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IAL 2021 Oct Q5

A Level / Edexcel / S1

IAL 2021 Oct Paper · Question 5

题目

Problem

The discrete random variable YY has the following probability distribution

y−9−5059P(Y=y)qrurq\begin{array}{c|ccccc} y&-9&-5&0&5&9\\ \hline P(Y=y)&q&r&u&r&q \end{array}

where qq, rr and uu are probabilities.

(a) Write down the value of E⁡(Y)\operatorname{E}(Y)

(1)

The cumulative distribution function of YY is F(y)F(y)

Given that F(0)=1930F(0)=\dfrac{19}{30}

(b) show that the value of uu is 415\dfrac{4}{15}

(3)

Given also that Var⁡(Y)=37\operatorname{Var}(Y)=37

(c) find the value of qq and the value of rr

(4)

The coordinates of a point PP are (12,Y)(12,Y) The random variable DD represents the length of OPOP

(d) Find the probability distribution of DD

(6)

解答

(a)

解法一

思路

展开

分布关于 00 对称，所以期望为 00。

答题过程

展开 E⁡(Y)=0.\begin{align*} \operatorname{E}(Y)=0. \end{align*}

(b)

解法一

思路

展开

F(0)=P(Y⩽0)=q+r+uF(0)=P(Y\leqslant0)=q+r+u。总概率给出 2(q+r)+u=12(q+r)+u=1。联立即可求 uu。

答题过程

展开

From F(0)=1930F(0)=\dfrac{19}{30},

q+r+u=1930.\begin{align*} q+r+u=\frac{19}{30}. \end{align*}

Also, the total probability is 11, so

2(q+r)+u=1.\begin{align*} 2(q+r)+u=1. \end{align*}

Using q+r=1930−uq+r=\dfrac{19}{30}-u,

2(1930−u)+u=1.\begin{align*} 2\left(\frac{19}{30}-u\right)+u=1. \end{align*}

Hence

3830−u=1,\begin{align*} \frac{38}{30}-u=1, \end{align*}

so

u=830=415.\begin{align*} u=\frac{8}{30}=\frac{4}{15}. \end{align*}

(c)

解法一

思路

展开

因为 E⁡(Y)=0\operatorname{E}(Y)=0，所以 Var⁡(Y)=E⁡(Y2)\operatorname{Var}(Y)=\operatorname{E}(Y^2)。再与 q+r=1130q+r=\dfrac{11}{30} 联立。

答题过程

展开

Since

u=415,\begin{align*} u=\frac{4}{15}, \end{align*}

we have

q+r=1−u2=1130.\begin{align*} q+r=\frac{1-u}{2}=\frac{11}{30}. \end{align*}

Also,

Var⁡(Y)= E⁡(Y2)= 81q+25r+0u+25r+81q= 162q+50r.\begin{align*} \operatorname{Var}(Y) =&\,\operatorname{E}(Y^2)\\[3mm] =&\,81q+25r+0u+25r+81q\\[3mm] =&\,162q+50r. \end{align*}

Given that Var⁡(Y)=37\operatorname{Var}(Y)=37,

162q+50r=37.\begin{align*} 162q+50r=37. \end{align*}

Using r=1130−qr=\dfrac{11}{30}-q,

162q+50(1130−q)=37.\begin{align*} 162q+50\left(\frac{11}{30}-q\right)=37. \end{align*}

So

112q=553.\begin{align*} 112q=\frac{55}{3}. \end{align*}

Therefore

q=16.\begin{align*} q=\frac{1}{6}. \end{align*}

Then

r=1130−16=15.\begin{align*} r=\frac{11}{30}-\frac{1}{6}=\frac{1}{5}. \end{align*}

(d)

解法一

思路

展开

点 P=(12,Y)P=(12,Y)，所以 D=122+Y2D=\sqrt{12^2+Y^2}。当 Y=0,±5,±9Y=0,\pm5,\pm9 时，DD 分别为 12,13,1512,13,15。

答题过程

展开 D=122+Y2.\begin{align*} D=\sqrt{12^2+Y^2}. \end{align*}

If Y=0Y=0,

D=12.\begin{align*} D=12. \end{align*}

If Y=±5Y=\pm5,

D=122+52=13.\begin{align*} D=\sqrt{12^2+5^2}=13. \end{align*}

If Y=±9Y=\pm9,

D=122+92=15.\begin{align*} D=\sqrt{12^2+9^2}=15. \end{align*}

Therefore the probability distribution of DD is

d121315P(D=d)4152513\begin{array}{c|ccc} d&12&13&15\\ \hline P(D=d)&\frac{4}{15}&\frac{2}{5}&\frac{1}{3} \end{array}