The Normal Distribution: Question 3

Worked solution

Setting up

$M\sim N(24,5^2)$, so $\mu=24$ and $\sigma=5$. To go from a probability back to a value of $M$, standardise as $z=\dfrac{M-\mu}{\sigma}$ and rearrange for $M$: $M=\mu+z\sigma$.

Part (a): Upper-tail probability

$P(M>m)=0.10$ means $m$ lies above the mean, so $P(Z>z)=0.10 \implies \Phi(z)=0.90$

From the standard normal (percentage points) table, $\Phi(1.2816)=0.90$, so $z=1.2816$.

$m=\mu+z\sigma=24+1.2816\times5=24+6.408=30.408$

So $m=30.4$ minutes (3 s.f.).

Part (b): Lower-tail probability

$P(M<x)=0.05$ means $x$ lies below the mean, so the corresponding $z$-value is negative. Since $\Phi(1.6449)=0.95$, the point with $\Phi(z)=0.05$ is $z=-1.6449$, by symmetry of the standard normal curve about $0$.

$x=\mu+z\sigma=24+(-1.6449)\times5=24-8.2245=15.7755$

So $x=15.8$ minutes (3 s.f.).

Check: $m=30.4$ is above the mean $24$, as expected since only $10\%$ of times exceed it, and $x=15.8$ is below the mean, as expected since only $5\%$ of times are less than it. Both lie within a couple of standard deviations of the mean, which is consistent with such tail probabilities.

Final answers

• (a) $m=\boxed{30.4}$ minutes (3 s.f.)
• (b) $x=\boxed{15.8}$ minutes (3 s.f.)