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IAL 2023 Oct Q5

A Level / Edexcel / S1

IAL 2023 Oct Paper · Question 5

题目

Problem

The weights, XX grams, of a particular variety of fruit are normally distributed with

X∼N(210,252)X\sim N(210,25^2)

A fruit of this variety is selected at random.

(a) Show that the probability that the weight of this fruit is less than 240240 grams is 0.88490.8849

(2)

(b) Find the probability that the weight of this fruit is between 190190 grams and 240240 grams.

(2)

(c) Find the value of kk such that P(210−k<X<210+k)=0.95P(210-k<X<210+k)=0.95

(3)

A wholesaler buys large numbers of this variety of fruit and classifies the lightest 15%15\% as small.

(d) Find the maximum weight of a fruit that is classified as small. You must show your working clearly.

(3)

The weights, YY grams, of a second variety of this fruit are normally distributed with

Y∼N(μ,σ2)Y\sim N(\mu,\sigma^2)

Given that 5%5\% of these fruit weigh less than 152152 grams and 40%40\% weigh more than 180180 grams,

(e) calculate the mean and standard deviation of the weights of this variety of fruit.

(5)

解答

(a)

解法一

思路

展开

先标准化 240240。这里方差写成 25225^2，所以标准差是 2525。

答题过程

展开 P(X<240)= P(Z<240−21025)= P(Z<1.2)= 0.8849.\begin{align*} P(X<240) =&\,P\left(Z<\frac{240-210}{25}\right)\\[3mm] =&\,P(Z<1.2)\\[3mm] =&\,0.8849. \end{align*}

This shows the required probability.

(b)

解法一

思路

展开

用上一小题的 P(X<240)P(X<240)，再减去 P(X<190)P(X<190)。

答题过程

展开 P(190<X<240)= P(X<240)−P(X<190)= 0.8849−P(Z<190−21025)= 0.8849−P(Z<−0.8)= 0.8849−0.2119= 0.6730.\begin{align*} P(190<X<240) =&\,P(X<240)-P(X<190)\\[3mm] =&\,0.8849-P\left(Z<\frac{190-210}{25}\right)\\[3mm] =&\,0.8849-P(Z<-0.8)\\[3mm] =&\,0.8849-0.2119\\[3mm] =&\,0.6730. \end{align*}

Therefore the probability is

0.673.\begin{align*} 0.673. \end{align*}

解法二

思路

展开

跨均值对称区间公式。当我们要求对称于均值两侧的区间概率 P(a<X<b)P(a < X < b)（其中 a<μ<ba < \mu < b）时，标准化后对应区间为 [−za,zb][-z_a, z_b]（其中 za,zb>0z_a, z_b > 0）。 利用标准正态分布的性质，该概率可以直接展开为：

P(−za<Z<zb)=Φ(zb)−Φ(−za)=Φ(zb)−(1−Φ(za))=Φ(zb)+Φ(za)−1\begin{align*} P(-z_a < Z < z_b) = \Phi(z_b) - \Phi(-z_a) = \Phi(z_b) - (1 - \Phi(z_a)) = \Phi(z_b) + \Phi(z_a) - 1 \end{align*}

在本题中，对应区间为 [−0.8,1.2][-0.8, 1.2]。因此：

P(−0.8<Z<1.2)=Φ(1.2)+Φ(0.8)−1\begin{align*} P(-0.8 < Z < 1.2) = \Phi(1.2) + \Phi(0.8) - 1 \end{align*}

我们查表直接得到两个累积概率，并相加再减去 11。这可以有效避免多重括号和减去补集的复杂步骤，是计算此类跨均值区间概率非常经典的快捷公式。

答题过程

展开

Standardise the boundaries 190190 and 240240:

P(190<X<240)=  P(190−21025<Z<240−21025)=  P(−0.8<Z<1.2).\begin{align*} P(190 < X < 240) =&\,\, P\left(\frac{190 - 210}{25} < Z < \frac{240 - 210}{25}\right)\\[3mm] =&\,\, P(-0.8 < Z < 1.2). \end{align*}

Using the interval identity for standard normal variables spanning across 00:

P(−a<Z<b)=Φ(b)+Φ(a)−1\begin{align*} P(-a < Z < b) = \Phi(b) + \Phi(a) - 1 \end{align*}

where a,b>0a, b > 0.

