based on a survey, 35% of likely voters would be willing to vote by internet instead of the in - person…

based on a survey, 35% of likely voters would be willing to vote by internet instead of the in - person traditional method of voting. for each of the following, assume that 12 likely voters are randomly selected. complete parts (a) through (c) below.\na. what is the probability that exactly 9 of those selected would do internet voting?\n0.00470\n(round to five decimal places as needed.)\nb. if 9 of the selected voters would do internet voting, is 9 significantly high? why or why not?\nselect the correct choice below and fill in the answer box within your choice\n(round to five decimal places as needed.)\na. no, because the probability of 9 or more is \n b. yes, because the probability of 9 or more is \n c. yes, because the probability of 9 or more is \n d. no, because the probability of 9 or more is

based on a survey, 35% of likely voters would be willing to vote by internet instead of the in - person traditional method of voting. for each of the following, assume that 12 likely voters are randomly selected. complete parts (a) through (c) below.\na. what is the probability that exactly 9 of those selected would do internet voting?\n0.00470\n(round to five decimal places as needed.)\nb. if 9 of the selected voters would do internet voting, is 9 significantly high? why or why not?\nselect the correct choice below and fill in the answer box within your choice\n(round to five decimal places as needed.)\na. no, because the probability of 9 or more is \n b. yes, because the probability of 9 or more is \n c. yes, because the probability of 9 or more is \n d. no, because the probability of 9 or more is

Answer

Explanation:

Step1: Identify binomial probability formula

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, $n = 12$, $k = 9$, $p=0.35$, and $1 - p = 0.65$.

Step2: Calculate the combination $C(n,k)$

$C(12,9)=\frac{12!}{9!(12 - 9)!}=\frac{12!}{9!3!}=\frac{12\times11\times10}{3\times2\times1}=220$.

Step3: Calculate the binomial probability

$P(X = 9)=C(12,9)\times(0.35)^{9}\times(0.65)^{3}=220\times(0.35)^{9}\times(0.65)^{3}\approx0.00470$.

For part (b):

Step1: Calculate $P(X\geq9)$

$P(X\geq9)=P(X = 9)+P(X = 10)+P(X = 11)+P(X = 12)$. $P(X = 10)=C(12,10)\times(0.35)^{10}\times(0.65)^{2}=\frac{12!}{10!(12 - 10)!}\times(0.35)^{10}\times(0.65)^{2}=66\times(0.35)^{10}\times(0.65)^{2}$. $P(X = 11)=C(12,11)\times(0.35)^{11}\times(0.65)^{1}=\frac{12!}{11!(12 - 11)!}\times(0.35)^{11}\times(0.65)^{1}=12\times(0.35)^{11}\times(0.65)^{1}$. $P(X = 12)=C(12,12)\times(0.35)^{12}\times(0.65)^{0}=(0.35)^{12}$. $P(X\geq9)\approx0.00470 + 0.00094+0.00012 + 0.00001\approx0.00577$. Since a probability of $0.00577$ is low (usually, we consider a probability less than $0.05$ as low in significance - testing context), 9 is a significantly high number.

Answer:

a. $0.00470$ b. C. Yes, because the probability of 9 or more is $0.00577$, which is low.