https://file.aiquickdraw.com/mathbot/file/1727236310546h7tle3me.jpg

Let's work through the questions from the image.

### Problem 4.3.1:
The problem involves a binomial distribution where:
- \( n = 20 \) (the sample size),
- \( p = 0.24 \) (the probability of success, i.e., the probability that a person has been told they have hypertension).

We need to calculate the probabilities for each part of the question.

In a binomial distribution, the probability mass function is given by:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
where \( k \) is the number of successes.

Let's handle each part:
- **(a) Exactly three**: We need to find \( P(X = 3) \).
- **(b) Three or more**: We need to find \( P(X \geq 3) \), which can be calculated as \( 1 - P(X < 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2)) \).
- **(c) Fewer than three**: This is \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \).
- **(d) Between three and seven, inclusive**: This is \( P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) \).

### Problem 4.3.2:
The expected number of adults in the sample who have been told they have hypertension is calculated using the expected value formula for a binomial distribution, which is:

\[
E(X) = n \cdot p
\]

Given \( n = 20 \) and \( p = 0.24 \), we can find the expected number.

I will now compute the required probabilities and expected value.Here are the results for the problems:

### Problem 4.3.1:
- **(a) Exactly three**: The probability that exactly 3 people have been told they have hypertension is approximately **0.1484** (14.84%).
- **(b) Three or more**: The probability that 3 or more people have been told they have hypertension is approximately **0.8915** (89.15%).
- **(c) Fewer than three**: The probability that fewer than 3 people have been told they have hypertension is approximately **0.1085** (10.85%).
- **(d) Between three and seven, inclusive**: The probability that between 3 and 7 people have been told they have hypertension is approximately **0.8080** (80.80%).

### Problem 4.3.2:
- The expected number of adults in a sample of 20 who have been told they have hypertension is **4.8**.

Would you like more details or have any questions about these solutions?

Here are 5 related questions to expand on:
1. How does the binomial distribution change with different sample sizes?
2. What happens if the probability of having hypertension changes to 0.3? How would the probabilities be affected?
3. Can you explain the difference between cumulative probability and point probability?
4. How would the expected number change if the sample size doubled?
5. Can we approximate this binomial distribution using a normal distribution? If so, how?

**Tip:** When working with binomial distributions, it’s useful to visualize the distribution to understand the likelihood of different outcomes.

Based on data collected by the National Center for Health Statistics and made available to the public in the Sample Adult database, an estimate of the percentage of adults who have at some point been told they have hypertension is 23.53 percent. If we select a simple random sample of 20 U.S. adults, and assume that the probability that each has been told that he or she has hypertension is 0.24, find the probability that the number of people in the sample who have been told they have hypertension will be: (a) Exactly three (b) Three or more (c) Fewer than three (d) Between three and seven, inclusive. Then, for question 4.3.2, find how many adults who have been told that they have hypertension would be expected in a sample of 20.

Solve probability problems using binomial distribution for a hypertension study with U.S. adults. Learn how to calculate exact probabilities and expected values for given sample sizes. Includes solutions for exactly three, fewer than three, and between three and seven outcomes.

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