The variance is given by:
[tex]\sigma^2=\sum ^{}_{}x^2p(x)-\mu^2[/tex]
In this case the mean is equal to 4.9 so lets calculate the sum first:
[tex]\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}[/tex]
Now that we have the sum we plug it in the expression for the variance and the value of the mean:
[tex]\begin{gathered} \sigma^2=26.3-4.9^2 \\ \sigma^2=2.29 \end{gathered}[/tex]
Therefore the variance is 2.3 (rounded to one decimal place)
The standard deviation is given by:
[tex]\sigma=\sqrt[]{\sigma^2}[/tex]
Then in this case we have:
[tex]\begin{gathered} \sigma=\sqrt[]{2.29} \\ \sigma=1.5 \end{gathered}[/tex]
Therefore the standard deviation is 1.5
