Question

An industry manufactures bus tyres that have an average lifetime of (42569) kilometers with a standard deviation of (2569) kilometers and the distribution of lifetime of the tyres is normal.

1-Find the probability of tyre’s lifetime is between (40069) km and (45069) km

2-What is the lifetime in km, that 69 % of tyres exceed?

Answer 1

$X\sim N\left( 42569,2569^2 \right) \\1:\\P\left( 40069\leqslant X\leqslant 45069 \right) =P\left( \frac{40069-42569}{2569}\leqslant \frac{X-42569}{2569}\leqslant \frac{45069-42569}{2569} \right) =\\=P\left( -0.973141\leqslant Z\leqslant 0.973141 \right) =2\varPhi \left( 0.97341 \right) -1=2\cdot 0.834825-1=0.66965\\2:\\P\left( X\geqslant C \right) =0.69\Rightarrow P\left( \frac{X-42569}{2569}\geqslant \frac{C-42569}{2569} \right) =0.69\Rightarrow \\\Rightarrow P\left( Z\geqslant \frac{C-42569}{2569} \right) =0.69\Rightarrow P\left( Z\leqslant \frac{C-42569}{2569} \right) =0.31\Rightarrow \\\Rightarrow \varPhi \left( \frac{C-42569}{2569} \right) =0.31\Rightarrow \frac{C-42569}{2569}=z_{0.31}=-0.49585\Rightarrow \\\Rightarrow C=-0.49585\cdot 2569+42569=41295.2$