The Normal Distribution: Question 10

Worked solution

Part (a): Finding σ

$S\sim N(500,\sigma^2)$, so $\mu=500$. $P(S>650)=0.05$ means $650$ lies above the mean, so $P(Z>z)=0.05 \implies \Phi(z)=0.95$

From the standard normal (percentage points) table, $\Phi(1.6449)=0.95$, so $z=1.6449$.

Standardising $650$: $\frac{650-500}{\sigma}=1.6449$

$\sigma=\frac{150}{1.6449}=91.191\ldots$

So $\sigma=91.2$ (3 s.f.).

Part (b): P(400 < S < 600)

Using $\sigma=91.191$ (unrounded, to minimise rounding error), the interval $(400,600)$ is symmetric about the mean $500$, with half-width $100$: $z=\frac{600-500}{91.191}=\frac{100}{91.191}=1.0966$

Rounding to 2 d.p. for the table, $z=1.10$.

$P(400<S<600)=P(-1.10<Z<1.10)=\Phi(1.10)-\Phi(-1.10)=2\Phi(1.10)-1$

From the table, $\Phi(1.10)=0.8643$, so $P(400<S<600)=2\times0.8643-1=1.7286-1=0.7286$

Check: $\sigma=91.2$ is smaller than the $150$-point gap to the $5\%$ cutoff at $650$ divided by $1.6449$ would suggest at a glance is roughly right ($150/91.2\approx1.64$), and a symmetric interval of $\pm1.10\sigma$ capturing about $73\%$ of the probability is a sensible order of magnitude (compared with $\pm1\sigma\approx68\%$).

Final answers

• (a) $\sigma\approx\boxed{91.2}$ (3 s.f.)
• (b) $P(400<S<600)\approx\boxed{0.7286}$