Substitute a=0.8a = 0.8 and b=1.2b = 1.2:

P(−0.8<Z<1.2)=  Φ(1.2)+Φ(0.8)−1=  0.8849+0.7881−1=  1.6730−1=  0.673.\begin{align*} P(-0.8 < Z < 1.2) =&\,\, \Phi(1.2) + \Phi(0.8) - 1\\[3mm] =&\,\, 0.8849 + 0.7881 - 1\\[3mm] =&\,\, 1.6730 - 1\\[3mm] =&\,\, 0.673. \end{align*}

Therefore, the probability is 0.6730.673.

(c)

解法一

思路

展开

区间 210−k<X<210+k210-k<X<210+k 以平均数为中心，中央概率为 0.950.95，所以两边尾部各 0.0250.025。对应的正标准正态值是 1.961.96。

答题过程

展开

For the central 95%95\% interval,

z=1.96.\begin{align*} z=1.96. \end{align*}

So

210+k−21025=1.96.\begin{align*} \frac{210+k-210}{25}=1.96. \end{align*}

Therefore

k=1.96(25)=49.\begin{align*} k=1.96(25)=49. \end{align*}

(d)

解法一

思路

展开

最轻的 15%15\% 表示左侧概率为 0.150.15。查 percentage points 表可得对应 z=−1.0364z=-1.0364。

答题过程

展开

Let the required maximum weight be SS.

Then

P(X<S)=0.15.\begin{align*} P(X<S)=0.15. \end{align*}

The corresponding standard normal value is

z=−1.0364.\begin{align*} z=-1.0364. \end{align*}

Hence

S−21025=−1.0364.\begin{align*} \frac{S-210}{25}=-1.0364. \end{align*}

So

S= 210−1.0364(25)= 184.09.\begin{align*} S =&\,210-1.0364(25)\\[3mm] =&\,184.09. \end{align*}

The maximum weight is approximately

184 grams.\begin{align*} 184\text{ grams}. \end{align*}

(e)

解法一

思路

展开

把两个百分比条件都转成标准正态方程。5%5\% 在左尾，对应 z=−1.6449z=-1.6449；40%40\% 在右尾，表示左侧概率 0.600.60，对应 z=0.2533z=0.2533。

答题过程

展开

Since 5%5\% weigh less than 152152 grams,

152−μσ=−1.6449.\begin{align*} \frac{152-\mu}{\sigma}=-1.6449. \end{align*}

Since 40%40\% weigh more than 180180 grams,

P(Y<180)=0.60.\begin{align*} P(Y<180)=0.60. \end{align*}

So

180−μσ=0.2533.\begin{align*} \frac{180-\mu}{\sigma}=0.2533. \end{align*}

Thus

152−μ= −1.6449σ,180−μ= 0.2533σ.\begin{align*} 152-\mu=&\,-1.6449\sigma,\\[3mm] 180-\mu=&\,0.2533\sigma. \end{align*}

Subtracting the first equation from the second gives

28= 1.8982σσ= 14.7508…\begin{align*} 28=&\,1.8982\sigma\\[3mm] \sigma=&\,14.7508\ldots \end{align*}

Then

180−μ= 0.2533(14.7508…)μ= 176.263…\begin{align*} 180-\mu=&\,0.2533(14.7508\ldots)\\[3mm] \mu=&\,176.263\ldots \end{align*}

Therefore

μ=176,σ=14.8\begin{align*} \mu=176,\qquad \sigma=14.8 \end{align*}

to 33 significant figures.