Question 1200573
**1. Check for Normal Approximation**

* **Conditions:**
    * **np ≥ 10:** (758)(0.035) = 26.53 ≥ 10
    * **n(1-p) ≥ 10:** (758)(0.965) = 731.47 ≥ 10

* **Conclusion:** Since both conditions are met, the normal approximation to the binomial distribution is appropriate.

**2. Calculate Mean and Standard Deviation**

* **Mean (μ):** μ = n * p = 758 * 0.035 = 26.53
* **Standard Deviation (σ):** σ = √(n * p * (1 - p)) = √(758 * 0.035 * 0.965) ≈ 5.07

**3. Calculate Z-score**

* **Continuity Correction:** For "30 or more," we use 29.5 as the lower bound.
* **Z-score:** z = (X - μ) / σ = (29.5 - 26.53) / 5.07 ≈ 0.586

**4. Find Probability**

* **Using a Standard Normal Distribution Table or Calculator:**
    * Find the area to the right of z = 0.586. 
    * P(X ≥ 30) ≈ 1 - 0.7219 = 0.2781

**Therefore, the probability that 30 or more of the 758 high school seniors will live beyond their 90th birthday is approximately 0.2781.**