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recall the defining formula used to compute the population standard dev…

Question

recall the defining formula used to compute the population standard deviation $sigma=sqrt{\frac{sum(x - mu)^2}{n}}$ where $x$ represents the individual values of the population, $n$ is the number of data values in the population, and $mu$ is the population mean. here the number of data values in the population is equal to the number of data values in the given data - set. recall that we previously determined that there are 5 values in the data set. therefore, $n = 5$. the population mean is calculated as $mu=\frac{sum x}{n}$. since $n = n$, we see that $mu=\frac{sum x}{n}=\frac{sum x}{n}=\bar{x}$. recall that we previously calculated $\bar{x}=4.8$. therefore, $mu = 4.8$, and our previous calculations for $(x-\bar{x})^2$ can be used for $(x - mu)^2$. now, with the information in the table compute the population standard deviation. simplify and round your final answer to two decimal places.

$x$$mu$$x-mu$$(x - mu)^2$
24.8- 2.87.84
54.80.20.04
74.82.24.84
94.84.217.64

$sigma=sqrt{\frac{sum(x - mu)^2}{n}}=sqrt{\frac{14.44 + 7.84+0.04 + 4.84+17.64}{5}}=sqrt{\frac{}{5}}=$

Explanation:

Step1: Identify values from the problem

We know $\mu = 4.8$ and $N=5$, and $\sum(x - \mu)^2=14.44 + 7.84+0.04 + 4.84+17.64$.

Step2: Calculate $\sum(x - \mu)^2$

$\sum(x - \mu)^2=14.44+7.84 + 0.04+4.84+17.64=44.8$.

Step3: Calculate the population standard - deviation $\sigma$

Using the formula $\sigma=\sqrt{\frac{\sum(x - \mu)^2}{N}}$, substitute $\sum(x - \mu)^2 = 44.8$ and $N = 5$ into it. So $\sigma=\sqrt{\frac{44.8}{5}}=\sqrt{8.96}\approx2.99$.

Answer:

$2.99$