Question

II. Find the expected value, variance, and standard deviation of the following discrete 

probability distributions. (3 items x 15 points) 

 

1.

T -3 5 7 

P(T) 0.27 0.4 0.33 

 

2.

X 0 1 2 3 4 

P(x=X) 0.21 0.44 0.06 0.11 0.18 

3.

Y 27 30 14 43 

P(Y=y) 0.3 0.4 0.15 0.15

Answer 1

1.

$\begin{array}{|c|c|c|c|}\hline T& -3& 5& 7\\ \hline P(T=t)& 0.27& 0.4& 0.33\\ \hline\end{array}$

Check the sum of probabilities

$0.27+ 0.4+ 0.33=1$

The expected value is

$\bold{E}[T]= -3\cdot0.27+ 5\cdot0.4+ 7\cdot0.33 =3.5$

The variance is

$\bold{Var}[T]=\bold{E}[T^2]-\bold{E}^2[T]=$ $\bigg((-3)^2\cdot0.27+ 5^2\cdot0.4+ 7^2\cdot0.33\bigg)-(3.5)^2 =$ $16.35$

The standard deviation is

$\sigma[T]=\sqrt{\bold{Var}[T]}=\sqrt{16.35}\approx4.04$

2.

$\begin{array}{|c|c|c|c|c|c|}\hline X &0& 1& 2& 3& 4 \\ \hline P(X=x) &0.21& 0.44& 0.06& 0.11& 0.18\\ \hline\end{array}$

Check the sum of probabilities

$0.21+ 0.44 +0.06+ 0.11+ 0.18=1$

The expected value is

$\bold{E}[X]=$ $0\cdot0.21+1\cdot0.44+2\cdot0.06+3\cdot0.11+4\cdot0.18=$ $1.61$

The variance is

$\bold{Var}[X]=$ $(0^2\cdot0.21+1^2\cdot0.44+2^2\cdot0.06+3^2\cdot0.11+4^2\cdot0.18)-(1.61)^2\approx1.96$

The standard deviation is

$\sigma[X]=\sqrt{1.96}\approx1.40$

3.

$\begin{array}{|c|c|c|c|c|c|}\hline Y& 27& 30& 14& 43 \\ \hline P(Y=y)& 0.3& 0.4& 0.15& 0.15\\ \hline\end{array}$

Check the sum of probabilities

$0.3 +0.4+ 0.15+ 0.15=1$

The expected value is

$\bold{E}[Y]=27\cdot0.3+ 30\cdot0.4 + 14\cdot0.15 + 43\cdot0.15=$ $28.65$

The variance is

$\bold{Var}[Y]=$ $(27^2\cdot0.3+ 30^2\cdot0.4 + 14^2\cdot0.15 + 43^2\cdot0.15)-(28.65)^2\approx64.63$

The standard deviation is

$\sigma[Y]=\sqrt{64.63}\approx8.04$