Question 1199567
To solve this, we approach each part of the problem systematically using probability theory and approximations.

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### **Part 1: Approximation for at least 5 persons with blood type O**

1. **Given data**:
   - Probability of having type O blood: \( p = \frac{1}{80} = 0.0125 \)
   - Sample size: \( n = 200 \)
   - Random variable: Let \( X \) be the number of students with blood type O. \( X \) follows a **binomial distribution**:
     \[
     X \sim \text{Binomial}(n = 200, p = 0.0125)
     \]

2. **Normal approximation**:
   Since \( n \) is large and \( p \) is small, we approximate \( X \) using a normal distribution with:
   - Mean: \( \mu = n \cdot p = 200 \cdot 0.0125 = 2.5 \)
   - Standard deviation: \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{200 \cdot 0.0125 \cdot 0.9875} \approx 1.567 \)
   - Approximation: \( X \sim N(\mu = 2.5, \sigma = 1.567) \)

3. **Find \( P(X \geq 5) \):**
   Using the normal approximation, apply continuity correction:
   \[
   P(X \geq 5) \approx P\left(Z \geq \frac{5 - 0.5 - \mu}{\sigma}\right)
   \]
   Substitute values:
   \[
   Z = \frac{5 - 0.5 - 2.5}{1.567} = \frac{2}{1.567} \approx 1.276
   \]
   From standard normal tables or a calculator:
   \[
   P(Z \geq 1.276) = 1 - P(Z \leq 1.276) \approx 1 - 0.8980 = 0.1020
   \]
   Thus:
   \[
   P(X \geq 5) \approx 0.1020
   \]

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### **Part 2: Number of donors for \( P(\text{at least one type O}) \geq 0.9 \)**

1. **Given data**:
   - Probability of having type O blood: \( p = 0.0125 \)
   - Required: Find \( n \) such that \( P(\text{at least one type O}) \geq 0.9 \).

2. **Complement rule**:
   \[
   P(\text{at least one type O}) = 1 - P(\text{none with type O})
   \]
   For \( P(\text{none with type O}) \), the probability is:
   \[
   P(\text{none with type O}) = (1 - p)^n
   \]
   Thus:
   \[
   1 - (1 - p)^n \geq 0.9 \quad \Rightarrow \quad (1 - p)^n \leq 0.1
   \]

3. **Solve for \( n \):**
   Taking the natural logarithm:
   \[
   \ln((1 - p)^n) \leq \ln(0.1) \quad \Rightarrow \quad n \cdot \ln(1 - p) \leq \ln(0.1)
   \]
   Substitute \( p = 0.0125 \):
   \[
   \ln(1 - 0.0125) \approx \ln(0.9875) \approx -0.0126
   \]
   \[
   n \cdot (-0.0126) \leq \ln(0.1) \approx -2.3026
   \]
   \[
   n \geq \frac{-2.3026}{-0.0126} \approx 183
   \]

Thus, at least **183 donors** are required to ensure the probability of including at least one donor with type O blood is \( \geq 0.9 \).

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### **Final Answers**:
1. Probability of at least 5 type O donors: **0.1020**.
2. Number of donors needed for \( P(\text{at least one type O}) \geq 0.9 \): **183**.