69% of U.S. adults have very littie conhidence in newspapers. You randomly select 10 U.S. adults.
(b) P(x≥6) =
(Round to three decimal places as needed.)

This problem involves a **binomial distribution**, where:

- The probability of success (having little confidence in newspapers) is \( p = 0.69 \),
- The probability of failure is \( q = 1 - p = 0.31 \),
- The number of trials (the sample of U.S. adults) is \( n = 10 \).

We are tasked with calculating \( P(x \geq 6) \), the probability that 6 or more adults out of 10 have little confidence in newspapers.

We can calculate this by finding the cumulative probability for \( x = 0 \) to \( x = 5 \) and subtracting it from 1:

\[
P(x \geq 6) = 1 - P(x \leq 5)
\]

The formula for the binomial probability of \( x \) successes is given by:

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

Where:
- \( \binom{n}{x} \) is the binomial coefficient.

Let me calculate this for you.The probability that 6 or more U.S. adults out of 10 have little confidence in newspapers is approximately \( P(x \geq 6) = 0.832 \) (rounded to three decimal places).

Would you like further details or have any other questions?

Here are 5 related questions you might find useful:
1. What is the probability that exactly 6 adults have little confidence in newspapers?
2. What is the expected number of adults out of 10 who have little confidence in newspapers?
3. How do you calculate binomial probabilities by hand?
4. What is the probability that fewer than 4 adults have little confidence in newspapers?
5. Can this problem be solved using normal approximation?

**Tip**: When working with binomial distributions, if \( n \) is large and \( p \) is not too close to 0 or 1, you can use a normal approximation for easier calculations.

69% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. (b) P(x≥6) = (Round to three decimal places as needed.)

Calculate the probability that 6 or more out of 10 U.S. adults have little confidence in newspapers using a binomial distribution. With a success probability of 0.69, this problem demonstrates key concepts in probability and binomial distributions. Suitable for students in Grades 10-12.

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