The discrete random variable X has the probability distribut

The discrete random variable X has the probability distribution shown here:

X -2 -1 0 1 2

P(X=x) 0.15 0.10 0.35 0.30 ?

(a) Find P(-1 < X < 1)

The discrete random variable X has the probability distribution shown here: X -2 -1 0 1 2 P(X=x) 0.15 0.10 0.35 0.30 ? (a) Find P(-1

Solution

(a) P(-1<X<1)=P(X=0)=0.35

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(b) E(x)=sum of x*f(x)

=(-2)*0.15+(-1)*0.1+0*0.35+1*0.3+2*0.1

=0.1

E(x^2)=sum of x^2*f(x)

=(-2)^2*0.15+(-1)^2*0.1+0*0.35+1*0.3+2^2*0.1

=1.4

So Var(X)=E(X^2)-[E(X)]^2

=1.4-0.1^2

=1.39

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(c) standard deviation =sqrt(1.39)=1.178983

So the probability is

P(mu-o <X<mu+o) = P(0.1-1.178983 <X<0.1+1.178983)

=P(-1.078983<X<1.278983)

=P(X=-1)+P(X=0)+P(X=1)

=0.1+0.35+0.3

=0.75