Question

What is the standard deviation of the following data set rounded to the nearest tenth? 52.1, 45.5, 51, 48.8, 43.6

Answer 1

The standard deviation of a set of data is given by
$\sigma = \sqrt{ \frac{1}{N} \sum (x_i - \mu)^2}$
where
N is the number of data (in this case, N=5)
$x_i$ are the data
$\mu$ is the average value

Let's calculate the average value:
$\mu = \frac{52.1+45.5+51+48.8+43.6}{5}=48.2$

And now we can apply the formula to calculate the standard deviation:
$\sigma = \sqrt{ \frac{1}{5} \sum ( x_i - 48.2 )^2} =$
$= \sqrt{ \frac{1}{5} ( (3.9)^2 + (-2.7)^2 + (2.8)^2 + (0.6)^2 + (-4.6)^2 ) } =$
$= \sqrt{ \frac{1}{5} (51.86) } =3.2$

Answer 2

Answer:

Standard deviation is 3.2.

Step-by-step explanation:

Given data is,

52.1, 45.5, 51, 48.8, 43.6,

Let x represents the data points,

Now, mean of the data is,

$\mu = \frac{52.1 + 45.5 + 51 + 48.8 + 43.6}{5}=48.2$

Population size, N = 5,

Hence, the standard deviation of the following data set is,

$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N} (x_i-\mu )^2$

$=\sqrt{\frac{1}{5}\sum_{i=1}^{5} (x_i-48.2 )^2$

$=\sqrt{\frac{1}{5} (15.21+7.29+7.84+0.36+21.16)}$

$=\sqrt{10.372}$

$=3.2205589577$

$\approx 3.2$