We can solve this problem using the normal approximation to the binomial distribution, which is appropriate because the number of trials ($n = 450$) is large.

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Problem Breakdown:

• $n = 450$: total number of students who receive an offer.
• $p = 0.3$: probability that a student will accept the offer.
• $X$: number of students who accept the offer, modeled as a Binomial random variable $X \sim \text{Binomial}(n, p)$.
• We want to calculate $P(X > 150)$.

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Step 1: Mean and Variance of $X$

For a binomial distribution, the mean ($\mu$) and variance ($\sigma^2$) are given by: $\mu = n \cdot p = 450 \cdot 0.3 = 135,$ $\sigma^2 = n \cdot p \cdot (1-p) = 450 \cdot 0.3 \cdot 0.7 = 94.5.$ The standard deviation ($\sigma$) is: $\sigma = \sqrt{\sigma^2} = \sqrt{94.5} \approx 9.72.$

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Step 2: Normal Approximation

Using the normal approximation, $X \sim N(\mu, \sigma^2)$. That is: $X \approx N(135, 9.72^2).$ We want $P(X > 150)$. Applying the continuity correction, this becomes: $P(X > 150) \approx P\left(Z > \frac{150 - \mu - 0.5}{\sigma}\right),$ where $Z \sim N(0, 1)$ is the standard normal variable.

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Step 3: Standardization

Substitute $\mu = 135$ and $\sigma = 9.72$: $Z = \frac{150 - 135 - 0.5}{9.72} = \frac{14.5}{9.72} \approx 1.49.$ We now need $P(Z > 1.49)$.

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Step 4: Find the Probability

From standard normal distribution tables or a calculator: $P(Z > 1.49) = 1 - P(Z \leq 1.49).$ Using cumulative distribution tables: $P(Z \leq 1.49) \approx 0.9319.$ Thus: $P(Z > 1.49) = 1 - 0.9319 = 0.0681.$

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Final Answer:

The approximate probability that more than 150 students accept the offer is: $\boxed{0.0681 \, \text{or about } 6.81\%}.$

Let me know if you'd like additional details or a more visual explanation!

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5 Related Questions:

1. How would the probability change if the acceptance rate ($p$) was 35% instead of 30%?
2. What if the college needed exactly 150 students, not more than 150?
3. How does the normal approximation compare to the exact binomial probability?
4. If the college only made 300 offers, what would the probability of exceeding 150 acceptances be?
5. How does increasing $n$ affect the accuracy of the normal approximation?

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Tip:

For binomial probabilities with large $n$, the normal approximation works well, especially when $n \cdot p$ and $n \cdot (1-p)$ are both greater than 5.