The quantity of wheat demanded, per day, in a midwestern market during a certain marketing period is represented by

\(Q=100,000-12,500 P+V\) for \(\mathrm{p} \in[2,6]\), where

\(Q\) is quantity demanded in bushels; \(p\) is price/bushel; and

\(V\) is approximately normally distributed.

You know that the expected quantity demanded is given by

\(E(Q)=100,000-12,500\) p for \(p \in[2,6]\), and thus is a function of \(p\), and the variance of quantity demanded is \(\operatorname{var}(Q)=16 \times 10^{6}\).

(a) What is the mean and variance of \(V\) ?

(b) If \(p=4\), what is the probability that more than 50,000 bushels of wheat will be demanded?

(c) If \(p=4.50\), what is the probability that more than 50,000 bushels of wheat will be demanded?

(d) For quantity demanded to be greater than 50,000 bushels with probability .95 , what does \(p\) have to be?

(e) Is it possible that \(V\) could actually be normally distributed instead of only approximately normally distributed? Explain.