Question 1070493
If it's a distribution, 
{{{0<=P(x)<=1}}}
{{{sum(P)=1}}}
First one, check.
Second one,
{{{.03+.15+.29+.26+.16+.11=1}}}
Check.
.
.
.

{{{mu=sum(x*P(x))=0*.03+1*.15+2*.29+3*.26+4*.16+5*.11=2.7}}}
{{{sigma^2=sum((x-mu)^2*P(x))=(0-2.7)^2*.03+(1-2.7)^2*.15+(2-2.7)^2*.29+(3-2.7)^2*.26+(4-2.7)^2*.16+(5-2.7)^2*.11=1.67}}}
{{{sigma=1.29}}}
So then a value of {{{X=0}}} would be 
{{{2.7-N*1.29=0}}}
{{{N=2.1}}}
This value is more than two standard deviations away from the mean. 
Yes, it's unusual.