Question

Find the variance and the standard deviation. Round your answer to one decimal place .

Answer 1

The variance is given by:

`\sigma^2=\sum ^{}_{}x^2p(x)-\mu^2`

In this case the mean is equal to 4.9 so lets calculate the sum first:

`\begin{gathered} \sum ^{}_{}x^2p(x)=3^2(0.3)^{}+4^2(0.1)^{}+5^2(0.2)^{}+6^2(0.2)^{}+7^2(0.2)^{} \\ =26.3 \end{gathered}`

Now that we have the sum we plug it in the expression for the variance and the value of the mean:

`\begin{gathered} \sigma^2=26.3-4.9^2 \\ \sigma^2=2.29 \end{gathered}`

Therefore the variance is 2.3 (rounded to one decimal place)

The standard deviation is given by:

`\sigma=\sqrt[]{\sigma^2}`

Then in this case we have:

`\begin{gathered} \sigma=\sqrt[]{2.29} \\ \sigma=1.5 \end{gathered}`

Therefore the standard deviation is 1.5