91影视

Probability of sample is:

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

Add all probabilities are:

$P(X鈮�6)=0.04+0.03+0.06+0.08+0.09+0.08+0.05$

$=0.43$

The sum of all probabilities should be equal to $1$.

Now, find the missing probabilities.

$P(X=7)=1-P(X鈮�6)$

$=1-0.43$

$=0.57$

Probability of sample is:

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

The sample mean $渭=鈭�xP\left(x\right)$

Calculation:

Find the mean, use the formula $渭=鈭�xP\left(x\right)$.

$渭=鈭�xP\left(x\right)$

$=0脳0.04+1脳0.03+2脳0.06+3脳0.08+4脳0.09+5脳0.08+6脳0.05+7脳0.57$

$=5.44$

Hence, the mean of the random variable $X$is $5.44$.

Probability of sample is

$\begin{array}{|ll|}\hline \text{Days X}& \text{Probabilities}\\ 0& 0.04\\ 1& 0.03\\ 2& 0.06\\ 3& 0.08\\ 4& 0.09\\ 5& 0.08\\ 6& 0.05\\ \hline\end{array}$

Find the standard deviation, use the formula $蟽=\sqrt{鈭�}(x-渭{)}^{2}P(x)$.

$蟽=\sqrt{鈭�(x-渭{)}^{2}P(x)}$

$=\sqrt{\left[\begin{array}{c}(0-5.44{)}^2脳0.04+(1-5.44{)}^2脳0.03+(2-5.44{)}^2脳0.06\\ +(3-5.44{)}^2脳0.08+(4-5.44{)}^2脳0.09+(5-5.44{)}^2脳0.08\\ +(6-5.44{)}^2脳0.05+(7-5.44{)}^2脳0.57\\ =\end{array}\right]}$

localid="1649851646550" $=2.1369$