Questions: Based on a survey, 32% 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 14 likely voters are randomly selected Complete parts (a) through (c) below. a. What is the probability that exactly 11 of those selected would do internet voting? (Round to five decimal places as needed.) b. If 11 of the selected voters would do internet voting, is 11 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. Yes, because the probability of 11 or more is which is low B. No, because the probability of 11 or more is which is low. C. No because the probability of 11 or more is which is not low. D. Yes, because the probability of 11 or more is which is not low

Solution Steps

To find the probability that exactly 11 out of 14 likely voters would choose to vote by internet, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 14 \) (total number of trials),
• \( x = 11 \) (number of successes),
• \( p = 0.32 \) (probability of success),
• \( q = 1 - p = 0.68 \) (probability of failure).

Calculating this gives:

\[ P(X = 11) = \binom{14}{11} \cdot (0.32)^{11} \cdot (0.68)^{3} \approx 0.00041 \]

Thus, the probability that exactly 11 would do internet voting is:

\[ \boxed{P(X = 11) = 0.00041} \]

Next, we need to determine the probability of 11 or more voters choosing internet voting, which is given by:

\[ P(X \geq 11) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) \]

Calculating these probabilities, we find:

• \( P(X = 11) \approx 0.00041 \)
• \( P(X = 12) \approx 5 \times 10^{-5} \)
• \( P(X = 13) \approx 0.0 \)
• \( P(X = 14) \approx 0.0 \)

Summing these probabilities gives:

\[ P(X \geq 11) \approx 0.00041 + 5 \times 10^{-5} + 0.0 + 0.0 \approx 0.00046 \]

Thus, the probability of 11 or more voters choosing internet voting is:

\[ \boxed{P(X \geq 11) = 0.00046} \]

To assess whether 11 is significantly high, we compare the probability \( P(X \geq 11) \) to a significance level, typically \( \alpha = 0.05 \). Since:

\[ P(X \geq 11) = 0.00046 < 0.05 \]

This indicates that the occurrence of 11 or more voters choosing internet voting is statistically rare. Therefore, we conclude:

• Yes, because the probability of 11 or more is \( 0.00046 \), which is low.

Final Answer