ $(26.9-16.37)^{2}=(10.53)^{2}=110.8809$ $(28.5-16.37)^{2}=(12.13)^{2}=147.1369$ $\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=103.4289+97.4169+29.9209+20.8849+5.1529+5.4289+53.7289+110.8809+147.1369 = 573.971$ $s=\sqrt{\frac{573.971}{9 - 1}}=\sqrt{\frac{573.971}{8}}\approx8.44$

Answer:

Mean: $16.37$ Median: $14.10$ Sample Standard Deviation: $8.44$