First check if the given table represents a probability distribution by verifying all probabilities add up to 1.
Sum all probabilities:
0.02+0.04+0.06+0.08+0.13+0.25+0.19+0.15+0.08=1.0........ok
(a) mean of the distribution equals the expected value, E[x], where
E[x]=sum xi*p(xi) for xi=x1.....x9
=(0.02+0.08+0.18+0.32+0.65+1.5+1.33+1.2+0.72)
=6.0
(b) Variance = E[x^2]-(E[x])^2
where
E[x^2] = sum xi^2*p(xi) for xi=x1.....x9
=(0.02+0.16+0.54+1.28+3.25+9.0+9.31+9.6+6.48)
=39.64
=>
Variance
=E[x^2]-(E[x])^2
=39.64-6^2
=3.64
(c) Standard deviation
= sqrt(variance)
=sqrt(3.64)
=1.908
(d) expected value, E[x] = mean, as calculated in (a)
