Question 2 of 8

Question 2 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Let $X=$ number of solar lights (out of $n$ sampled) lasting more than two years, with per-light success probability $p=0.70$.

Find. (a) $P(X>7)$ for $n=10$; (b) $P(5<X<9)$ for $n=12$; (c) $P(X<83)$ for $n=1{,}220$ via a Normal approximation; (d) $P(X<3)$ for $n=4$.

Approach. Parts (a), (b), (d) are exact Binomial$(n,p)$ tail sums; part (c) has $n$ too large for a direct Binomial table, so the Normal approximation to the Binomial (with continuity correction) is used instead.

1. (a) Binomial upper tail, $n=10$. $$P(X>7)=P(X=8)+P(X=9)+P(X=10)=\sum_{x=8}^{10}\binom{10}{x}(0.7)^x(0.3)^{10-x}=\boxed{0.3828}$$
2. (b) Binomial middle range, $n=12$. "More than five but fewer than nine" means $X=6,7,8$. $$P(6\le X\le 8)=\sum_{x=6}^{8}\binom{12}{x}(0.7)^x(0.3)^{12-x}=\boxed{0.4689}$$
3. (c) Normal approximation with continuity correction, $n=1{,}220$. Check $np=854$ and $n(1-p)=366$ are both large, so the approximation is valid. $$\mu=np=1220(0.7)=854, \qquad \sigma=\sqrt{np(1-p)}=\sqrt{1220(0.7)(0.3)}=16.006$$ "Fewer than 83" is $X<83$, corrected to $X<82.5$: $$Z=\frac{82.5-854}{16.006}=-48.2$$ $$P(X<83)\approx P(Z<-48.2)\approx\boxed{0\ (\text{effectively impossible})}$$
4. (d) Binomial lower tail, $n=4$. $$P(X<3)=P(X=0)+P(X=1)+P(X=2)=\sum_{x=0}^{2}\binom{4}{x}(0.7)^x(0.3)^{4-x}=\boxed{0.3483}$$