Question

Practice exercise for GED Can I find the correct forms doing mean variance and standard deviation discrete probability

Answer 1

Given

Probability distribution table

Find

Mean , variance and Standard Deviation

Explanation

As we have given

X = 0 , 1 , 2 , 3 , 4 , 5

P(X) = 0.15 , 0.20 , 0.35 , 0.22 , 0.07 , 0.01

Mean

Formula is given by

$\mu=\sum_^X\cdot P(X)$

first we find the values of X.P(X)

$\begin{gathered} \sum_^X\cdot P(X)=0+0.20+0.70+0.66+0.28+0.05 \\ \sum_^X\cdot P(X)=1.89 \end{gathered}$

Variance =

$\begin{gathered} \sum_^(x-\mu)^2P(x) \\ 0.54+0.16+0+0.27+0.31+0.10 \\ 1.3779\approx1.38 \end{gathered}$

standard deviation =

$\begin{gathered} \sqrt{variance} \\ \sqrt{1.38} \\ 1.174\approx1.17 \end{gathered}$

Final Answer

Therefore ,

Mean = 1.89

Variance = 1.38

Standard Deviation = 1.17