Question

calculating the standard deviation of a data set
december temperatures
what is the sample mean of the data set? round to the nearest hundredth.

what is the sample standard deviation of the sample? round to the nearest hundredth.

Explanation:

Step1: Calculate the sample mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here, $n = 9$, and $x_{1}=33,x_{2}=34,x_{3}=42,x_{4}=36,x_{5}=39,x_{6}=36,x_{7}=39,x_{8}=40,x_{9}=44$.
$\sum_{i=1}^{9}x_{i}=33 + 34+42+36+39+36+39+40+44=343$
$\bar{x}=\frac{343}{9}\approx38.11$

Step2: Calculate the sample standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$.
$(x_{1}-\bar{x})^{2}=(33 - 38.11)^{2}=(-5.11)^{2}=26.1121$
$(x_{2}-\bar{x})^{2}=(34 - 38.11)^{2}=(-4.11)^{2}=16.8921$
$(x_{3}-\bar{x})^{2}=(42 - 38.11)^{2}=(3.89)^{2}=15.1321$
$(x_{4}-\bar{x})^{2}=(36 - 38.11)^{2}=(-2.11)^{2}=4.4521$
$(x_{5}-\bar{x})^{2}=(39 - 38.11)^{2}=(0.89)^{2}=0.7921$
$(x_{6}-\bar{x})^{2}=(36 - 38.11)^{2}=(-2.11)^{2}=4.4521$
$(x_{7}-\bar{x})^{2}=(39 - 38.11)^{2}=(0.89)^{2}=0.7921$
$(x_{8}-\bar{x})^{2}=(40 - 38.11)^{2}=(1.89)^{2}=3.5721$
$(x_{9}-\bar{x})^{2}=(44 - 38.11)^{2}=(5.89)^{2}=34.6921$
$\sum_{i = 1}^{9}(x_{i}-\bar{x})^{2}=26.1121+16.8921 + 15.1321+4.4521+0.7921+4.4521+0.7921+3.5721+34.6921=106.88$
$s=\sqrt{\frac{106.88}{9 - 1}}=\sqrt{\frac{106.88}{8}}=\sqrt{13.36}\approx3.66$

Answer:

The sample mean is approximately $38.11$.
The sample standard deviation is approximately $3.66$.