since ∑ x = 0 4 P ( X = x ) = 1 \sum\limits_{x=0}^{4}P(X=x) =1 x = 0 ∑ 4 P ( X = x ) = 1
therefore P ( X = 4 ) = 1 − ( 0.1 + 0.15 + 0.2 + 0.25 ) = 0.3 P(X=4)=1-(0.1+0.15+0.2+0.25)=0.3 P ( X = 4 ) = 1 − ( 0.1 + 0.15 + 0.2 + 0.25 ) = 0.3
therefore ,
In Question 01,
answer is (2)
P ( X = 4 ) = 0 P(X=4)=0 P ( X = 4 ) = 0 is incorrect and correct one is P ( X = 4 ) = 0.3 P(X=4)=0.3 P ( X = 4 ) = 0.3
In Question 02,
expected number E[X],
E [ X ] = ∑ i = 0 4 x P ( X = x ) E [ X ] = 0 ∗ 0.1 + 1 ∗ 0.15 + 2 ∗ 0.2 + 3 ∗ 0.25 + 4 ∗ 0.3 E [ X ] = 2.5 E[X]=\sum\limits_{i=0}^{4}xP(X=x)\\
E[X]=0*0.1+1*0.15+2*0.2\\\hspace {4 em}+3*0.25+4*0.3\\
E[X]=2.5 E [ X ] = i = 0 ∑ 4 x P ( X = x ) E [ X ] = 0 ∗ 0.1 + 1 ∗ 0.15 + 2 ∗ 0.2 + 3 ∗ 0.25 + 4 ∗ 0.3 E [ X ] = 2.5
answer is (3)
In Question 03,
standard deviation =v a r i a n c e = V ( X ) \sqrt{variance}=\sqrt{V(X)} v a r ian ce = V ( X )
V ( X ) = E [ X 2 ] − E [ X ] 2 E [ X 2 ] = ∑ i = 0 4 x 2 P ( X = x ) E [ X 2 ] = 0 2 ∗ 0.1 + 1 2 ∗ 0.15 + 2 2 ∗ 0.2 + 3 2 ∗ 0.25 + 4 2 ∗ 0.3 E [ X 2 ] = 8 V ( X ) = 8 − 2. 5 2 = 1.75 s t a n d a r d d e v i a t i o n = 1.75 = 1.322 V(X)=E[X^2]-E[X]^2\\
E[X^2]=\sum\limits_{i=0}^{4}x^2P(X=x)\\
E[X^2]=0^2*0.1+1^2*0.15+2^2*0.2\\\hspace {4 em}+3^2*0.25+4^2*0.3\\
E[X^2]=8\\
V(X)=8-2.5^2=1.75\\
standard\ deviation=\sqrt{1.75}=1.322 V ( X ) = E [ X 2 ] − E [ X ] 2 E [ X 2 ] = i = 0 ∑ 4 x 2 P ( X = x ) E [ X 2 ] = 0 2 ∗ 0.1 + 1 2 ∗ 0.15 + 2 2 ∗ 0.2 + 3 2 ∗ 0.25 + 4 2 ∗ 0.3 E [ X 2 ] = 8 V ( X ) = 8 − 2. 5 2 = 1.75 s t an d a r d d e v ia t i o n = 1.75 = 1.322
answer is (1)
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