The problem provided can be described as follows:

Let $X$ be the number of people who make a pledge during a drive for contributions to a school. There are 500 people contacted in the drive, and each has a probability of 0.15 of making a pledge, independent of one another. The task is to estimate $P(80 < X)$.

Step 1: Model Selection

Since we are dealing with independent trials (500 people) and each trial has a constant probability of success (0.15), $X$, the number of people making pledges, follows a binomial distribution. That is: $X \sim \text{Binomial}(n = 500, p = 0.15)$ Where:

• $n = 500$ is the number of people contacted,
• $p = 0.15$ is the probability of making a pledge.

Step 2: Approximation with Normal Distribution

For large $n$, the binomial distribution can be approximated by a normal distribution using the Central Limit Theorem. The parameters of the approximating normal distribution are: $\mu = np = 500 \times 0.15 = 75$ $\sigma^2 = np(1 - p) = 500 \times 0.15 \times 0.85 = 63.75$ $\sigma = \sqrt{63.75} \approx 7.98$

Thus, $X$ can be approximated by $\mathcal{N}(75, 7.98^2)$, a normal distribution with mean 75 and standard deviation 7.98.

Step 3: Standardization and Probability Calculation

We are asked to find $P(80 < X)$. To compute this, we standardize the value of 80: $Z = \frac{80 - \mu}{\sigma} = \frac{80 - 75}{7.98} \approx 0.626$

Now, we need to find $P(Z > 0.626)$. Using standard normal distribution tables or a calculator: $P(Z > 0.626) \approx 0.265$

Final Answer:

$P(80 < X) \approx 0.265$

Let me know if you would like more details on the steps or calculations.

Here are 5 related questions to expand your understanding:

1. How would the result change if the probability of pledging was 0.20 instead of 0.15?
2. Can you explain why the normal approximation works well for large sample sizes in binomial distributions?
3. How would you calculate the exact binomial probability without using the normal approximation?
4. What would happen to the probability if we wanted $P(X > 90)$?
5. How do we interpret the result of $P(80 < X) \approx 0.265$ in terms of the school's fundraising campaign?

Tip: For binomial problems with large $n$ and moderate $p$, normal approximations can simplify calculations greatly.