The Normal Distribution: Question 9

Worked solution

Part (a): Justifying and stating the approximation

$X\sim B(200,0.05)$, so $n=200$ and $p=0.05$, giving $q=1-p=0.95$.

Check the conditions for a normal approximation: $np=200\times0.05=10>5 \qquad nq=200\times0.95=190>5$

Both conditions hold, so $X$ may be approximated by a normal distribution with $\text{mean}=np=10, \qquad \text{variance}=npq=200\times0.05\times0.95=9.5$

$X\approx N(10,9.5)$

Part (b): P(8 ≤ X ≤ 14) with a continuity correction

$X$ takes only whole-number values, so the continuity correction extends the required region outward by $0.5$ at each boundary: $P(8\le X\le14)\approx P(7.5<Y<14.5), \quad \text{where } Y\sim N(10,9.5)$

Standardise using $\sigma=\sqrt{9.5}=3.0822$ (4 d.p.): $z_1=\frac{7.5-10}{3.0822}=\frac{-2.5}{3.0822}=-0.81 \text{ (2 d.p.)}$ $z_2=\frac{14.5-10}{3.0822}=\frac{4.5}{3.0822}=1.46 \text{ (2 d.p.)}$

$P(7.5<Y<14.5)=P(-0.81<Z<1.46)=\Phi(1.46)-\Phi(-0.81)$

From the table, $\Phi(1.46)=0.9279$. By symmetry, $\Phi(-0.81)=1-\Phi(0.81)=1-0.7910=0.2090$.

$P(8\le X\le14)\approx0.9279-0.2090=0.7189$

Check: the mean is $10$, and the interval $[8,14]$ extends about $0.81$ standard deviations below and $1.46$ standard deviations above it, so a central probability of roughly $72\%$ is a sensible order of magnitude.

Final answers

• (a) $X\approx N(10,9.5)$, since $np=10>5$ and $nq=190>5$
• (b) $P(8\le X\le14)\approx\boxed{0.7189}$