46% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four. (a) P(5) = |(Round to three decimal places as needed.)

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### Probability and Statistics: Confidence in Newspapers

**Scenario:**
46% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is:

a) Exactly five  
b) At least six  
c) Less than four

**Solution:**

(a) \( P(5) = \) __________ (Round to three decimal places as needed.)

**Enter your answer in the answer box and then click Check Answer.**

[Textbox]

**Graphical Representation:**
There are no graphs or diagrams presented in this scenario.

**Explanation:**
- This problem involves a binomial probability distribution where:
  - The number of trials \( n \) is 10 (since 10 U.S. adults are randomly selected).
  - The probability of success \( p \) (an adult having very little confidence in newspapers) is 46%, or 0.46.
- You will use the binomial probability formula or a binomial distribution table to find the probability for each specified scenario (exactly five, at least six, and less than four).

**Key Points to Remember:**
- **Binomial Probability Formula:** 
  \[
  P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
  \]
  where \(\binom{n}{k}\) is the binomial coefficient.

- **Cumulative Probabilities:**
  - **At least six** indicates the sum of probabilities from P(6) to P(10).
  - **Less than four** indicates the sum of probabilities from P(0) to P(3).

By solving these, you can determine the respective probabilities for each scenario concerning the confidence in newspapers among selected U.S. adults.

**Note:** This page contains interactive elements, such as a textbox to input your calculated result and a "Check Answer" button to validate it. Be sure to have a calculator handy to compute these probabilities accurately.
Transcribed Image Text:### Probability and Statistics: Confidence in Newspapers **Scenario:** 46% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is: a) Exactly five b) At least six c) Less than four **Solution:** (a) \( P(5) = \) __________ (Round to three decimal places as needed.) **Enter your answer in the answer box and then click Check Answer.** [Textbox] **Graphical Representation:** There are no graphs or diagrams presented in this scenario. **Explanation:** - This problem involves a binomial probability distribution where: - The number of trials \( n \) is 10 (since 10 U.S. adults are randomly selected). - The probability of success \( p \) (an adult having very little confidence in newspapers) is 46%, or 0.46. - You will use the binomial probability formula or a binomial distribution table to find the probability for each specified scenario (exactly five, at least six, and less than four). **Key Points to Remember:** - **Binomial Probability Formula:** \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(\binom{n}{k}\) is the binomial coefficient. - **Cumulative Probabilities:** - **At least six** indicates the sum of probabilities from P(6) to P(10). - **Less than four** indicates the sum of probabilities from P(0) to P(3). By solving these, you can determine the respective probabilities for each scenario concerning the confidence in newspapers among selected U.S. adults. **Note:** This page contains interactive elements, such as a textbox to input your calculated result and a "Check Answer" button to validate it. Be sure to have a calculator handy to compute these probabilities accurately.
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