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IAL 2024 Oct Q6

A Level / Edexcel / S1

IAL 2024 Oct Paper · Question 6

题目

Problem

A biased die with six faces is rolled. The discrete random variable XX represents the score which is uppermost. The cumulative distribution function of XX is shown in the table below.

xx112233445566
F(x)F(x)0.10.10.20.23k3k5k5k7k7k10k10k

(a) Find the value of the constant kk.

(1)

(b) Find the probability distribution of XX.

(3)

A biased die with eight faces is rolled. The discrete random variable YY represents the score which is uppermost. The probability distribution of YY is shown in the table below, where aa and bb are constants.

yy1122334455667788
P(Y=y)P(Y=y)aaaaaabbbbbb0.110.110.050.05

Given that E(Y)=4.02E(Y)=4.02

(c) form and solve two equations in aa and bb to show that a=0.15a=0.15. You must show your working.

Solutions relying on calculator technology are not acceptable.

(3)

(d) Show that E(Y2)=20.7E(Y^2)=20.7.

(2)

(e) Find Var⁡(5−2Y)\operatorname{Var}(5-2Y).

(3)

These dice are each rolled once. The scores on the two dice are independent.

(f) Find the probability that the sum of these two scores is 33.

(2)

解答

(a)

解法一

思路

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CDF 最后一个值必须是 11，因为 P(X⩽6)=1P(X\leqslant6)=1。

答题过程

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Since XX can only take the values 1,2,3,4,5,61,2,3,4,5,6,

F(6)=1.\begin{align*} F(6)=1. \end{align*}

So

10k=1\begin{align*} 10k=1 \end{align*}

and hence

k=0.1.\begin{align*} k=0.1. \end{align*}

(b)

解法一

思路

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由 CDF 转成概率分布时，用相邻的累计概率相减：

P(X=x)=F(x)−F(x−1).\begin{align*} P(X=x)=F(x)-F(x-1). \end{align*}

这里先把 k=0.1k=0.1 代入。

答题过程

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Using k=0.1k=0.1, the cumulative probabilities are

0.1, 0.2, 0.3, 0.5, 0.7, 1.\begin{align*} 0.1,\ 0.2,\ 0.3,\ 0.5,\ 0.7,\ 1. \end{align*}

Therefore

x123456P(X=x)0.10.10.10.20.20.3\begin{array}{c|cccccc} x&1&2&3&4&5&6\\ \hline P(X=x)&0.1&0.1&0.1&0.2&0.2&0.3 \end{array}

(c)

解法一

思路

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第一条方程来自总概率等于 11。第二条方程来自期望 E(Y)=4.02E(Y)=4.02。再联立解出 aa 和 bb。

答题过程

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Since the probabilities sum to 11,

3a+3b+0.11+0.05= 13a+3b= 0.84.\begin{align*} 3a+3b+0.11+0.05=&\,1\\[3mm] 3a+3b=&\,0.84. \end{align*}

Using E(Y)=4.02E(Y)=4.02,

4.02= a+2a+3a+4b+5b+6b +7(0.11)+8(0.05)6a+15b+1.17= 4.026a+15b= 2.85.\begin{align*} 4.02 =&\,a+2a+3a+4b+5b+6b\\[3mm] &\,\hspace{2pt}+7(0.11)+8(0.05)\\[3mm] 6a+15b+1.17=&\,4.02\\[3mm] 6a+15b=&\,2.85. \end{align*}

From

3a+3b=0.84,\begin{align*} 3a+3b=0.84, \end{align*}

we get

6a+6b=1.68.\begin{align*} 6a+6b=1.68. \end{align*}

Subtracting this from 6a+15b=2.856a+15b=2.85,

9b= 1.17b= 0.13.\begin{align*} 9b=&\,1.17\\[3mm] b=&\,0.13. \end{align*}

Then

3a+3(0.13)= 0.843a= 0.45a= 0.15.\begin{align*} 3a+3(0.13)=&\,0.84\\[3mm] 3a=&\,0.45\\[3mm] a=&\,0.15. \end{align*}

So

a=0.15.\begin{align*} a=0.15. \end{align*}

(d)

解法一

思路

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先由上一小题得到 a=0.15a=0.15，b=0.13b=0.13。然后用

E(Y2)=∑y2P(Y=y).\begin{align*} E(Y^2)=\sum y^2P(Y=y). \end{align*}

答题过程

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Using a=0.15a=0.15 and b=0.13b=0.13,

E(Y2)= 12(0.15)+22(0.15)+32(0.15) +42(0.13)+52(0.13)+62(0.13) +72(0.11)+82(0.05)= 0.15+0.60+1.35 +2.08+3.25+4.68 +5.39+3.20= 20.7.\begin{align*} E(Y^2) =&\,1^2(0.15)+2^2(0.15)+3^2(0.15)\\[3mm] &\,\hspace{2pt}+4^2(0.13)+5^2(0.13)+6^2(0.13)\\[3mm] &\,\hspace{4pt}+7^2(0.11)+8^2(0.05)\\[3mm] =&\,0.15+0.60+1.35\\[3mm] &\,\hspace{2pt}+2.08+3.25+4.68\\[3mm] &\,\hspace{4pt}+5.39+3.20\\[3mm] =&\,20.7. \end{align*}

(e)

