Question

79% of all bald eagles survive their first year of life. give your answers as decimals, not percents. if 48 bald eagles are randomly selected, find the probability that

a. exactly 38 of them survive their first year of life.

b. at most 36 of them survive their first year of life.

c. at least 39 of them survive their first year of life.

d. between 35 and 43 (including 35 and 43) of them survive their first year of life.

Explanation:

This is a binomial probability problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 48$, $p=0.79$, and $1 - p = 0.21$.

Step1: Calculate combination for part a

For $n = 48$ and $k = 38$, calculate $C(48,38)=\frac{48!}{38!(48 - 38)!}=\frac{48!}{38!10!}=\frac{48\times47\times\cdots\times39}{10\times9\times\cdots\times1}=654071589408$. Then $P(X = 38)=C(48,38)\times(0.79)^{38}\times(0.21)^{10}$.
$P(X = 38)=654071589408\times(0.79)^{38}\times(0.21)^{10}\approx0.139$

Step2: Calculate cumulative probability for part b

$P(X\leq36)=\sum_{k = 0}^{36}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}$. Using a binomial probability calculator or software (e.g., Excel's BINOM.DIST function: =BINOM.DIST(36,48,0.79,TRUE)), we get $P(X\leq36)\approx0.247$

Step3: Calculate cumulative probability for part c

$P(X\geq39)=1 - P(X\leq38)$. First, $P(X\leq38)=\sum_{k = 0}^{38}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}$. Using a binomial probability calculator, $P(X\leq38)\approx0.544$. So $P(X\geq39)=1 - 0.544 = 0.456$

Step4: Calculate cumulative probability for part d

$P(35\leq X\leq43)=\sum_{k = 35}^{43}C(48,k)\times(0.79)^{k}\times(0.21)^{48 - k}=P(X\leq43)-P(X\leq34)$. Using a binomial probability calculator, $P(X\leq43)\approx0.977$ and $P(X\leq34)\approx0.067$. So $P(35\leq X\leq43)=0.977-0.067 = 0.910$

Answer:

a. $0.139$
b. $0.247$
c. $0.456$
d. $0.910$