x − 2 − 1 0 1 2 3 p ( x ) 0.1 k 0.2 2 k 0.3 3 k \begin{matrix}
x & -2 & -1& 0 & 1 & 2 & 3 \\
p(x) & 0.1 & k & 0.2 & 2k & 0.3 & 3k
\end{matrix} x p ( x ) − 2 0.1 − 1 k 0 0.2 1 2 k 2 0.3 3 3 k (i)
∑ i p ( x i ) = 1 \sum_ip(x_i)=1 i ∑ p ( x i ) = 1
0.1 + k + 0.2 + 2 k + 0.3 + 3 k = 1 0.1+k+0.2+2k+0.3+3k=1 0.1 + k + 0.2 + 2 k + 0.3 + 3 k = 1
6 k = 0.4 6k=0.4 6 k = 0.4
k = 1 15 k=\dfrac{1}{15} k = 15 1 (ii)
x − 2 − 1 0 1 2 3 p ( x ) 0.1 1 15 0.2 2 15 0.3 0.2 \begin{matrix}
x & -2 & -1& 0 & 1 & 2 & 3 \\ \\
p(x) & 0.1 & \dfrac{1}{15} & 0.2 & \dfrac{2}{15} & 0.3 & 0.2
\end{matrix} x p ( x ) − 2 0.1 − 1 15 1 0 0.2 1 15 2 2 0.3 3 0.2
P ( X < 2 ) = 1 − P ( X = 2 ) − P ( X = 3 ) P(X<2)=1-P(X=2)-P(X=3) P ( X < 2 ) = 1 − P ( X = 2 ) − P ( X = 3 )
= 1 − 0.3 − 0.2 = 0.5 =1-0.3-0.2=0.5 = 1 − 0.3 − 0.2 = 0.5
P ( − 2 < X < 2 ) = P ( X = − 1 ) + P ( X = 0 ) + P ( X = 1 ) P(-2<X<2)=P(X=-1)+P(X=0)+P(X=1) P ( − 2 < X < 2 ) = P ( X = − 1 ) + P ( X = 0 ) + P ( X = 1 )
= 1 15 + 0.2 + 2 15 = 0.4 =\dfrac{1}{15}+0.2+\dfrac{2}{15}=0.4 = 15 1 + 0.2 + 15 2 = 0.4
(iii)
F ( x ) = { 0 , for x < − 2 0.1 , for − 2 ≤ x < − 1 1 / 6 , for − 1 ≤ x < 0 11 / 30 , for 0 ≤ x < 1 0.5 , for 1 ≤ x < 2 0.8 , for 2 ≤ x < 3 1 , for x ≥ 3 F(x) = \begin{cases}
0, &\text{for } x<-2 \\
0.1, &\text{for } -2\leq x<-1 \\
1/6, &\text{for } -1\leq x<0\\
11/30, &\text{for } 0\leq x<1\\
0.5, &\text{for } 1\leq x<2 \\
0.8, &\text{for } 2\leq x<3 \\
1, &\text{for } x\geq3
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , 0.1 , 1/6 , 11/30 , 0.5 , 0.8 , 1 , for x < − 2 for − 2 ≤ x < − 1 for − 1 ≤ x < 0 for 0 ≤ x < 1 for 1 ≤ x < 2 for 2 ≤ x < 3 for x ≥ 3 (iv)
E ( X ) = − 2 ( 0.1 ) + ( − 1 ) ( 1 15 ) + 0 ( 0.2 ) + 1 ( 2 15 ) E(X)=-2(0.1)+(-1)(\dfrac{1}{15})+0(0.2)+1(\dfrac{2}{15}) E ( X ) = − 2 ( 0.1 ) + ( − 1 ) ( 15 1 ) + 0 ( 0.2 ) + 1 ( 15 2 )
+ 2 ( 0.3 ) + 3 ( 0.2 ) = 16 15 +2(0.3)+3(0.2)=\dfrac{16}{15} + 2 ( 0.3 ) + 3 ( 0.2 ) = 15 16 E ( X ) = 16 15 E(X)=\dfrac{16}{15} E ( X ) = 15 16
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