Question

calculate the mean, variance, and standard deviation of the data shown. give your answers to two decimal places. (make sure you can do these computations by hand in case it shows up on an exam or quiz!) mean = variance = standard deviation = next question

Explanation:

Step1: Calculate the mean

The formula for the mean $\bar{x}$ of a set of data $x_1,x_2,\cdots,x_n$ is $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$.
Here, $n = 6$, $x_1=12.1$, $x_2 = 25.7$, $x_3=14.6$, $x_4=21.2$, $x_5=21.8$, $x_6=26.3$.
$\sum_{i=1}^{6}x_i=12.1 + 25.7+14.6+21.2+21.8+26.3=121.7$.
$\bar{x}=\frac{121.7}{6}\approx20.28$.

Step2: Calculate the variance

The formula for the variance $s^{2}$ is $s^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^{2}}{n - 1}$.
$(x_1-\bar{x})^{2}=(12.1 - 20.28)^{2}=(-8.18)^{2}=66.9124$.
$(x_2-\bar{x})^{2}=(25.7-20.28)^{2}=(5.42)^{2}=29.3764$.
$(x_3-\bar{x})^{2}=(14.6 - 20.28)^{2}=(-5.68)^{2}=32.2624$.
$(x_4-\bar{x})^{2}=(21.2-20.28)^{2}=(0.92)^{2}=0.8464$.
$(x_5-\bar{x})^{2}=(21.8-20.28)^{2}=(1.52)^{2}=2.3104$.
$(x_6-\bar{x})^{2}=(26.3-20.28)^{2}=(6.02)^{2}=36.2404$.
$\sum_{i = 1}^{6}(x_i-\bar{x})^{2}=66.9124+29.3764+32.2624+0.8464+2.3104+36.2404 = 167.9484$.
$s^{2}=\frac{167.9484}{6 - 1}=\frac{167.9484}{5}=33.59$.

Step3: Calculate the standard deviation

The formula for the standard deviation $s$ is $s=\sqrt{s^{2}}$.
Since $s^{2}=33.59$, then $s=\sqrt{33.59}\approx5.80$.

Answer:

Mean = $20.28$
Variance = $33.59$
Standard Deviation = $5.80$