题目
Problem
The random variable Y ∼ B ( 225 , p ) Y\sim\operatorname{B}(225,p) Y ∼ B ( 225 , p )
Using a normal approximation, the probability that Y Y Y
(i) Show that p p p
145 p 2 − 241 p + 100 = 0 \begin{align*}
&\,145p^2-241p \\[4mm]
&\,\hspace{2pt}+100=0
\end{align*} 145 p 2 − 241 p + 100 = 0 when the normal probability tables are used.
(ii) Hence find the value of p p p
(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 \begin{align*}
&\,145p^2-241p \\[4mm]
&\,\hspace{2pt}+100=0
\end{align*} 145 p 2 − 241 p + 100 = 0 (ii) 据此求 p p p
(10)
（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, write the mean and variance as
μ = 225 p σ 2 = 225 p ( 1 − p ) . \begin{align*}
\mu=&\,225p \\[4mm]
\sigma^2=&\,225p(1-p).
\end{align*} μ = σ 2 = 225 p 225 p ( 1 − p ) . Let
X ∼ N ( μ , σ 2 ) . X\sim N(\mu,\sigma^2). X ∼ N ( μ , σ 2 ) . Applying the continuity correction,
P ( Y ⩾ 188 ) ≈ P ( X > 187.5 ) = 0.1056. \begin{align*}
&\,P(Y\geqslant188) \\[4mm]
&\,\approx P(X>187.5) \\[4mm]
&\,=0.1056.
\end{align*} 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 − μ σ = 1.25. \frac{187.5-\mu}{\sigma}=1.25. σ 187.5 − μ = 1.25. The standard deviation is
σ = 225 p ( 1 − p ) = 15 p ( 1 − p ) . \begin{align*}
\sigma=&\,\sqrt{225p(1-p)} \\[4mm]
=&\,15\sqrt{p(1-p)}.
\end{align*} σ = = 225 p ( 1 − p ) 15 p ( 1 − p ) . Hence,
187.5 − 225 p = 1.25 σ . 187.5-225p=1.25\sigma. 187.5 − 225 p = 1.25 σ . Dividing by 37.5 37.5 37.5 σ \sigma σ
187.5 − 225 p 37.5 = 1.25 × 15 37.5 p ( 1 − p ) . \begin{align*}
&\,\frac{187.5-225p}{37.5} \\[4mm]
&\,=\frac{1.25\times15}{37.5}\sqrt{p(1-p)}.
\end{align*} 37.5 187.5 − 225 p = 37.5 1.25 × 15 p ( 1 − p ) . Thus,
5 − 6 p = 1 2 p ( 1 − p ) . \begin{align*}
&\,5-6p \\[4mm]
&\,=\frac12\sqrt{p(1-p)}.
\end{align*} 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 \\[4mm]
&\,=p(1-p) \\[4mm]
&\,4(25-60p+36p^2) \\[4mm]
&\,=p-p^2 \\[4mm]
&\,100-240p+144p^2 \\[4mm]
&\,=p-p^2 \\[4mm]
&\,145p^2-241p+100 \\[4mm]
&\,=0,
\end{align*} 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 , 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 \\[4mm]
&\,=0 \\[4mm]
&\,(29p-25)(5p-4) \\[4mm]
&\,=0.
\end{align*} 145 p 2 − 241 p + 100 = 0 ( 29 p − 25 ) ( 5 p − 4 ) = 0. Hence,
p = 25 29 = 0.862068 … \begin{align*}
p=&\,\frac{25}{29} \\[4mm]
=&\,0.862068\ldots
\end{align*} p = = 29 25 0.862068 … or
p = 4 5 = 0.8. p=\frac45=0.8. 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 . 