# Quod Erat Demonstrandum

## 2008/04/14

### 解說 Applied Math 小題

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 8:22 下午

Nothing special, just answer a question from some students in my class.

Ex. 5(a) Q.35

Rods are made of nomial length of 8 cm but in fact they form a normal distribution with mean 8.02 cm and standard derivation 0.03 cm. Each rod costs $9 to make and may be used immediately if its length lies between 7.96 cm and 8.04 cm. If its length is less than 7.96 cm, the rod is useless but has but has a scrap value of$1. If its length exceeds 8.04 cm it may be shortened and used at a further cost \$2. Find the average cost per usable rod.

Let $X$ ~ $N(8.02 , 0.03^2)$. Students don’t see why the way of calculation is given by

$((9 - 1)P(X \le 7.96) + 9P(7.96 \le X \le 8.04) + (9 + 2)P(X \ge 8.04)) \div p$

where $p = P(7.96 \le X \le 8.04) + P(X \ge 8.04)$.

Actually, they don’t understand why “$\div p$“.

There is nothing special, it is about the phrase “per usable rod".

Let me put it this way. Suppose $N$ is the total number of rods produced. Then,

the total expected cost of too-short rods = $N\times (9-1)P(X \le 7.96)$;
the total expected cost of acceptable rods (usable) = $N\times 9P(7.96 \le X \le 8.04)$;
the total expected cost of too-long rods (also usable) = $N\times (9+2)P(X \ge 8.04)$.

Then, if the question askes: average cost “per rod", we just sum up the above 3 numbers and $\div N$. But now, the question askes about “per usable rod", and the expected number of usable rods is $N\times (P(7.96 \le X \le 8.04) + P(X \ge 8.04))$, and hence the required answer is

$N \times ((9 - 1)P(X \le 7.96) + 9P(7.96 \le X \le 8.04) + (9 + 2)P(X \ge 8.04)) \div Np$

where $p = P(7.96 \le X \le 8.04) + P(X \ge 8.04)$

A cost is a bit “higher" than the overall in the whole batch.