what is the average cost of a product, given the cost of each case and its probability

Question

for the 33rd sum, part a, I did

average cost = 5[P(x<8.04)] + 6.5[P(X>8.04)]

i got 5.46 but the text says 5.11

where did i go wrong

also, the answer for part b says 7.39. Is that possible, given that its supposed to be 'average' but 7.39 is greater than 6.50 (the maximum cost)?

Answer 1

Let $X$ denote the length of a rod and let $Z=(X-8.02)/0.04$.

For (a) you should be calculating:

$$4\text{Pr}(X<8.00)+5\text{Pr}(8.00<X<8.04)+6.5\text{Pr}(X>8.04)$$

Now, $$\text{Pr}(X<8.00)=\text{Pr}(X>8.04)=\text{Pr}(Z>0.5)=0.5-0.19146=0.30854$$

and

$$\text{Pr}(8.00<X<8.04)=2(0.19146)=0.38292.$$

Substituting:

$$4(0.30854)+5(0.38292)+6.5(0.30854)=5.15427$$

(The difference in answers is probably due to rounding errors.)

For (b) the average cost is:

$$\begin{align*}\frac{4\text{Pr}(X<8.00)+5\text{Pr}(8.00<X<8.04)+6.5\text{Pr}(X>8.04)}{\text{Pr}(8.00<X<8.04)+\text{Pr}(X>8.04)}\end{align*}$$

The denominator is $1-\text{Pr}(X<8.00)=1-0.30854=0.69146$

So the average cost is the answer from (a) divided by $0.69146$, i.e. $7.45$