Question

# For certain bird species, with appropriate assumptions, the number of nests escaping predation has a binomial distribution. Suppose the probability of success (that is, a nest escaping predation) is 0.3. Find the probability that at least half of 24 nests escape predation.

Solution

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Let $x$ = the number of nests escaping predation.

Then , P(at least half escape predation) = $P(x\geq 12)$.

To include 12, we will find $P(x\geq 11.5).$

If we approximate the binomial distribution with the normal,

$\mu=np\quad$ and $\quad \sigma=\sqrt{np(1-p)}$

We find the z-score for x=$11.5$

\begin{align*} n&=24, p=0.3 \\ \mu&=np \\ &= 24(0.3) \\ &=7.2 \\\\ \sigma&=\sqrt{np(1-p)} \\ & =\sqrt{24(0.3)(0.7)}\\ &=\sqrt{504} \\ &\approx 2.245 \\ \\ z&=\displaystyle \frac{x-\mu}{\sigma} \\ & =\displaystyle \frac{11.5-7.2}{2.245}\\ & \approx 1.92 \end{align*}

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