To solve this, we need to compute the probability that a majority of the sampled 100 people say "yes". This means we want the number of people who say "yes" to be greater than 50 out of the 100.

Step-by-Step Approach:

1. Define the Problem:

   • Probability of a single person saying "yes" is $p = 0.41$.
   • Probability of a single person saying "no" is $1 - p = 0.59$.
   • We sample 100 people, so the total number of people in the sample is $n = 100$.
   • We want the probability that more than 50 people say "yes", i.e., we need $X > 50$, where $X$ is the number of people saying "yes".

2. Use the Binomial Distribution: Since each person's response is independent, the number of "yes" responses follows a binomial distribution: $X \sim \text{Binomial}(n = 100, p = 0.41)$ The binomial distribution has the following probability mass function (PMF): $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ However, calculating this probability directly for all values of $X > 50$ can be cumbersome. So, we use an approximation.

3. Normal Approximation to the Binomial Distribution: The binomial distribution can be approximated by a normal distribution when $n$ is large, and this approximation works well here because $n = 100$ is sufficiently large.

   The parameters for the normal distribution are:

   • Mean ($\mu$): $\mu = np = 100 \times 0.41 = 41$
   • Variance ($\sigma^2$): $\sigma^2 = np(1 - p) = 100 \times 0.41 \times 0.59 = 24.19$
   • Standard deviation ($\sigma$): $\sigma = \sqrt{24.19} \approx 4.92$

   So, $X$ can be approximated by: $X \sim \mathcal{N}(\mu = 41, \sigma = 4.92)$

4. Apply the Normal Approximation: To find the probability that more than 50 people say "yes", we need to compute: $P(X > 50)$ First, we standardize this to a Z-score using: $Z = \frac{X - \mu}{\sigma}$ For $X = 50$, we compute the Z-score: $Z = \frac{50 - 41}{4.92} = \frac{9}{4.92} \approx 1.83$

5. Find the Probability: Using the standard normal distribution, we find the probability corresponding to a Z-score of 1.83. The cumulative probability for $Z = 1.83$ is approximately 0.9664.

   So, the probability of $X > 50$ is: $P(X > 50) = 1 - P(Z \leq 1.83) = 1 - 0.9664 = 0.0336$

Final Answer:

The probability that a majority of the 100 people say "yes" is approximately 0.0336, or 3.36%.

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Would you like more details on any of the steps?

Here are a few follow-up questions:

1. How does the normal approximation compare to the exact binomial distribution for this problem?
2. What would happen if the probability of "yes" was closer to 0.5?
3. How can we use the continuity correction in the normal approximation?
4. What are other scenarios where the binomial distribution is approximated by a normal distribution?
5. How does increasing the sample size affect the probability of a majority?

Tip: In problems with large $n$, always consider using the normal approximation to simplify calculations, especially when the binomial distribution becomes tedious to compute directly.