The Normal Distribution: Question 5

Worked solution

Part (a): Justifying and stating the approximation

$X\sim B(150,0.4)$, so $n=150$ and $p=0.4$, giving $q=1-p=0.6$.

Check the conditions for a normal approximation: $np=150\times0.4=60>5 \qquad nq=150\times0.6=90>5$

Both conditions hold, so $X$ may be approximated by a normal distribution with $\text{mean}=np=60, \qquad \text{variance}=npq=150\times0.4\times0.6=36$

$X\approx N(60,36)$

Part (b): P(X > 70) with a continuity correction

$X$ takes only whole-number values, so "$X>70$" means $X\ge71$. To use the continuous normal distribution, the continuity correction extends the required region down to $70.5$: $P(X>70)=P(X\ge71)\approx P(Y>70.5), \quad \text{where } Y\sim N(60,36)$

Standardise using $\sigma=\sqrt{36}=6$: $z=\frac{70.5-60}{6}=\frac{10.5}{6}=1.75$

$P(Y>70.5)=P(Z>1.75)=1-\Phi(1.75)$

From the table, $\Phi(1.75)=0.9599$, so $P(X>70)\approx1-0.9599=0.0401$

Check: $70$ bullseyes is $\dfrac{70-60}{6}\approx1.67$ standard deviations above the mean of $60$, so a right-tail probability of a few percent is a sensible size.

Final answers

• (a) $X\approx N(60,36)$, since $np=60>5$ and $nq=90>5$
• (b) $P(X>70)\approx\boxed{0.0401}$