Question #346920

3 suppose ‘X’ is normally distributed with mean 20 and variance 8 i.e X-NID


Find


A P(25≤X≤30)


B P(X≥25)


4 Given the following discrete probability destruction.


X 1 5 7 9


F(x) 1/6 2/6 2/6 1/6



Find


A the expected value of x E(X)


The variance of x V(X)


6 the probabilities that Abebe passes microeconomics is 2/3 and the probability that he passed statics 4/9. If the probability of passing both courses is 1/4 ,what is the probability that Abebe will pass at lease one of these courses

1
Expert's answer
2022-06-03T13:18:12-0400

3.

A.


P(25X30)=P(Z30208)P(25\le X\le30)=P(Z\le\dfrac{30-20}{\sqrt{8}})

P(Z<25208)P(Z3.5355)-P(Z<\dfrac{25-20}{\sqrt{8}})\approx P(Z\le3.5355)

P(Z<1.7678)0.0383-P(Z<1.7678)\approx0.0383

B.


P(X25)=1P(Z<25208)P( X\ge25)=1-P(Z<\dfrac{25-20}{\sqrt{8}})

1P(Z<1.7678)0.0385\approx1-P(Z<1.7678)\approx0.0385

4.

A.


E(X)=16(1)+26(5)+26(7)+16(9)E(X)=\dfrac{1}{6}(1)+\dfrac{2}{6}(5)+\dfrac{2}{6}(7)+\dfrac{1}{6}(9)


=173=\dfrac{17}{3}

B.


E(X2)=16(1)2+26(5)2+26(7)2+16(9)2=1153E(X^2)=\dfrac{1}{6}(1)^2+\dfrac{2}{6}(5)^2+\dfrac{2}{6}(7)^2+\dfrac{1}{6}(9)^2=\dfrac{115}{3}

Var(X)=E(X2)(E(X))2=1153(173)2Var(X)=E(X^2)-(E(X))^2=\dfrac{115}{3} -(\dfrac{17}{3})^2

=569=\dfrac{56}{9}

6.


P(MS)=P(M)+P(S)P(MS)P(M\cup S)=P(M)+P(S)-P(M\cap S)

=23+4914=3136=\dfrac{2}{3}+\dfrac{4}{9}-\dfrac{1}{4}=\dfrac{31}{36}



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