题目
Problem
The random variable Y ~ B(225, p)
Using a normal approximation, the probability that Y is at least 188 is 0.1056 to 4 decimal places.
(i) Show that p satisfies 145p^2 – 241p + 100 = 0 when the normal probability tables are used.
(ii) Hence find the value of p, justifying your answer.
(10)
(Total for Question 6 is 10 marks)
题目中文翻译
随机变量 Y ∼ B ( 225 , p ) Y \sim B(225,p) Y ∼ B ( 225 , p )
使用正态近似，Y Y Y
(i) 证明在使用正态分布表时，p p p 145 p 2 − 241 p + 100 = 0 145p^2-241p+100=0 145 p 2 − 241 p + 100 = 0
(ii) 据此求 p p p
（第 6 题共 10 分）
解答
(i)
解法一
思路
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用均值 225 p 225p 225 p 225 p ( 1 − p ) 225p(1-p) 225 p ( 1 − p ) 187.5 187.5 187.5 0.1056 0.1056 0.1056 0.8944 0.8944 0.8944 z = 1.25 z=1.25 z = 1.25
答题过程
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Using a normal approximation, let
X ∼ N ( 225 p , 225 p ( 1 − p ) ) . X\sim N\big(225p,225p(1-p)\big). X ∼ N ( 225 p , 225 p ( 1 − p ) ) . Applying the continuity correction,
P ( Y ⩾ 188 ) ≈ P ( X > 187.5 ) = 0.1056. P(Y\geqslant188)\approx P(X>187.5)=0.1056. P ( Y ⩾ 188 ) ≈ P ( X > 187.5 ) = 0.1056. Thus
P ( X ⩽ 187.5 ) = 0.8944. P(X\leqslant187.5)=0.8944. P ( X ⩽ 187.5 ) = 0.8944. From the normal probability tables,
Φ ( 1.25 ) = 0.8944. \Phi(1.25)=0.8944. Φ ( 1.25 ) = 0.8944. Therefore,
187.5 − 225 p 225 p ( 1 − p ) = 1.25. \frac{187.5-225p}{\sqrt{225p(1-p)}}=1.25. 225 p ( 1 − p ) 187.5 − 225 p = 1.25. Since 225 p ( 1 − p ) = 15 p ( 1 − p ) \sqrt{225p(1-p)}=15\sqrt{p(1-p)} 225 p ( 1 − p ) = 15 p ( 1 − p )
187.5 − 225 p = 18.75 p ( 1 − p ) . 187.5-225p=18.75\sqrt{p(1-p)}. 187.5 − 225 p = 18.75 p ( 1 − p ) . Dividing by 37.5 37.5 37.5
5 − 6 p = 1 2 p ( 1 − p ) . 5-6p=\frac12\sqrt{p(1-p)}. 5 − 6 p = 2 1 p ( 1 − p ) . Squaring and simplifying,
4 ( 5 − 6 p ) 2 = p ( 1 − p ) 4 ( 25 − 60 p + 36 p 2 ) = p − p 2 100 − 240 p + 144 p 2 = p − p 2 145 p 2 − 241 p + 100 = 0 , \begin{align*}
4(5-6p)^2=&\,p(1-p) \\[4mm]
4(25-60p+36p^2)=&\,p-p^2 \\[4mm]
100-240p+144p^2=&\,p-p^2 \\[4mm]
145p^2-241p+100=&\,0,
\end{align*} 4 ( 5 − 6 p ) 2 = 4 ( 25 − 60 p + 36 p 2 ) = 100 − 240 p + 144 p 2 = 145 p 2 − 241 p + 100 = p ( 1 − p ) p − p 2 p − p 2 0 , as required.
(ii)
解法一
思路
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因式分解 (i) 的二次方程会得到两个位于 [ 0 , 1 ] [0,1] [ 0 , 1 ] 187.5 − 225 p > 0 187.5-225p>0 187.5 − 225 p > 0
答题过程
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From part (i),
145 p 2 − 241 p + 100 = 0 ( 29 p − 25 ) ( 5 p − 4 ) = 0. \begin{align*}
145p^2-241p+100=&\,0 \\[4mm]
(29p-25)(5p-4)=&\,0.
\end{align*} 145 p 2 − 241 p + 100 = ( 29 p − 25 ) ( 5 p − 4 ) = 0 0. Hence,
p = 25 29 = 0.862069 … or p = 4 5 = 0.8. p=\frac{25}{29}=0.862069\ldots
\quad\text{or}\quad
p=\frac45=0.8. p = 29 25 = 0.862069 … or p = 5 4 = 0.8. However, the standardised value in part (i) is positive, so
187.5 − 225 p > 0 , 187.5-225p>0, 187.5 − 225 p > 0 , which requires p < 5 / 6 p<5/6 p < 5/6 25 / 29 25/29 25/29 0.5 0.5 0.5 0.1056 0.1056 0.1056
Therefore,
p = 0.8 . \boxed{p=0.8}. p = 0.8 . 