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IAL 2020 Oct Q6

A Level / Edexcel / S1

IAL 2020 Oct Paper · Question 6

题目

Problem

The random variable AA represents the score when a spinner is spun. The probability distribution for AA is given in the following table.

aa1457
P(A=a)P(A=a)0.400.200.250.15

(a) Show that E(A)=3.5E(A)=3.5

(2)

(b) Find Var⁡(A)\operatorname{Var}(A)

(3)

The random variable BB represents the score on a 4-sided die. The probability distribution for BB is given in the following table where kk is a positive integer.

bb134kk
P(B=b)P(B=b)0.250.250.250.25

(c) Write down the name of the probability distribution of BB.

(1)

(d) Given that E(B)=E(A)E(B)=E(A) state, giving a reason, the value of kk.

(1)

The random variable X∼N(μ,σ2)X\sim N(\mu,\sigma^2)

Sam and Tim are playing a game with the spinner and the die. They each spin the spinner once to obtain their value of AA and each roll the die once to obtain their value of BB. Their value of AA is taken as their value of μ\mu and their value of BB is taken as their value of σ\sigma. The person with the larger value of P(X>3.5)P(X>3.5) is the winner.

(e) Given that Sam obtained values of a=4a=4 and b=3b=3 and Tim obtained b=4b=4 find, giving a reason, the probability that Tim wins.

(2)

(f) Find the largest value of P(X>3.5)P(X>3.5) achievable in this game.

(4)

(g) Find the probability of achieving this value.

(2)

解答

(a)

解法一

思路

展开

这是 show that，直接按期望公式算到 3.53.5。

答题过程

展开 E(A)= 1(0.40)+4(0.20)+5(0.25)+7(0.15)= 0.40+0.80+1.25+1.05= 3.5.\begin{aligned} E(A) =&\,1(0.40)+4(0.20)+5(0.25)+7(0.15)\\ =&\,0.40+0.80+1.25+1.05\\ =&\,3.5. \end{aligned}

Hence E(A)=3.5E(A)=3.5, as required.

(b)

解法一

思路

展开

先求 E(A2)E(A^2)，再用 Var⁡(A)=E(A2)−[E(A)]2\operatorname{Var}(A)=E(A^2)-[E(A)]^2。

答题过程

展开 E(A2)= 12(0.40)+42(0.20)+52(0.25)+72(0.15)= 0.40+3.20+6.25+7.35= 17.2.\begin{aligned} E(A^2) =&\,1^2(0.40)+4^2(0.20)+5^2(0.25)+7^2(0.15)\\ =&\,0.40+3.20+6.25+7.35\\ =&\,17.2. \end{aligned}

Therefore

Var⁡(A)=17.2−3.52=4.95.\begin{align*} \operatorname{Var}(A)=17.2-3.5^2=4.95. \end{align*}

解法二

思路

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方差定义式法。 除了使用常用的计算简化公式 Var⁡(A)=E⁡(A2)−[E⁡(A)]2\operatorname{Var}(A) = \operatorname{E}(A^2) - [\operatorname{E}(A)]^2 之外，也可以直接利用方差的定义式：

Var⁡(A)=E⁡[(A−E⁡(A))2]=∑(ai−E⁡(A))2P(A=ai)\begin{align*} \operatorname{Var}(A) = \operatorname{E}[(A - \operatorname{E}(A))^2] = \sum (a_i - \operatorname{E}(A))^2 P(A = a_i) \end{align*}

由于在第一问中我们已经求得（并证明了） E⁡(A)=3.5\operatorname{E}(A) = 3.5，所以每一个偏差值 ai−3.5a_i - 3.5 都是容易计算的：

  • 对于 a=1a=1，偏差平方为 (1−3.5)2=6.25(1 - 3.5)^2 = 6.25；
  • 对于 a=4a=4，偏差平方为 (4−3.5)2=0.25(4 - 3.5)^2 = 0.25；
  • 对于 a=5a=5，偏差平方为 (5−3.5)2=2.25(5 - 3.5)^2 = 2.25；
  • 对于 a=7a=7，偏差平方为 (7−3.5)2=12.25(7 - 3.5)^2 = 12.25。 把这些偏差平方乘以各自对应的概率后相加，即可直接求出方差。这种方法对于深入理解方差的“均方偏差”本质非常有帮助，且在均值 E⁡(A)\operatorname{E}(A) 为比较整齐的数值（如本题中的 3.53.5）时，计算量其实也非常小。

答题过程

展开

Use the definition of variance:

