Math Problem Statement

Based on a​ survey,

3838​%

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

1414

likely voters are randomly selected. Complete parts​ (a) through​ (c) below.

Question content area bottom

Part 1

a. What is the probability that exactly

1111

of those selected would do internet​ voting?

In a binomial probability​ distribution, probabilities can be calculated by using​ technology, a table of binomial​ probabilities, or the binomial probability​ formula, shown below where n is the number of​ trials, x is the number of successes among n​ trials, p is the probability of success in any one​ trial, and q is the probability of failure in any one trial

​(qequals=1minus−​p).

​P(x)equals=StartFraction n exclamation mark Over left parenthesis n minus x right parenthesis exclamation mark x exclamation mark EndFraction times p Superscript x Baseline times left parenthesis 1 minus p right parenthesis Superscript n minus xn!(n−x)!x!•px•(1−p)n−x​,

for

xequals=​0,

​1, 2,​ ..., n

While either technology or the binomial probability formula can be used to find the simple binomial​ probabilities, for this​ exercise, use technology.

Part 2

First find the values of​ n, x, and p.

nequals=1414

Part 3

xequals=1111

Part 4

pequals=0.380.38

​(Type an integer or a decimal. Do not​ round.)

Part 5

Use these values to calculate

​P(1111​).

​P(1111​)equals=0.002070.00207

​(Round to five decimal places as​ needed.)

Part 6

b. If

1111

of the selected voters would do internet​ voting, is

1111

significantly​ high? Why or why​ not?

Having x successes among n trials is a significantly high number of successes if the probability of x or more successes is 0.05 or less. That​ is, x is a significantly high number of successes if​ P(x or

​more)less than or equals≤0.05.

Part 7

Use technology to calculate

​P(1111

or​ more).

​P(1111

or​ more) = 0.002420.00242

​(Round to five decimal places as​ needed.)

Part 8

Is this probability less than or equal to​ 0.05?

Yes

Your answer is correct.

No

Part 9

Use this information to determine whether

1111

voters willing to vote by internet is significantly high.

Part 10

c. Find the probability that at least one of the selected likely voters would do internet voting.

Use technology to calculate​ P(1 or​ more).

​P(1 or​ more) = enter your response here

​(Round to three decimal places as​ needed.)

Solution

Let's go through each part and clarify the steps in solving this.

Part 1: Finding P(1111)P(1111)

We need to calculate the binomial probability that exactly 1111 voters out of 1414 will vote by internet. The formula is:

P(x)=(nx)px(1p)nxP(x) = \binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}

Given:

  • n=1414n = 1414 (number of trials),
  • x=1111x = 1111 (number of successes),
  • p=0.38p = 0.38 (probability of success).

Using technology or software, we already found:

P(1111)=0.00207P(1111) = 0.00207

This is the probability that exactly 1111 voters will choose internet voting.

Part 2: Determining if 1111 voters is significantly high

To determine whether 1111 voters willing to vote by internet is a significantly high number, we need to calculate the probability of 1111 or more successes:

P(x1111)=P(1111 or more)P(x \geq 1111) = P(1111 \text{ or more})

We already calculated:

P(1111 or more)=0.00242P(1111 \text{ or more}) = 0.00242

Since 0.002420.050.00242 \leq 0.05, the number of 1111 voters is indeed considered significantly high because the probability of this happening is very small (less than 5%).

Part 3: Finding the probability of at least 1 voter choosing internet voting

Now, we need to calculate the probability that at least one of the selected likely voters would choose internet voting. The probability of at least one success can be found by subtracting the probability of zero successes from 1:

P(at least 1)=1P(0)P(\text{at least 1}) = 1 - P(0)

Where P(0)P(0) is the probability that no voters choose internet voting.

Using technology or software, you can calculate:

P(0)=(10.38)1414=(0.62)1414P(0) = (1 - 0.38)^{1414} = (0.62)^{1414}

Once computed, you get the result for P(0)P(0), and then:

P(at least 1)=1P(0)P(\text{at least 1}) = 1 - P(0)

After calculation, you should round your answer to three decimal places.


Would you like more details or have any further questions about any of the steps? Here are five related questions for further practice:

  1. How do you calculate the expected value of a binomial distribution?
  2. What does it mean for an event to be "significantly low" in terms of probability?
  3. How would the results change if the probability of success, pp, was higher?
  4. What is the variance of the binomial distribution in this scenario?
  5. Can you explain how to interpret cumulative binomial probabilities?

Tip: When dealing with binomial distributions, technology tools (like a calculator or software) are incredibly helpful for complex probabilities, but understanding the underlying formulas ensures deeper comprehension.

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Math Problem Analysis

Mathematical Concepts

Binomial Probability
Probability Theory
Significance Testing

Formulas

P(x) = (n! / (n - x)! x!) * p^x * (1 - p)^(n - x)
P(at least 1) = 1 - P(0)

Theorems

Binomial Distribution
Significance Testing (P-value < 0.05)

Suitable Grade Level

College Level or Advanced High School (Grades 11-12)