Question 1200573: In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.
It is estimated that 3.5% of the general population will live past their 90th birthday. In a graduating class of 758 high school seniors, find the following probabilities. (Round your answers to four decimal places)
what is the probability that 30 or more will live beyond their 90th birthday?
Answer by GingerAle(43)   (Show Source):
You can put this solution on YOUR website! **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.**
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