Answer:
Expected Grade=2 i.e., C
Variance=1.2
Step-by-step explanation:
[tex]Expected\ value=E\left [ x \right ]=\sum _{i=1}^{k} x_{i}p_{i}[/tex]
The x values are
A = 4
B = 3
C = 2
D = 1
F = 0
Probability of each of the events
P(4)=0.1
P(3)=0.2
P(2)=0.4
P(1)=0.2
P(0)=0.1
[tex]E\left [ x \right ]=4\times 0.1+3\times 0.2+2\times 0.4+1\times 0.2+0\times 0.1\\\therefore E\left [ x \right ]=2[/tex]
Variance
[tex]Var\left ( x\right)=E\left [ x^2 \right ]-E\left [ x \right ]^2[/tex]
[tex]E\left [ x^2 \right ]=4^2 \times 0.1+3^2 \times 0.2+2^2 \times 0.4+1^2 \times 0.2+0^2 \times 0.1\\\Rightarrow E\left [ x^2 \right ]=5.2\\E\left [ x \right ]^2=2^2=4\\\therefore Var\left ( x\right)=5.2-4=1.2\\[/tex]
