N2023 P2 Q9

Let L and R be lengths in metres of independently selected Long and Regular planks. Use \(L\sim\mathrm N(1.82,0.2^2)\) and \(R\sim\mathrm N(1.22,0.3^2)\). Lengths of distinct randomly selected planks are treated as independent.

1. \(P(L<1.79)=\Phi\left(\frac{1.79-1.82}{0.2}\right)=0.440382\ldots\approx0.440\)
2. The total is a sum of eight independent lengths, not eight times one random length.
   \(T\)\({}=L_1+\cdots+L_8\sim\mathrm N(8(1.82),8(0.2^2))\)\(T\sim\mathrm N(14.56,0.32)\)
   \(P(T>14.5)=1-\Phi\left(\frac{14.5-14.56}{\sqrt{0.32}}\right)=0.542235\ldots\approx0.542\)
3. The expected count is the number of planks multiplied by the probability that one qualifies.
   \(P(R\)\({}>1.25)=1-\Phi\left(\frac{1.25-1.22}{0.3}\right)=0.460172\ldots\)\(E(\text{count})\)\({}=120(0.460172\ldots)=55.2207\ldots\)
   \(E(\text{count})\approx55.2\ \text{planks}\)
4. Let D be the total of ten Long lengths minus the total of sixteen Regular lengths. Variances add for independent differences.
   \(E(D)\)\({}=10(1.82)-16(1.22)=-1.32\)\(\operatorname{Var}(D)\)\({}=10(0.2^2)+16(0.3^2)=1.84\)\(D\sim\mathrm N(-1.32,1.84)\)
   \(P(|D|<0.65)=P(-0.65<D<0.65)=0.237469\ldots\approx0.237\)
5. One Short length from a Long plank is one third of a single random length; one from a Regular plank is one half of a single random length. Their variances scale by the squares of these factors.
   \(U\)\({}=\frac L3\sim\mathrm N\left(\frac{91}{150},\frac1{225}\right)\)\(V\)\({}=\frac R2\sim\mathrm N\left(\frac{61}{100},\frac9{400}\right)\)\(U-V\sim\mathrm N\left(-\frac1{300},\frac{97}{3600}\right)\)
   \(P(U>V)=P(U-V>0)=0.491899\ldots\approx0.492\)
6. Under the intended model of the same small length lost at each cut, making three pieces requires two cuts whereas making two pieces requires one. Each Long-derived piece therefore loses more length than each Regular-derived piece. The difference shifts down, so the probability in (e) decreases.