Question

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.
a. what is the probability that exactly 9 of those selected would do internet voting?
0.00476
(round to five decimal places as needed.)
b. if 9 of the selected voters would do internet voting, is 9 significantly high? why or why not?
select the correct choice below and fill in the answer box within your choice
(round to five decimal places as needed.)
a. no, because the probability of 9 or more is , which is not low.
b. yes, because the probability of 9 or more is , which is not low.
c. yes, because the probability of 9 or more is , which is low.
d. no, because the probability of 9 or more is , which is low.

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, $n = 12$, $k = 9$, and $p=0.35$.

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

$P(X = 9)=C(12,9)\times(0.35)^{9}\times(1 - 0.35)^{12 - 9}=220\times(0.35)^{9}\times(0.65)^{3}$.
$(0.35)^{9}\approx0.0000789067$, $(0.65)^{3}=0.274625$.
$P(X = 9)=220\times0.0000789067\times0.274625\approx0.00476$.

Step4: Determine if 9 is significantly high

To determine if 9 is significantly high, we need to find $P(X\geq9)=P(X = 9)+P(X = 10)+P(X = 11)+P(X = 12)$.
$P(X = 10)=C(12,10)\times(0.35)^{10}\times(0.65)^{2}=\frac{12!}{10!(12 - 10)!}\times(0.35)^{10}\times(0.65)^{2}=66\times(0.35)^{10}\times(0.65)^{2}$.
$(0.35)^{10}\approx0.0000276173$, $(0.65)^{2}=0.4225$, $P(X = 10)=66\times0.0000276173\times0.4225\approx0.00077$.
$P(X = 11)=C(12,11)\times(0.35)^{11}\times(0.65)^{1}=\frac{12!}{11!(12 - 11)!}\times(0.35)^{11}\times(0.65)^{1}=12\times(0.35)^{11}\times0.65$.
$(0.35)^{11}\approx0.0000096661$, $P(X = 11)=12\times0.0000096661\times0.65\approx0.00007$.
$P(X = 12)=C(12,12)\times(0.35)^{12}\times(0.65)^{0}=(0.35)^{12}\approx0.0000033831$.
$P(X\geq9)=P(X = 9)+P(X = 10)+P(X = 11)+P(X = 12)\approx0.00476+0.00077 + 0.00007+0.0000033831\approx0.00560$.

Since $P(X\geq9)\approx0.00560$ is low (usually, a probability less than 0.05 is considered low in significance testing), 9 is significantly high.

Answer:

a. $0.00476$
b. C. Yes, because the probability of 9 or more is $0.00560$, which is low.