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

Standard deviation is 3.2.

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 = (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{(1)/(N)\sum_(i=1)^(N) (x_i-\mu )^2`

`=\sqrt{(1)/(5)\sum_(i=1)^(5) (x_i-48.2 )^2`

`=\sqrt{(1)/(5) (15.21+7.29+7.84+0.36+21.16)}`

`=√(10.372)`

`=3.2205589577`

`\approx 3.2`

Answer 2

The standard deviation of a set of data is given by

`\sigma = \sqrt{ (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 = (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{ (1)/(5) \sum ( x_i - 48.2 )^2} =`

`= \sqrt{ (1)/(5) ( (3.9)^2 + (-2.7)^2 + (2.8)^2 + (0.6)^2 + (-4.6)^2 ) } =`

`= \sqrt{ (1)/(5) (51.86) } =3.2`