解法一

思路

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先求 Var⁡(Y)\operatorname{Var}(Y)。线性变换的方差公式是

Var⁡(a+bY)=b2Var⁡(Y).\begin{align*} \operatorname{Var}(a+bY)=b^2\operatorname{Var}(Y). \end{align*}

所以 Var⁡(5−2Y)=4Var⁡(Y)\operatorname{Var}(5-2Y)=4\operatorname{Var}(Y)。

答题过程

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First,

Var⁡(Y)= E(Y2)−[E(Y)]2= 20.7−(4.02)2= 4.5396.\begin{align*} \operatorname{Var}(Y) =&\,E(Y^2)-[E(Y)]^2\\[3mm] =&\,20.7-(4.02)^2\\[3mm] =&\,4.5396. \end{align*}

Therefore

Var⁡(5−2Y)= (−2)2Var⁡(Y)= 4(4.5396)= 18.1584.\begin{align*} \operatorname{Var}(5-2Y) =&\,(-2)^2\operatorname{Var}(Y)\\[3mm] =&\,4(4.5396)\\[3mm] =&\,18.1584. \end{align*}

So

Var⁡(5−2Y)=18.2\begin{align*} \operatorname{Var}(5-2Y)=18.2 \end{align*}

to 33 significant figures.

解法二

思路

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期望与方差定义展开法。我们可以直接利用方差的定义式 Var⁡(W)=E(W2)−[E(W)]2\operatorname{Var}(W) = E(W^2) - [E(W)]^2，令 W=5−2YW = 5-2Y。 首先，通过期望的线性性质计算 E(5−2Y)E(5-2Y)：

E(5−2Y)=5−2E(Y)\begin{align*} E(5-2Y) = 5 - 2E(Y) \end{align*}

其次，展开平方项 (5−2Y)2=25−20Y+4Y2(5-2Y)^2 = 25 - 20Y + 4Y^2，进而利用期望的线性性质计算 E((5−2Y)2)E((5-2Y)^2)：

E((5−2Y)2)=25−20E(Y)+4E(Y2)\begin{align*} E((5-2Y)^2) = 25 - 20E(Y) + 4E(Y^2) \end{align*}

最后代入 E(Y)=4.02E(Y) = 4.02 和 E(Y2)=20.7E(Y^2) = 20.7 求解。这种方法非常巧妙，可以直接避开计算 YY 的方差 Var⁡(Y)\operatorname{Var}(Y)。

答题过程

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Let W=5−2YW = 5 - 2Y. We will find Var⁡(W)=E(W2)−[E(W)]2\operatorname{Var}(W) = E(W^2) - [E(W)]^2.

First, calculate the expectation E(5−2Y)E(5-2Y) using E(Y)=4.02E(Y) = 4.02:

E(5−2Y)=  5−2E(Y)=  5−2(4.02)=  −3.04\begin{align*} E(5-2Y) =&\,\, 5 - 2E(Y)\\[3mm] =&\,\, 5 - 2(4.02)\\[3mm] =&\,\, -3.04 \end{align*}

Thus, the square of the expectation is:

[E(5−2Y)]2=(−3.04)2=9.2416.\begin{align*} [E(5-2Y)]^2 = (-3.04)^2 = 9.2416. \end{align*}

Next, find E((5−2Y)2)E((5-2Y)^2) using E(Y)=4.02E(Y) = 4.02 and E(Y2)=20.7E(Y^2) = 20.7:

E((5−2Y)2)=  E(25−20Y+4Y2)=  25−20E(Y)+4E(Y2)=  25−20(4.02)+4(20.7)=  25−80.4+82.8=  27.4.\begin{align*} E((5-2Y)^2) =&\,\, E(25 - 20Y + 4Y^2)\\[3mm] =&\,\, 25 - 20E(Y) + 4E(Y^2)\\[3mm] =&\,\, 25 - 20(4.02) + 4(20.7)\\[3mm] =&\,\, 25 - 80.4 + 82.8\\[3mm] =&\,\, 27.4. \end{align*}

Finally, calculate the variance:

Var⁡(5−2Y)=  E((5−2Y)2)−[E(5−2Y)]2=  27.4−9.2416=  18.1584.\begin{align*} \operatorname{Var}(5-2Y) =&\,\, E((5-2Y)^2) - [E(5-2Y)]^2\\[3mm] =&\,\, 27.4 - 9.2416\\[3mm] =&\,\, 18.1584. \end{align*}

So the variance is 18.218.2 (to 33 significant figures).

(f)

解法一

思路

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两个骰子的和为 33，只有两种情况：

(X,Y)=(1,2)or(2,1).\begin{align*} (X,Y)=(1,2)\quad\text{or}\quad (2,1). \end{align*}

两次掷骰独立，所以概率相乘后相加。

答题过程

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The possible pairs are

(X,Y)=(1,2)or(2,1).\begin{align*} (X,Y)=(1,2)\quad\text{or}\quad (2,1). \end{align*}

Since the scores are independent,

P(X+Y=3)= P(X=1)P(Y=2)+P(X=2)P(Y=1)= 0.1(0.15)+0.1(0.15)= 0.03.\begin{align*} P(X+Y=3) =&\,P(X=1)P(Y=2)+P(X=2)P(Y=1)\\[3mm] =&\,0.1(0.15)+0.1(0.15)\\[3mm] =&\,0.03. \end{align*}