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)	0.0116	0.0834	0.2399	0.3462	0.2484	0.0715

$mu = 2.9499$ (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 $\mu$

$\mu=\sum [x\cdot P(x)]=0\times0.0116 + 1\times0.0834+2\times0.2399 + 3\times0.3462+4\times0.2484+5\times0.0715$
$=0 + 0.0834+0.4798+1.0386+0.9936+0.3575 = 2.9429$

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

$\sum [x^{2}\cdot P(x)]=0^{2}\times0.0116 + 1^{2}\times0.0834+2^{2}\times0.2399 + 3^{2}\times0.3462+4^{2}\times0.2484+5^{2}\times0.0715$
$=0\times0.0116+1\times0.0834 + 4\times0.2399+9\times0.3462+16\times0.2484+25\times0.0715$
$=0 + 0.0834+0.9596+3.1158+3.9744+1.7875=9.9207$

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

$\sigma^{2}=\sum [x^{2}\cdot P(x)]-\mu^{2}=9.9207-(2.9429)^{2}$
$=9.9207 - 8.6507=1.27$

Step4: Calculate the standard - deviation $\sigma$

$\sigma=\sqrt{\sigma^{2}}=\sqrt{1.27}\approx1.1269$

Answer:

$1.1269$