Question

mom week 13: box plots, frequency tables, standard deviation
score: 3/15 answered: 3/15
question 4
we are going to calculate the standard deviation for the following set of sample data (see the x values in the table for data values).

1. first, calculate the mean.

(\bar{x} =)

1. fill in the table below. fill in the differences of each data value from the mean, then the squared differences.
2. calculate the standard deviation.

(x)	(x - \bar{x})	((x - \bar{x})^2)
4
9
4
1
	total

standard deviation: (s = sqrt{\frac{sum (x - \bar{x})^2}{n - 1}}) (round to two decimal places)

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum x}{n}\). Here \(x = [6,4,9,4,1]\) and \(n = 5\).
\(\sum x=6 + 4+9+4+1=24\)
\(\bar{x}=\frac{24}{5}=4.8\)

Step2: Calculate \(x-\bar{x}\) and \((x - \bar{x})^2\)

• For \(x = 6\):

\(x-\bar{x}=6 - 4.8 = 1.2\)
\((x-\bar{x})^2=(1.2)^2 = 1.44\)

• For \(x = 4\):

\(x-\bar{x}=4 - 4.8=-0.8\)
\((x-\bar{x})^2=(-0.8)^2 = 0.64\)

• For \(x = 9\):

\(x-\bar{x}=9 - 4.8 = 4.2\)
\((x-\bar{x})^2=(4.2)^2 = 17.64\)

• For \(x = 4\):

\(x-\bar{x}=4 - 4.8=-0.8\)
\((x-\bar{x})^2=(-0.8)^2 = 0.64\)

• For \(x = 1\):

\(x-\bar{x}=1 - 4.8=-3.8\)
\((x-\bar{x})^2=(-3.8)^2 = 14.44\)

Step3: Calculate the sum of \((x-\bar{x})^2\)

\(\sum(x - \bar{x})^2=1.44+0.64 + 17.64+0.64+14.44=34.8\)

Step4: Calculate the standard deviation

The formula for the sample standard deviation \(s=\sqrt{\frac{\sum(x-\bar{x})^2}{n - 1}}\). Here \(n = 5\), so \(n-1=4\)
\(s=\sqrt{\frac{34.8}{4}}=\sqrt{8.7}\approx2.95\)

Answer:

1. \(\bar{x}=4.8\)

2)

\(x\)	\(x-\bar{x}\)	\((x - \bar{x})^2\)
\(4\)	\(-0.8\)	\(0.64\)
\(9\)	\(4.2\)	\(17.64\)
\(4\)	\(-0.8\)	\(0.64\)
\(1\)	\(-3.8\)	\(14.44\)

1. \(s\approx2.95\)