Let π~π(40,144) and if π = 2π β 1
Find the following probabilities:
(i) π(π β€ 40)
(ii) π(π β₯ 50)
(iii)π(35 β€ π β€ 45)
(iv)π(β45 β€ π β€ 45)
(v)π(β50 β€ π β€ 100)
If π~π(40,144) then the mathematical expectation is "a(X)=40", variance is "{\\sigma ^2}(X) = 144 \\Rightarrow \\sigma (X) = 12" .
Then
"a(Y) = a(2X - 1) = 2a(X) - a(1) = 2 \\cdot 40 - 1 = 79"
"{\\sigma ^2}(Y) = {\\sigma ^2}(2X - 1) = {2^2}{\\sigma ^2}(X) - {\\sigma ^2}(1) = 4{\\sigma ^2}(X) = 4 \\cdot 144 = {\\rm{576}} \\Rightarrow \\sigma (Y) = 24"
(i) Let's find
"P(X \\le 40) = \\Phi \\left( {\\frac{{\\beta - a(X)}}{{\\sigma (X)}}} \\right) - \\Phi \\left( { - \\infty } \\right) = \\Phi \\left( {\\frac{{40 - 40}}{{12}}} \\right) + 0.5 = 0 + 0.5 = 0.5"
Answer: "P(X \\le 40) = 0.5"
(ii) "P(Y \\ge 50) = \\Phi \\left( \\infty \\right) - \\Phi \\left( {\\frac{{\\alpha - a(Y)}}{{\\sigma (Y)}}} \\right) = 0.5 - \\Phi \\left( {\\frac{{50 - 79}}{{24}}} \\right) = 0.5 + \\Phi \\left( {1.2} \\right) = 0.5 + 0.3849 = 0.8849"
Answer: "P(Y \\ge 50) = 0.8849"
(iii) "P(35 \\le X \\le 45) = \\Phi \\left( {\\frac{{\\beta - a(X)}}{{\\sigma (X)}}} \\right) - \\Phi \\left( {\\frac{{\\alpha - a(X)}}{{\\sigma (X)}}} \\right) = \\Phi \\left( {\\frac{{45 - 40}}{{12}}} \\right) + \\Phi \\left( {\\frac{{35 - 40}}{{12}}} \\right) = \\Phi (0.42) + \\Phi (0.42) = 2 \\cdot 0.1628 = 0.3256"
Answer: "P(35 \\le X \\le 45) = 0.3256"
(iv) "P( - 45 \\le Y \\le 45) = \\Phi \\left( {\\frac{{\\beta - a(Y)}}{{\\sigma (Y)}}} \\right) - \\Phi \\left( {\\frac{{\\alpha - a(Y)}}{{\\sigma (Y)}}} \\right) = \\Phi \\left( {\\frac{{45 - 79}}{{24}}} \\right) - \\Phi \\left( {\\frac{{ - 45 - 79}}{{24}}} \\right) = - \\Phi (1.42) + \\Phi (5.17) = - 0.4222 + 0.5 = 0.0778"
Answer: "P( - 45 \\le Y \\le 45) = 0.0778"
(v) "P( - 50 \\le Y \\le 100) = \\Phi \\left( {\\frac{{100 - 79}}{{24}}} \\right) - \\Phi \\left( {\\frac{{ - 50 - 79}}{{24}}} \\right) = \\Phi (0.875) + \\Phi (5.375) = 0.3092 + 0.5 = 0.8092"
Answer: "P( - 50 \\le Y \\le 100) = 0.8092"
Comments
Leave a comment