题目
Problem
A spinner can land on the numbers 10, 12, 14 and 16 only and the probability of the spinner landing on each number is the same.
The random variable X X X
(a) State the name of the probability distribution of X X X
(1)
(b) (i) Write down the value of E ( X ) E(X) E ( X )
(1)
(ii) Find Var ( X ) \operatorname{Var}(X) Var ( X )
(2)
A second spinner can land on the numbers 1, 2, 3, 4 and 5 only.
The random variable Y Y Y Y Y Y
y y y 1 2 3 4 5 P ( Y = y ) P(Y=y) P ( Y = y ) 4 30 \frac4{30} 30 4 9 30 \frac9{30} 30 9 6 30 \frac6{30} 30 6 5 30 \frac5{30} 30 5 6 30 \frac6{30} 30 6
(c) Find
(i) E ( Y ) E(Y) E ( Y )
(2)
(ii) Var ( Y ) \operatorname{Var}(Y) Var ( Y )
(3)
The random variable W = a X + b W=aX+b W = a X + b a a a b b b a > 0 a>0 a > 0
Given that E ( W ) = E ( Y ) E(W)=E(Y) E ( W ) = E ( Y ) Var ( W ) = Var ( Y ) \operatorname{Var}(W)=\operatorname{Var}(Y) Var ( W ) = Var ( Y )
(d) find the value of a a a b b b
(5)
Each of the two spinners is spun once.
(e) Find P ( W = Y ) P(W=Y) P ( W = Y )
(2)
题目中文翻译
第一个转盘只可能停在 10、12、14 和 16，且停在每个数字上的概率相同。随机变量 X X X
(a) 写出 X X X
(b) (i) 写出 E ( X ) E(X) E ( X )
(ii) 求 Var ( X ) \operatorname{Var}(X) Var ( X )
第二个转盘只可能停在 1、2、3、4 和 5。随机变量 Y Y Y
(c) 求：
(i) E ( Y ) E(Y) E ( Y )
(ii) Var ( Y ) \operatorname{Var}(Y) Var ( Y )
随机变量 W = a X + b W=aX+b W = a X + b a a a b b b a > 0 a>0 a > 0 E ( W ) = E ( Y ) E(W)=E(Y) E ( W ) = E ( Y ) Var ( W ) = Var ( Y ) \operatorname{Var}(W)=\operatorname{Var}(Y) Var ( W ) = Var ( Y )
(d) 求 a a a b b b
两个转盘各转动一次。
(e) 求 P ( W = Y ) P(W=Y) P ( W = Y )
解答
(a)
解法一
思路
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四个可能结果概率相同，所以是离散均匀分布。
答题过程
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X X X
(b)
(i)
解法一
思路
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四个取值关于 13 对称，因此期望为 13。
答题过程
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By symmetry,
E ( X ) = 13. \begin{align*}
E(X)=13.
\end{align*} E ( X ) = 13.
(ii)
解法一
思路
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先求 E ( X 2 ) E(X^2) E ( X 2 ) Var ( X ) = E ( X 2 ) − [ E ( X ) ] 2 \operatorname{Var}(X)=E(X^2)-[E(X)]^2 Var ( X ) = E ( X 2 ) − [ E ( X ) ] 2
答题过程
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E ( X 2 ) = 10 2 + 12 2 + 14 2 + 16 2 4 = 174. \begin{align*}
E(X^2)
=&\,\frac{10^2+12^2+14^2+16^2}{4}\\
=&\,174.
\end{align*} E ( X 2 ) = = 4 1 0 2 + 1 2 2 + 1 4 2 + 1 6 2 174. Therefore
Var ( X ) = 174 − 13 2 = 5. \begin{align*}
\operatorname{Var}(X)=174-13^2=5.
\end{align*} Var ( X ) = 174 − 1 3 2 = 5.
解法二
思路
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也可以从均值 13 直接计算平方离差。
答题过程
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Since E ( X ) = 13 E(X)=13 E ( X ) = 13
Var ( X ) = ( 10 − 13 ) 2 + ( 12 − 13 ) 2 4 + ( 14 − 13 ) 2 + ( 16 − 13 ) 2 4 = 9 + 1 + 1 + 9 4 = 5. \begin{align*}
\operatorname{Var}(X)
=&\,\frac{(10-13)^2+(12-13)^2}{4}\\
&\,\hspace{2pt}+\frac{(14-13)^2+(16-13)^2}{4}\\
=&\,\frac{9+1+1+9}{4}\\
=&\,5.
\end{align*} Var ( X ) = = = 4 ( 10 − 13 ) 2 + ( 12 − 13 ) 2 + 4 ( 14 − 13 ) 2 + ( 16 − 13 ) 2 4 9 + 1 + 1 + 9 5.
(c)
(i)
解法一
思路
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把每个取值乘以对应概率后相加。
答题过程
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E ( Y ) = 1 ( 4 ) + 2 ( 9 ) + 3 ( 6 ) 30 + 4 ( 5 ) + 5 ( 6 ) 30 = 90 30 = 3. \begin{align*}
E(Y)
=&\,\frac{1(4)+2(9)+3(6)}{30}\\
&\,\hspace{2pt}+\frac{4(5)+5(6)}{30}\\
=&\,\frac{90}{30}=3.
\end{align*} E ( Y ) = = 30 1 ( 4 ) + 2 ( 9 ) + 3 ( 6 ) + 30 4 ( 5 ) + 5 ( 6 ) 30 90 = 3.
