Question

check here for instructional material to complete this problem. find $mu$ if $mu=sum xcdot p(x)$. then, find $sigma$ if $sigma^{2}=sum x^{2}cdot p(x)-mu^{2}$.

x	0	1	2	3	4	5
-	-	-	-	-	-	-
p(x)	0.5584	0.3451	0.0853	0.0105	0.0007	0.0000

$mu=square$ (simplify your answer. round to four decimal places as needed.) $sigma=square$ (simplify your answer. round to four decimal places as needed.)

Explanation:

Step1: Calculate the mean $\mu$

$\mu=\sum [x\cdot P(x)]=0\times0.5584 + 1\times0.3451+2\times0.0853 + 3\times0.0105+4\times0.0007+5\times0.0000$
$=0 + 0.3451+0.1706+0.0315 + 0.0028+0$
$=0.54999\approx0.5500$

Step2: Calculate $\sum [x^{2}\cdot P(x)]$

$\sum [x^{2}\cdot P(x)]=0^{2}\times0.5584 + 1^{2}\times0.3451+2^{2}\times0.0853 + 3^{2}\times0.0105+4^{2}\times0.0007+5^{2}\times0.0000$
$=0+0.3451 + 0.3412+0.0945+0.0112+0$
$=0.792$

Step3: Calculate the variance $\sigma^{2}$

$\sigma^{2}=\sum [x^{2}\cdot P(x)]-\mu^{2}=0.792-(0.5500)^{2}$
$=0.792 - 0.3025$
$=0.4895$

Step4: Calculate the standard - deviation $\sigma$

$\sigma=\sqrt{\sigma^{2}}=\sqrt{0.4895}\approx0.6996$

Answer:

$\mu = 0.5500$
$\sigma = 0.6996$