Given
The data,
To find:
The variance, standard deviation, P(X ≥ -1), and P(X ≤ -3).
Explanation:
It is given that,
Then,
The variance is,
[tex]\begin{gathered} Var[x]=(-4-(-2.1))^2\times0.2+(-3-(-2.1))^2\times0.3+(-2-(-2.1))^2 \\ \times0.1+(-1-(-2.1))^2\times0.2+(0-(-2.1))^2\times0.2 \\ =(-4+2.1)^2\times0.2+(-3+2.1)^2\times0.3+(-2+2.1)^2\times0.1+(-1+2.1)^2 \\ \times0.2+(2.1)^2\times0.2 \\ =3.61\times0.2+0.81\times0.3+0.01\times0.1+1.21\times0.2+4.41\times0.2 \\ =0.722+0.243+0.001+0.242+0.882 \\ =2.09 \end{gathered}[/tex]
And the standard deviation is,
[tex]\begin{gathered} SD=\sqrt{Var[x]} \\ =\sqrt{2.09} \\ =1.45 \end{gathered}[/tex]
Also,
[tex]\begin{gathered} P\left(X≥-1\right)=P(X=-1)+P(X=0) \\ =0.2+0.2 \\ =0.4 \\ P\left(X≤-3\right)=P(-4)+P(-3) \\ =0.2+0.3 \\ =0.5 \end{gathered}[/tex]
Hence, the answers are,
Variance is 2.09
Standard deviation is 1.45
P(X ≥ -1) is 0.4
P(X ≤ -3) is 0.5.
