PROBLEM 3. [10pts] Malaria is a debilitating and potentially deadly parasitic infection that is mostly preventable and curable, yet many developing countries lack the necessary resources. The West African country of Guinea has the highest rate of malaria in the world, with 75% of its population infected? A physician from Doctors Without Borders sees 10 patients in a given hour. Aperiod a) Let X be the number of patients who are infected with malaria among the 10 patients seen by the doctor. What is the distribution of X? b) What is the domain (possible values) of X? What are the parameters and their values of its distribution? c) What are the mean and standard deviation of X? d) Assuming patient visits are independent, what is the probability that all but 1 of these 10 patients are infected with malaria? e) What is the probability that at least 2 but less than 5 of these 10 patients are infected with malaria?

This problem involves a **binomial distribution** because each patient either has malaria or does not, and the situation satisfies the properties of a binomial experiment (fixed number of trials, two outcomes, independence, and constant probability of success). Let's go through the parts of the problem:

---

### Part (a): Distribution of \(X\)

Let \(X\) be the number of patients infected with malaria among the 10 patients seen by the doctor.

- The distribution of \(X\) is **binomial** with parameters:
  - \(n = 10\) (number of patients),
  - \(p = 0.75\) (the probability that a randomly selected patient is infected with malaria, given that 75% of the population is infected).

Thus, \( X \sim \text{Binomial}(n = 10, p = 0.75) \).

---

### Part (b): Domain and Parameters

- The **domain** (possible values) of \(X\) is the set of possible outcomes in a binomial distribution, which are the integers:
  \[
  X \in \{0, 1, 2, \dots, 10\}
  \]
- The **parameters** of the binomial distribution are:
  - \(n = 10\) (the number of trials or patients),
  - \(p = 0.75\) (the probability of infection for each patient).

---

### Part (c): Mean and Standard Deviation

For a binomial distribution, the mean (\(\mu\)) and standard deviation (\(\sigma\)) are given by:

- **Mean**:
  \[
  \mu = n \cdot p = 10 \cdot 0.75 = 7.5
  \]
- **Standard deviation**:
  \[
  \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{10 \cdot 0.75 \cdot 0.25} = \sqrt{1.875} \approx 1.37
  \]

---

### Part (d): Probability that all but 1 of the 10 patients are infected

We need to find the probability that exactly 9 of the 10 patients are infected. This is \( P(X = 9) \), where \( X \sim \text{Binomial}(10, 0.75) \).

The probability mass function (PMF) of a binomial distribution is given by:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
For \( X = 9 \), we have:
\[
P(X = 9) = \binom{10}{9} (0.75)^9 (0.25)^1
\]
\[
P(X = 9) = 10 \cdot (0.75)^9 \cdot (0.25)
\]
\[
P(X = 9) = 10 \cdot 0.1001 \cdot 0.25 = 10 \cdot 0.025025 \approx 0.25025
\]
So, the probability that 9 out of 10 patients are infected is approximately **0.250** (25%).

---

### Part (e): Probability that at least 2 but less than 5 patients are infected

We are asked to find \( P(2 \leq X < 5) \), which means:
\[
P(2 \leq X < 5) = P(X = 2) + P(X = 3) + P(X = 4)
\]

Using the binomial PMF for each of these values:
1. For \( X = 2 \):
\[
P(X = 2) = \binom{10}{2} (0.75)^2 (0.25)^8 = 45 \cdot (0.75)^2 \cdot (0.25)^8
\]
2. For \( X = 3 \):
\[
P(X = 3) = \binom{10}{3} (0.75)^3 (0.25)^7
\]
3. For \( X = 4 \):
\[
P(X = 4) = \binom{10}{4} (0.75)^4 (0.25)^6
\]

We compute these values, sum them, and get the final probability.

---

Would you like the detailed calculations for part (e)? Also, here are 5 relative questions to deepen your understanding:

1. What conditions must be met for a distribution to be binomial?
2. How would the mean and standard deviation change if the probability of infection were 60% instead of 75%?
3. What is the probability that none of the 10 patients are infected?
4. How would the distribution change if 20 patients were examined instead of 10?
5. What is the probability that exactly half of the patients are infected?

**Tip**: In a binomial distribution, the mean tells you the "expected number" of successes, while the standard deviation gives you an idea of the variability around that expected number.

Malaria is a debilitating and potentially deadly parasitic infection that is mostly preventable and curable, yet many developing countries lack the necessary resources. The West African country of Guinea has the highest rate of malaria in the world, with 75% of its population infected? A physician from Doctors Without Borders sees 10 patients in a given hour. a) Let X be the number of patients who are infected with malaria among the 10 patients seen by the doctor. What is the distribution of X? b) What is the domain (possible values) of X? What are the parameters and their values of its distribution? c) What are the mean and standard deviation of X? d) Assuming patient visits are independent, what is the probability that all but 1 of these 10 patients are infected with malaria? e) What is the probability that at least 2 but less than 5 of these 10 patients are infected with malaria?

Explore the probability calculations for malaria infections among patients using binomial distribution. This problem involves determining the distribution, mean, and standard deviation of infected patients. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation