题目

Problem

A machine makes bolts such that the length, $L$ cm, of a bolt has distribution

L\sim N(4.1,0.125^2)

A bolt is selected at random.

(a) Find the probability that the length of this bolt is more than 4.3 cm.

(3)

(b) Show that $P(3.9<L<4.3)$ is 0.890 correct to 3 decimal places.

(1)

The machine makes 500 bolts.

The cost to make each bolt is 5 pence.

Only bolts with length between 3.9 cm and 4.3 cm can be used.

These are sold for 9 pence each.

All the bolts that cannot be used are recycled with a scrap value of 1 pence each.

(4)

Following adjustments to the machine, the length of a bolt, $B$ cm, made by the machine is such that $B\sim N(\mu,\sigma^2)$

Given that $P(B>4.198)=0.025$ and $P(B<4.065)=0.242$

(d) find the value of $\mu$ and the value of $\sigma$

(6)

(e) State, giving a reason, whether the adjustments to the machine will result in a decrease or an increase in the profit made on 500 bolts.

(2)

解答

解法一

思路

展开

先用标准化求可用概率和利润。调整后用两个分位点建立关于 $\mu,\sigma$ 的方程。

答题过程

展开

For $L\sim N(4.1,0.125^2)$ ,

P(L>4.3)=P\left(Z>\frac{4.3-4.1}{0.125}\right) =P(Z>1.6).

\begin{align*} P(L>4.3)=1-0.9452=0.0548. \end{align*}

Also,

\frac{3.9-4.1}{0.125}=-1.6,\qquad \frac{4.3-4.1}{0.125}=1.6.

Therefore

P(3.9<L<4.3)=P(-1.6<Z<1.6) =0.9452-0.0548 =0.8904.

This is $0.890$ to 3 decimal places.

The estimated number of usable bolts is

\begin{align*} 500(0.8904)=445.2. \end{align*}

The estimated number of unusable bolts is

\begin{align*} 500-445.2=54.8. \end{align*}

The estimated profit, in pence, is

\begin{align*} 445.2(9-5)+54.8(1-5)=1561.6. \end{align*}

So the estimated profit is approximately

\begin{align*} 1560\text{ pence}. \end{align*}

For the adjusted machine,

\begin{align*} P(B>4.198)=0.025 \end{align*}

gives

\begin{align*} \frac{4.198-\mu}{\sigma}=1.960. \end{align*}

Also,

\begin{align*} P(B<4.065)=0.242 \end{align*}

gives

\begin{align*} \frac{4.065-\mu}{\sigma}=-0.70\quad\text{approximately}. \end{align*}

\begin{align*} \mu+1.960\sigma=4.198 \end{align*}

and

\begin{align*} \mu-0.70\sigma=4.065. \end{align*}

Subtracting,

\begin{align*} 2.660\sigma=0.133. \end{align*}

Hence

\begin{align*} \sigma=0.0500. \end{align*}

Then

\begin{align*} \mu=4.065+0.70(0.0500)=4.10. \end{align*}

The adjustments decrease the standard deviation while keeping the mean approximately the same, so a greater proportion of bolts will be between 3.9 cm and 4.3 cm.

Therefore the profit will increase.

IAL 2018 Oct Q6

题目

解答

解法一

思路

答题过程