Question

1. calculate the population standard deviation (σ).

σ=√(∑(x−μ)^2/n)=5.35 (please round your answer to two decimal places)
question 9
a sample of size n=10 is drawn from a population. the data is shown below.
44.7 55.8 76.6 46 54.4
67.1 76.6 76.6 55.6 37.6
what is the range of this data set?
range = 39
what is the standard deviation of this data set? (remember, it is a sample.) please report the
with appropriate rounding, reporting 2 more decimal places than the original data.
stdev = 14.411
question 10

Explanation:

Step1: Calculate the mean ($\bar{x}$)

The formula for the mean of a sample is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$.
Sum of data: $44.7+55.8+76.6+46+54.4+67.1+76.6+76.6+55.6+37.6 = 590.6$
$n = 10$
$\bar{x}=\frac{590.6}{10}=59.06$

Step2: Calculate the squared differences $(x_{i}-\bar{x})^{2}$

For $x = 44.7$: $(44.7 - 59.06)^{2}=(-14.36)^{2}=206.2096$
For $x = 55.8$: $(55.8 - 59.06)^{2}=(-3.26)^{2}=10.6276$
For $x = 76.6$: $(76.6 - 59.06)^{2}=(17.54)^{2}=307.6516$ (do this for all 10 data points)

Sum of squared differences $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=206.2096+10.6276+307.6516+(46 - 59.06)^{2}+(54.4 - 59.06)^{2}+(67.1 - 59.06)^{2}+307.6516+307.6516+(55.6 - 59.06)^{2}+(37.6 - 59.06)^{2}$
$=206.2096+10.6276+307.6516 + 170.5636+21.7156+64.6416+307.6516+307.6516+11.9716+460.5316$
$=1868.216$

Step3: Calculate the sample variance $s^{2}$

The formula for sample variance is $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$
$n=10$, so $s^{2}=\frac{1868.216}{9}\approx207.58$

Step4: Calculate the sample standard deviation $s$

The formula for sample standard deviation is $s=\sqrt{s^{2}}$
$s=\sqrt{207.58}\approx14.41$

Answer:

$14.41$