Var⁡(A)=∑(ai−E⁡(A))2P(A=ai).\begin{align*} \operatorname{Var}(A) = \sum (a_i - \operatorname{E}(A))^2 P(A = a_i). \end{align*}

From part (a), E⁡(A)=3.5\operatorname{E}(A) = 3.5. Compute the squared deviation for each value of AA:

  • For a=1a=1: (1−3.5)2=(−2.5)2=6.25(1 - 3.5)^2 = (-2.5)^2 = 6.25
  • For a=4a=4: (4−3.5)2=(0.5)2=0.25(4 - 3.5)^2 = (0.5)^2 = 0.25
  • For a=5a=5: (5−3.5)2=(1.5)2=2.25(5 - 3.5)^2 = (1.5)^2 = 2.25
  • For a=7a=7: (7−3.5)2=(3.5)2=12.25(7 - 3.5)^2 = (3.5)^2 = 12.25

Multiply each squared deviation by its probability:

Var⁡(A)=  6.25(0.40)+0.25(0.20)+2.25(0.25)+12.25(0.15)=  2.50+0.05+0.5625+1.8375=  4.95.\begin{align*} \operatorname{Var}(A) =&\,\, 6.25(0.40) + 0.25(0.20) + 2.25(0.25) + 12.25(0.15)\\[3mm] =&\,\, 2.50 + 0.05 + 0.5625 + 1.8375\\[3mm] =&\,\, 4.95. \end{align*}

(c)

解法一

思路

展开

四个结果概率相同，所以是离散均匀分布。

答题过程

展开

BB has a discrete uniform distribution.

(d)

解法一

思路

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因为 BB 是四个等可能数值的均匀分布，平均值是四个数的平均。让它等于 3.53.5 即可。

答题过程

展开

Since E(B)=E(A)=3.5E(B)=E(A)=3.5,

1+3+4+k4=3.5.\begin{align*} \frac{1+3+4+k}{4}=3.5. \end{align*}

So

8+k=14,\begin{align*} 8+k=14, \end{align*}

and therefore

k=6.\begin{align*} k=6. \end{align*}

(e)

解法一

思路

展开

Sam 的参数是 μ=4,σ=3\mu=4,\sigma=3。Tim 已知 σ=4\sigma=4，需要他的 μ=A\mu=A 让 P(X>3.5)P(X>3.5) 比 Sam 大。对同一个临界值 3.53.5，标准化后的 zz 越小，右尾概率越大。

答题过程

展开

For Sam,

z=3.5−43=−16.\begin{align*} z=\frac{3.5-4}{3}=-\frac16. \end{align*}

For Tim to win, Tim needs

3.5−A4<−16.\begin{align*} \frac{3.5-A}{4}<-\frac16. \end{align*}

So

3.5−A<−23,\begin{align*} 3.5-A<-\frac23, \end{align*}

which gives

A>4.166….\begin{align*} A>4.166\ldots. \end{align*}

Thus Tim wins if A=5A=5 or A=7A=7.

Therefore

P(Tim wins)=0.25+0.15=0.40.\begin{align*} P(\text{Tim wins})=0.25+0.15=0.40. \end{align*}

(f)

解法一

思路

展开

要最大化 P(X>3.5)P(X>3.5)，需要让平均数 μ\mu 尽可能大、标准差 σ\sigma 尽可能小。所以取 μ=7,σ=1\mu=7,\sigma=1。

答题过程

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The largest value is obtained by taking

μ=7,σ=1.\begin{align*} \mu=7,\qquad \sigma=1. \end{align*}

Then

P(X>3.5)=P(Z>3.5−71)=P(Z>−3.5).P(X>3.5) =P\left(Z>\frac{3.5-7}{1}\right) =P(Z>-3.5).

Using symmetry,

P(Z>−3.5)=P(Z<3.5).\begin{align*} P(Z>-3.5)=P(Z<3.5). \end{align*}

Therefore

P(X>3.5)=0.9998.\begin{align*} P(X>3.5)=0.9998. \end{align*}

(g)

解法一

思路

展开

要达到 (f) 的最大值，必须 spinner 得到 A=7A=7，die 得到 B=1B=1。两个结果相乘。

答题过程

展开

The largest value requires

A=7andB=1.\begin{align*} A=7\quad\text{and}\quad B=1. \end{align*}

So

P(achieving this value)=P(A=7)P(B=1)=0.15(0.25).P(\text{achieving this value}) =P(A=7)P(B=1) =0.15(0.25).

Therefore

P(achieving this value)=0.0375.\begin{align*} P(\text{achieving this value})=0.0375. \end{align*}