Question

how to find the variance and round to one decimal place

Answer 1

The variance of a probability distribution is given as:

$\sigma^2=E(x^2)-\lbrack E(x)\rbrack^2$

Where,

$E(x^2)=\sum ^{\text{ }}_ix^2_iP(x_i)\text{ and }E(x)=\sum ^{\text{ }}_ix^{}_iP(x_i)$

From the table, the expressions are:

$E(x^2)=7^2(0.1)+8^2(0.2)+9^2(0.3)+10^2(0.2)+11^2(0.2)=86.2$

For E(x), it follows:

$E(x)=7(0.1)+8(0.2)+9(0.3)+10(0.2)+11(0.2)=9.2$

Substitute these values into the formula for variance:

$\sigma^2=86.2-9.2^2=86.2-84.64=1.56\approx1.6$

The variance is approximately 1.6