The Normal Distribution: Question 8

Worked solution

Setting up

$R\sim N(48,4^2)$, so $\mu=48$ and $\sigma=4$. “Fastest $10\%$” means the shortest times, so $t$ lies below the mean and satisfies $P(R<t)=0.10$.

Finding z

Since $t$ is below the mean, the corresponding $z$-value is negative. From the standard normal (percentage points) table, $\Phi(1.2816)=0.90$, so by symmetry $\Phi(-1.2816)=1-0.90=0.10$. Hence $z=-1.2816$.

Solving for t

$t=\mu+z\sigma=48+(-1.2816)\times4=48-5.1264=42.8736$

So $t=42.9$ minutes (3 s.f.).

Why the other options are wrong

• $53.1$ comes from using $z=+1.2816$ instead of $-1.2816$, placing the cutoff above the mean rather than below it.
• $48.0$ is just the mean, obtained by forgetting to standardise at all.
• $41.4$ comes from using $z=-1.6449$ (the percentage point for $\Phi(z)=0.95$) instead of the correct $z=-1.2816$ (for $\Phi(z)=0.90$).

Final answer

$t=\boxed{42.9}$ minutes, so the correct option is A.