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. (A) $p=0.70$ of adults are against privatisation. (a) $n=12$; (b) $n=10$; (c) $n=2{,}000$, using the count FOR privatisation ($p_{\text{for}}=0.30$). (B) letter-loss probability $p=0.00001$, $n=300{,}000$.

Find. (a) $P(X>9)$, $X\sim\text{Bin}(12,0.70)$; (b) $P(3<X<7)$, $X\sim\text{Bin}(10,0.70)$; (c) $P(Y<620)$, $Y$= number FOR, via normal approximation; (B) $P(W<3)$ via Poisson approximation.

Approach. Exact binomial sums for small $n$ ((a),(b)); normal approximation with continuity correction for large $n$ with moderate $p$ (c); Poisson approximation for large $n$, tiny $p$, with $\lambda=np$ (B).

1. (a) Exact binomial, more than nine of twelve. $P(X>9)=\sum_{k=10}^{12}\binom{12}{k}(0.70)^k(0.30)^{12-k}=0.0851+0.1585+0.0138\times\ldots$ Summing directly in the tail gives $\boxed{P(X>9)=0.2528}$.
2. (b) Exact binomial, strictly between three and seven. $P(3<X<7)=P(X=4)+P(X=5)+P(X=6)=\sum_{k=4}^{6}\binom{10}{k}(0.70)^k(0.30)^{10-k}$, giving $\boxed{P(3<X<7)=0.3398}$.
3. (c) Normal approximation, n=2,000. Let $Y$ count those FOR privatisation, $Y\sim\text{Bin}(2{,}000,\,0.30)$: $\mu_Y=np=600$, $\sigma_Y=\sqrt{np(1-p)}=\sqrt{2{,}000(0.30)(0.70)}=20.49$. With the continuity correction, $z=\dfrac{619.5-600}{20.49}=0.95$, so $P(Y<620)\approx\Phi(0.95)=0.8293$, giving $\boxed{P(Y<620)=0.8293}$.
4. (B) Poisson approximation to the binomial, n=300,000. $\lambda=np=300{,}000(0.00001)=3.0$. $P(W<3)=P(W\le2)=e^{-3}\left(1+3+\dfrac{3^2}{2!}\right)=e^{-3}(8.5)=0.4232$, giving $\boxed{P(W<3)=0.4232}$.