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IAL 2019 Oct Q7

A Level / Edexcel / S1

IAL 2019 Oct Paper · Question 7

题目

Problem

The number of cakes, XX, bought by customers at a particular shop has probability distribution

xx123456>6>6
P(X=x)P(X=x)0.35aaaa0.15bbbb0

where aa and bb are constants.

Given that E(X)=2.5E(X)=2.5

(a) (i) show that a=0.2a=0.2

(ii) find the value of bb

(5)

(b) Calculate Var⁡(4X+3)\operatorname{Var}(4X+3)

(4)

The cost to produce each cake is 20 cents and the shopkeeper sells each cake for 80 cents.

(c) Find the expected profit made by the shopkeeper for a randomly selected customer buying cakes.

(2)

The shopkeeper decides to run a promotion where she gives 1 free cake to all customers who buy 4 or more cakes. During the promotion the number of cakes, YY, taken away by a customer buying cakes has the following probability distribution

yy123456
P(Y=y)P(Y=y)340\frac3{40}440\frac4{40}340\frac3{40}02240\frac{22}{40}840\frac8{40}

(d) Find the expected profit made by the shopkeeper for a randomly selected customer buying cakes during the promotion.

(4)

解答

(a)

解法一

思路

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用总概率为 1 得到一个方程，再用 E(X)=2.5E(X)=2.5 得到另一个方程，联立求 a,ba,b。

答题过程

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From the sum of probabilities,

0.35+a+a+0.15+b+b=1.\begin{align*} 0.35+a+a+0.15+b+b=1. \end{align*}

So

2a+2b=0.5.\begin{align*} 2a+2b=0.5. \end{align*}

Hence

a+b=0.25.\begin{align*} a+b=0.25. \end{align*}

Using E(X)=2.5E(X)=2.5,

1(0.35)+2a+3a+4(0.15)+5b+6b=2.5.\begin{align*} 1(0.35)+2a+3a+4(0.15)+5b+6b=2.5. \end{align*}

So

5a+11b=1.55.\begin{align*} 5a+11b=1.55. \end{align*}

Since b=0.25−ab=0.25-a,

5a+11(0.25−a)=1.55.\begin{align*} 5a+11(0.25-a)=1.55. \end{align*}

Thus

5a+2.75−11a=1.55.\begin{align*} 5a+2.75-11a=1.55. \end{align*}

So

6a=1.20.\begin{align*} 6a=1.20. \end{align*}

Therefore

a=0.20.\begin{align*} a=0.20. \end{align*}

Then

b=0.25−0.20=0.05.\begin{align*} b=0.25-0.20=0.05. \end{align*}

(b)

解法一

思路

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先求 Var⁡(X)\operatorname{Var}(X)，再使用 Var⁡(4X+3)=16Var⁡(X)\operatorname{Var}(4X+3)=16\operatorname{Var}(X)。

答题过程

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First,

E(X2)= 12(0.35)+22(0.2)+32(0.2)+42(0.15)+52(0.05)+62(0.05)= 8.4.\begin{aligned} E(X^2) =&\,1^2(0.35)+2^2(0.2)+3^2(0.2)\\ &\quad+4^2(0.15)+5^2(0.05)+6^2(0.05)\\ =&\,8.4. \end{aligned}

Therefore

Var⁡(X)=8.4−2.52=2.15.\begin{align*} \operatorname{Var}(X)=8.4-2.5^2=2.15. \end{align*}

So

Var⁡(4X+3)=42Var⁡(X)=16(2.15)=34.4.\begin{align*} \operatorname{Var}(4X+3)=4^2\operatorname{Var}(X)=16(2.15)=34.4. \end{align*}

(c)

解法一

思路

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每个 cake 利润 80−20=6080-20=60 cents，平均卖出 E(X)=2.5E(X)=2.5 个。

答题过程

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The profit per cake is

80−20=60\begin{align*} 80-20=60 \end{align*}

cents.

Therefore the expected profit is

2.5(60)=150\begin{align*} 2.5(60)=150 \end{align*}

cents, or

$1.50.\begin{align*} \$1.50. \end{align*}

(d)

解法一

思路

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促销时 YY 是顾客带走的蛋糕数，但若 Y≥5Y\geq5，其中 1 个是免费的。也就是说 Y=5Y=5 时顾客付 4 个，利润 4×80−5×20=2204\times80-5\times20=220 cents；Y=6Y=6 时利润 5×80−6×20=2805\times80-6\times20=280 cents。

答题过程

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The profit, in cents, for each value of YY is

Y12356profit60120180220280\begin{array}{c|ccccc} Y&1&2&3&5&6\\ \hline \text{profit}&60&120&180&220&280 \end{array}

