Question

68% of owned dogs in the united states are spayed or neutered. round your answers to four decimal places. if 37 owned dogs are randomly selected, find the probability that

a. exactly 23 of them are spayed or neutered.

b. at most 27 of them are spayed or neutered.

c. at least 23 of them are spayed or neutered.

d. between 24 and 32 (including 24 and 32) of them are spayed or neutered.

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, $n = 37$, $p=0.68$, and $1 - p = 0.32$.

Step1: Calculate binomial coefficient for part a

For $k = 23$, calculate $C(37,23)=\frac{37!}{23!(37 - 23)!}=\frac{37!}{23!14!}$
\[

$$\begin{align*} C(37,23)&=\frac{37\times36\times\cdots\times15}{23\times22\times\cdots\times1\times14\times13\times\cdots\times1}\\ \end{align*}$$

\]
Then $P(X = 23)=C(37,23)\times(0.68)^{23}\times(0.32)^{14}$
\[

$$\begin{align*} P(X = 23)&=\frac{37!}{23!14!}\times(0.68)^{23}\times(0.32)^{14}\\ &\approx0.1292 \end{align*}$$

\]

Step2: Calculate cumulative probability for part b

$P(X\leq27)=\sum_{k = 0}^{27}C(37,k)\times(0.68)^{k}\times(0.32)^{37 - k}$. This can be calculated using a binomial cumulative - distribution function on a calculator or software. Using a TI - 84 Plus: binomcdf(37,0.68,27) $\approx0.7793$

Step3: Calculate cumulative probability for part c

$P(X\geq23)=1 - P(X\lt23)=1-\sum_{k = 0}^{22}C(37,k)\times(0.68)^{k}\times(0.32)^{37 - k}$. Using a TI - 84 Plus: 1 - binomcdf(37,0.68,22) $\approx0.8944$

Step4: Calculate cumulative probability for part d

$P(24\leq X\leq32)=\sum_{k = 24}^{32}C(37,k)\times(0.68)^{k}\times(0.32)^{37 - k}=binomcdf(37,0.68,32)-binomcdf(37,0.68,23)$ $\approx0.7369$

Answer:

a. $0.1292$
b. $0.7793$
c. $0.8944$
d. $0.7369$