(ii)
解法一
思路
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先算 E ( Y 2 ) E(Y^2) E ( Y 2 ) [ E ( Y ) ] 2 [E(Y)]^2 [ E ( Y ) ] 2
答题过程
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E ( Y 2 ) = 1 2 ( 4 ) + 2 2 ( 9 ) + 3 2 ( 6 ) 30 + 4 2 ( 5 ) + 5 2 ( 6 ) 30 = 324 30 = 10.8. \begin{align*}
E(Y^2)
=&\,\frac{1^2(4)+2^2(9)+3^2(6)}{30}\\
&\,\hspace{2pt}+\frac{4^2(5)+5^2(6)}{30}\\
=&\,\frac{324}{30}=10.8.
\end{align*} E ( Y 2 ) = = 30 1 2 ( 4 ) + 2 2 ( 9 ) + 3 2 ( 6 ) + 30 4 2 ( 5 ) + 5 2 ( 6 ) 30 324 = 10.8. Therefore
Var ( Y ) = 10.8 − 3 2 = 1.8. \begin{align*}
\operatorname{Var}(Y)=10.8-3^2=1.8.
\end{align*} Var ( Y ) = 10.8 − 3 2 = 1.8.
解法二
思路
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也可以直接计算 E [ ( Y − μ ) 2 ] E[(Y-\mu)^2] E [( Y − μ ) 2 ] μ = 3 \mu=3 μ = 3
答题过程
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Using E ( Y ) = 3 E(Y)=3 E ( Y ) = 3
Var ( Y ) = ( 1 − 3 ) 2 4 30 + ( 2 − 3 ) 2 9 30 + ( 3 − 3 ) 2 6 30 + ( 4 − 3 ) 2 5 30 + ( 5 − 3 ) 2 6 30 = 16 + 9 + 0 + 5 + 24 30 = 1.8. \begin{align*}
\operatorname{Var}(Y)
=&\,(1-3)^2\frac4{30}+(2-3)^2\frac9{30}\\
&\,\hspace{2pt}+(3-3)^2\frac6{30}\\
&\,\hspace{4pt}+(4-3)^2\frac5{30}+(5-3)^2\frac6{30}\\
=&\,\frac{16+9+0+5+24}{30}\\
=&\,1.8.
\end{align*} Var ( Y ) = = = ( 1 − 3 ) 2 30 4 + ( 2 − 3 ) 2 30 9 + ( 3 − 3 ) 2 30 6 + ( 4 − 3 ) 2 30 5 + ( 5 − 3 ) 2 30 6 30 16 + 9 + 0 + 5 + 24 1.8.
(d)
解法一
思路
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用线性变换公式建立方程。方差方程先给出 a a a b b b
答题过程
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Since W = a X + b W=aX+b W = a X + b
E ( W ) = a E ( X ) + b = 13 a + b . \begin{align*}
E(W)=aE(X)+b=13a+b.
\end{align*} E ( W ) = a E ( X ) + b = 13 a + b . Given E ( W ) = E ( Y ) = 3 E(W)=E(Y)=3 E ( W ) = E ( Y ) = 3
13 a + b = 3. \begin{align*}
13a+b=3.
\end{align*} 13 a + b = 3. Also,
Var ( W ) = a 2 Var ( X ) = 5 a 2 . \begin{align*}
\operatorname{Var}(W)=a^2\operatorname{Var}(X)=5a^2.
\end{align*} Var ( W ) = a 2 Var ( X ) = 5 a 2 . Given Var ( W ) = Var ( Y ) = 1.8 \operatorname{Var}(W)=\operatorname{Var}(Y)=1.8 Var ( W ) = Var ( Y ) = 1.8
5 a 2 = 1.8. \begin{align*}
5a^2=1.8.
\end{align*} 5 a 2 = 1.8. Since a > 0 a>0 a > 0
a = 1.8 5 = 0.6. \begin{align*}
a=\sqrt{\frac{1.8}{5}}=0.6.
\end{align*} a = 5 1.8 = 0.6. Then
13 ( 0.6 ) + b = 3 , \begin{align*}
13(0.6)+b=3,
\end{align*} 13 ( 0.6 ) + b = 3 , so
b = − 4.8. \begin{align*}
b=-4.8.
\end{align*} b = − 4.8.
(e)
解法一
思路
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把 X = 10 , 12 , 14 , 16 X=10,12,14,16 X = 10 , 12 , 14 , 16 W = 0.6 X − 4.8 W=0.6X-4.8 W = 0.6 X − 4.8 W W W Y Y Y
答题过程
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The possible values of W W W
0.6 ( 10 ) − 4.8 = 1.2 , 0.6 ( 12 ) − 4.8 = 2.4 , 0.6 ( 14 ) − 4.8 = 3.6 , 0.6 ( 16 ) − 4.8 = 4.8. \begin{align*}
0.6(10)-4.8=&\,1.2,\\
0.6(12)-4.8=&\,2.4,\\
0.6(14)-4.8=&\,3.6,\\
0.6(16)-4.8=&\,4.8.
\end{align*} 0.6 ( 10 ) − 4.8 = 0.6 ( 12 ) − 4.8 = 0.6 ( 14 ) − 4.8 = 0.6 ( 16 ) − 4.8 = 1.2 , 2.4 , 3.6 , 4.8. Since Y Y Y 1 , 2 , 3 , 4 , 5 1,2,3,4,5 1 , 2 , 3 , 4 , 5 W = Y W=Y W = Y
Therefore
P ( W = Y ) = 0. \begin{align*}
P(W=Y)=0.
\end{align*} P ( W = Y ) = 0. 