Therefore

E(profit)= 140{60(3)+120(4)+180(3)+220(22)+280(8)}= 207.\begin{aligned} E(\text{profit}) =&\,\frac{1}{40}\{60(3)+120(4)+180(3)\\ &\quad+220(22)+280(8)\}\\ =&\,207. \end{aligned}

So the expected profit is

207 cents.\begin{align*} 207\text{ cents}. \end{align*}

解法二

思路

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期望线性性质法。 设促销期间每位顾客带来的利润为随机变量 WW（单位为 cents），顾客带走的蛋糕数为随机变量 YY。 我们知道每个蛋糕成本为 20 cents，售价为 80 cents。

  • 如果顾客带走的蛋糕 Y⩽3Y \leqslant 3，他们付 YY 个蛋糕的钱，利润为 80Y−20Y=60Y80 Y - 20 Y = 60 Y。
  • 如果顾客带走的蛋糕 Y⩾5Y \geqslant 5，因为促销送了 1 个免费蛋糕，所以顾客实际付 Y−1Y - 1 个蛋糕的钱，利润为 80(Y−1)−20Y=60Y−8080(Y - 1) - 20 Y = 60 Y - 80。 我们可以引入指示变量 I{Y⩾5}I_{\{Y \geqslant 5\}}（表示带走蛋糕数是否不小于 5。若是则为 11，否则为 00），将利润 WW 统一表达为关于 YY 的线性组合：
W=60Y−80I{Y⩾5}\begin{align*} W = 60 Y - 80 I_{\{Y \geqslant 5\}} \end{align*}

根据期望的线性性质，预期的利润期望即为：

E⁡(W)=60E⁡(Y)−80E⁡(I{Y⩾5})=60E⁡(Y)−80P(Y⩾5)\begin{align*} \operatorname{E}(W) = 60 \operatorname{E}(Y) - 80 \operatorname{E}\left(I_{\{Y \geqslant 5\}}\right) = 60 \operatorname{E}(Y) - 80 P(Y \geqslant 5) \end{align*}

这种方法将复杂的分类讨论利润转化为对 YY 期望的线性修正，极大减少了算术运算量，且不易出错。

答题过程

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Let WW be the profit in cents from a randomly selected customer. Let YY be the number of cakes taken away. The relationship between WW and YY is:

  • If Y⩽3Y \leqslant 3: W=80Y−20Y=60YW = 80Y - 20Y = 60Y
  • If Y⩾5Y \geqslant 5: W=80(Y−1)−20Y=60Y−80W = 80(Y - 1) - 20Y = 60Y - 80

This can be written compactly using the indicator variable for {Y⩾5}\{Y \geqslant 5\}:

W=60Y−80I{Y⩾5}.\begin{align*} W = 60Y - 80 I_{\{Y \geqslant 5\}}. \end{align*}

By the linearity of expectation:

E⁡(W)=  60E⁡(Y)−80P(Y⩾5).\begin{align*} \operatorname{E}(W) =&\,\, 60 \operatorname{E}(Y) - 80 P(Y \geqslant 5). \end{align*}

First, calculate the expected number of cakes taken away, E⁡(Y)\operatorname{E}(Y):

E⁡(Y)=  1(340)+2(440)+3(340)+5(2240)+6(840)=  3+8+9+110+4840=  17840=  4.45.\begin{align*} \operatorname{E}(Y) =&\,\, 1\left(\frac{3}{40}\right) + 2\left(\frac{4}{40}\right) + 3\left(\frac{3}{40}\right) + 5\left(\frac{22}{40}\right) + 6\left(\frac{8}{40}\right)\\[3mm] =&\,\, \frac{3 + 8 + 9 + 110 + 48}{40}\\[3mm] =&\,\, \frac{178}{40}\\[3mm] =&\,\, 4.45. \end{align*}

Next, find P(Y⩾5)P(Y \geqslant 5):

P(Y⩾5)=  P(Y=5)+P(Y=6)=  2240+840=  3040=  0.75.\begin{align*} P(Y \geqslant 5) =&\,\, P(Y=5) + P(Y=6)\\[3mm] =&\,\, \frac{22}{40} + \frac{8}{40}\\[3mm] =&\,\, \frac{30}{40}\\[3mm] =&\,\, 0.75. \end{align*}

Substitute E⁡(Y)=4.45\operatorname{E}(Y) = 4.45 and P(Y⩾5)=0.75P(Y \geqslant 5) = 0.75 into the expectation formula:

E⁡(W)=  60(4.45)−80(0.75)=  267−60=  207.\begin{align*} \operatorname{E}(W) =&\,\, 60(4.45) - 80(0.75)\\[3mm] =&\,\, 267 - 60\\[3mm] =&\,\, 207. \end{align*}

So the expected profit is

207 cents.\begin{align*} 207\text{ cents}. \end{align*}