Answer to Question #219511 in Statistics and Probability for mimi

Question #219511

Random variable X has the following probability distribution.

x 0 1 3 4 5 6

P .X D x/ 0.1 0.1 0.3 0.2 0.1 0.2

(a) Find P .3  X  5/. (3)

(b) Find E .X/: (5)

(c) Find V ar .X/ (7)


1
Expert's answer
2021-07-22T08:48:04-0400
x013456P(X=x)0.10.10.30.20.10.2\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 3 & 4 & 5 & 6 \\ \hline P(X=x) & 0.1 & 0.1 & 0.3 & 0.2 & 0.1 & 0.2 \\ \end{array}


Check


0.1+0.1+0.3+0.2+0.1+0.2=10.1+0.1+0.3+0.2+0.1+0.2=1

(a)


P(3X5)=P(X=3)+P(X=4)+P(X=5)P(3\leq X\leq 5)=P(X=3)+P(X=4)+P(X=5)

=0.3+0.2+0.1=0.6=0.3+0.2+0.1=0.6

P(3<X<5)=P(X=4)=0.2P(3< X<5)=P(X=4)=0.2



(b)


E[X]=0(0.1)+1(0.1)+3(0.3)+4(0.2)E[X]=0(0.1)+1(0.1)+3(0.3)+4(0.2)

+5(0.1)+6(0.2)=3.5+5(0.1)+6(0.2)=3.5



(c)


E[X2]=02(0.1)+12(0.1)+32(0.3)+42(0.2)E[X^2]=0^2(0.1)+1^2(0.1)+3^2(0.3)+4^2(0.2)

+52(0.1)+62(0.2)=15.7+5^2(0.1)+6^2(0.2)=15.7




Var(X)=E[X2](E[X])2Var(X)=E[X^2]-(E[X])^2

=15.7(3.5)2=3.45=15.7-(3.5)^2=3.45


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