Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in
the original data.
-$$\begin{array} { l l l l l l l l l }
18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15
\end{array}$$
A) 1.6
B) 5.8
C) 5.4
D) 5.1

Question

Find the standard deviation for the given sample data. Round your answer to one more decimal place than is present in
the original data.
-$$\begin{array} { l l l l l l l l l } 
18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15
\end{array}$$
A) 1.6
B) 5.8
C) 5.4
D) 5.1

Quizplus · Accepted Answer

First, we need to find the mean of the data set:

$\frac{18+18+18+9+15+5+10+5+15}{9} = 12$

Next, we find each deviation from the mean:

$18-12=6$
$18-12=6$
$18-12=6$
$9-12=-3$
$15-12=3$
$5-12=-7$
$10-12=-2$
$5-12=-7$
$15-12=3$

We square each deviation:

$6^2=36$
$6^2=36$
$6^2=36$
$(-3)^2=9$
$3^2=9$
$(-7)^2=49$
$(-2)^2=4$
$(-7)^2=49$
$3^2=9$

We add the squared deviations:

$36+36+36+9+9+49+4+49+9=247$

We divide by the sample size, and take the square root:

$\sqrt{\frac{247}{9}} \approx 5.4$

Therefore, the answer is choice C, 5.4.

Find the Standard Deviation for the Given Sample Data 181818915510515\begin{array} { l l l l l l l l l } 18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15\end{array}18​18​18​9​15​5​10​5​15​

Find the Standard Deviation for the Given Sample Data $\begin{array} { l l l l l l l l l } 18 & 18 & 18 & 9 & 15 & 5 & 10 & 5 & 15\end{array}$