I was given this exercise:
I did this but i am not sure if my reasoning or even my answer is okay.
Given cheeses (weight in Kg)X with N(10,2^2)
u = 10, σ = 2
more than 9 kg, paid $20
less than 9 kg, lost $5
$$P (x > g) = P ({\frac{(x - 10)}{2} > \frac{(g - 10)}{2}})$$
$$= P( Z > -0.5) = 1 - P(Z <= -0.5)$$ $$= 1 - P(Z <= -0.5)$$ $$= 1 - [1 - P(Z <= -0.5)]$$ $$= P(Z < 0.5)$$
$$P(x <= g) = P(\frac{(x - 10)}{2} <= \frac{(g - 10)}{2})$$
$$=P(Z <= -0.5)$$
$$=1 - P (Z < 0.5)$$
$$P(x > h) >= 0.991$$
$$P({\frac{(x - 10)}{1}} > \frac{(h - 10)}{2}) >= 0.991$$
$$P(Z > \frac{(h -10)}{2}) >= 0.991$$
$$(h - 10)/2 = 2.37$$
$$h = 4.74 + 10$$
so, the answer should be 14 cheeses.
Am i right?
We first determine the probability $p$ that a cheese is good (weight > $9$ kilo). Now the standard deviation is $2$ kilo, and a cheese $1$ kilo lighter than the mean ($10$ kilo) is not good. A table with normal distribution values tells us for $z = -0.5$ that $p = 0.6915$.
Next we derive expressions for the average value $A$ (in dollars) of one cheese; and for the standard deviation $\sigma$ in the value (in dollars) of one a cheese:
$$A = p * 20 + (1-p) * -5 = 12.2875$$
$$\sigma = (20 + 5) * \sqrt {p(1-p)} = 11.5469$$
Now the question is how many cheeses must be produced to guarantee a total value of $45000$ dollar with a probability of at least $0.991$. Using a table with normal distribution values we see that the probabilty of $0.991$ corresponds with a $z$ value of $-2,37$.
The equation for the required value $V$ of $N$ cheeses at a $0.991$ confidence level thus becomes:
$$V = A * N - 2.37 * \sigma * \sqrt{N} > 45000$$
This is a quadratic equation in $N$, or rather in $\sqrt{N}$, that can be solved by standard methods. The result is $N = 3800$. Had we asked for the number of cheeses to yield an average value of $45000$ dollar, the result would have been $N = 3663$. Hence the effect of fluctuations on the required number of cheeses is fairly small.