Question 1194131: Yuen Zhi is running a ring toss event at the school fair. In the event, each attempt has a 15% chance of winning a prize. She has 45 prizes and believes that 250 people will attempt the event. She is worried that she won't have enough prizes. Based on the information provided, can Yuen be at least 95% confident that she will she have enough prizes for the event?
Answer by parmen(42)   (Show Source):
You can put this solution on YOUR website! **1. Define the Variables:**
* **n:** Number of attempts = 250
* **p:** Probability of winning a prize = 0.15
* **X:** Number of prizes won
**2. Calculate the Expected Number of Prizes Won:**
* Expected number of prizes = n * p = 250 * 0.15 = 37.5
**3. Determine the Standard Deviation:**
* Standard deviation (σ) = √(n * p * (1 - p)) = √(250 * 0.15 * 0.85) ≈ 5.63
**4. Use Normal Approximation (Assuming n*p and n*(1-p) are both greater than 5):**
* Since n*p = 37.5 and n*(1-p) = 212.5, both are greater than 5, we can use the normal approximation to the binomial distribution.
* **Calculate the Z-score for 95% Confidence:**
* For a 95% confidence level, the Z-score is approximately 1.96 (from the standard normal distribution table).
* **Calculate the Upper Bound for the Number of Prizes:**
* Upper Bound = Expected Value + (Z-score * Standard Deviation)
= 37.5 + (1.96 * 5.63) ≈ 48.6
**5. Conclusion:**
* Based on the normal approximation, Yuen can be at least 95% confident that the number of prizes won will be less than or equal to 48.6.
* Since she has 45 prizes, there is a possibility that she might not have enough prizes for all the winners at the 95% confidence level.
**Important Note:**
* This analysis uses a normal approximation, which may have some slight inaccuracies. For more precise calculations, you could use the binomial distribution directly (using statistical software or tables).
Let me know if you'd like a more detailed explanation of any of the steps or if you have any other questions.
|
|
|
