Question

3.2 measures of spread: standard deviation and variance
for the sample data shown, answer the questions. round to 2 decimal places.
x
3.1
6.1
6.6
11.2
29.1
find the mean:
find the median:
find the sample standard deviation:
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Explanation:

Step1: Calculate the mean

The mean $\bar{x}$ of a sample $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here $n = 5$, $x_1=3.1$, $x_2 = 6.1$, $x_3=6.6$, $x_4=11.2$, $x_5=29.1$. So $\bar{x}=\frac{3.1 + 6.1+6.6 + 11.2+29.1}{5}=\frac{56.1}{5}=11.22$.

Step2: Calculate the median

First, order the data: $3.1,6.1,6.6,11.2,29.1$. Since $n = 5$ (odd - numbered data set), the median is the middle - value. So the median is $6.60$.

Step3: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}}$.
$(x_1-\bar{x})^2=(3.1 - 11.22)^2=(-8.12)^2 = 65.9344$
$(x_2-\bar{x})^2=(6.1 - 11.22)^2=(-5.12)^2 = 26.2144$
$(x_3-\bar{x})^2=(6.6 - 11.22)^2=(-4.62)^2 = 21.3444$
$(x_4-\bar{x})^2=(11.2 - 11.22)^2=(-0.02)^2 = 0.0004$
$(x_5-\bar{x})^2=(29.1 - 11.22)^2=(17.88)^2 = 319.6944$
$\sum_{i = 1}^{5}(x_i-\bar{x})^2=65.9344 + 26.2144+21.3444 + 0.0004+319.6944=433.188$.
$s=\sqrt{\frac{433.188}{4}}=\sqrt{108.297}\approx10.41$

Answer:

Mean: $11.22$
Median: $6.60$
Sample Standard Deviation: $10.41$