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kunne noen vær så snill regnet denne oppgaven for meg! har ikke peiling



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goorgoor wrote:kunne noen vær så snill regnet denne oppgaven for meg! har ikke peiling

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Binomisk fordeling, Bin (n, k);
gjelder generelt for disse kastene:

[tex]P(X=k)=(4Ck)\cdot ({1\over 6})^k\cdot ({5\over 6})^{4-k}[/tex]

a)
[tex]P(X=0)=(4C0)\cdot ({1\over 6})^0\cdot ({5\over 6})^4=[/tex][tex]{625\over 1296}[/tex][tex]\approx\;0,482[/tex]

b)
[tex]P(X=1)=(4C1)\cdot ({1\over 6})^1\cdot ({5\over 6})^3[/tex][tex]={500\over 1296}[/tex][tex]\approx\;0,386[/tex]

c)
[tex]P(X=2)=(4C2)\cdot ({1\over 6})^2\cdot ({5\over 6})^2[/tex][tex]={25\over 216}[/tex][tex]\approx\;0,116[/tex]

d)
[tex]P(X=3)=(4C3)\cdot ({1\over 6})^3\cdot ({5\over 6})^1[/tex][tex]={5\over 324}[/tex][tex]\approx1.54\cdot 10^{-2}[/tex]

e)
[tex]P(X=4)=(4C4)\cdot ({1\over 6})^4\cdot ({5\over 6})^0[/tex][tex]={1\over 1296}[/tex][tex]\approx\;7,72\cdot 10^{-4}[/tex]

f)
[tex]E(X)=\mu=x\cdot P(X=k)=[/tex][tex]{500\over 1296}+{50\over 216}+{15\over 324}+{4\over 1296}={2\over 3}[/tex]

[tex]Var(X)=\sigma^2 = \Sigma_{k=0}^4(k-\mu)^2\cdot P(X=k)\;=\;{2\over 9}[/tex]

[tex]SD(X)=\sigma={sqrt 2\over 3}[/tex]

g)
[tex]\mu=4\cdot {1\over 6}={2\over 3}[/tex]

[tex]\sigma^2=4\cdot{1\over 6}\cdot {1\over 3}={2\over 9}[/tex]