Question #298304

Suppose that a pair of fair dice are to be tossed and let x the random variable denote the sum of the points.

compute the expected value

the variance of X

the standard deviation of x


1
Expert's answer
2022-02-16T17:21:54-0500
12345611+11+21+31+41+51+622+12+22+32+42+52+633+13+23+33+43+53+644+14+24+34+44+54+655+15+25+35+45+55+666+16+26+36+46+56+6\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c:c} & 1 & 2 &3 & 4 & 5 & 6 \\ \hline 1 & 1+1 & 1+2 & 1+3 & 1+ 4& 1+5 & 1+6 \\ \hdashline 2 & 2+1 & 2+2 & 2+3 & 2+4 & 2+5 & 2+6 \\ \hdashline 3 & 3+1 & 3+2 & 3+3 & 3+4 & 3+5 & 3+6 \\ \hdashline 4 & 4+1 & 4+2 & 4+3 & 4+4 & 4+5 & 4+6 \\ \hdashline 5 & 5+1 & 5+2 & 5+3 & 5+ 4& 5+5 & 5+6 \\ \hdashline 6 & 6+1 & 6+2 & 6+3 & 6+4 & 6+5 & 6+6 \\ \hdashline \end{array}


XfXfX2X2f21/362/3644/3632/366/36918/3643/3612/361648/3654/3620/3625100/3665/3630/3636180/3676/3642/3649294/3685/3640/3664320/3694/3636/3681324/36103/3630/36100300/36112/3622/36121242/36121/3612/36144144/36Sum17329/6\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & f & Xf &X^2 & X^2f\\ \hline & 2 & 1/36 & 2/36 & 4 & 4/36 \\ \hdashline & 3 & 2/36 & 6/36 & 9 & 18/36 \\ \hdashline & 4 & 3/36 & 12/36 & 16 & 48/36 \\ \hdashline & 5 & 4/36 & 20/36 & 25 & 100/36 \\ \hdashline & 6 & 5/36 & 30/36 & 36 & 180/36 \\ \hdashline & 7 & 6/36 & 42/36 & 49 & 294/36 \\ \hdashline & 8 & 5/36 & 40/36 & 64 & 320/36 \\ \hdashline & 9 & 4/36 & 36/36 & 81 & 324/36 \\ \hdashline & 10 & 3/36 & 30/36 & 100 & 300/36 \\ \hdashline & 11 & 2/36 & 22/36 & 121 & 242/36 \\ \hdashline & 12 & 1/36 & 12/36 & 144 & 144/36 \\ \hdashline Sum & & 1 & 7 & & 329/6 \end{array}

i)


E(X)=ixip(xi)=7E(X)=\sum_ix_ip(x_i)=7

ii)


Var(X)=σ2=E(X2)(E(X))2Var(X)=\sigma^2=E(X^2)-(E(X))^2

=329/672=35/65.83333=329/6-7^2=35/6\approx5.83333

iii)


σ=σ2=35/62.41523\sigma=\sqrt{\sigma^2}=\sqrt{35/6}\approx2.41523



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