We can approach this problem as a binomial distribution problem because we are given a probability of success (receiving financial aid) and we are interested in the number of successes (students who receive financial aid) in a fixed number of trials (4 students).

Part a: What constitutes a success on a single trial?

Success on a single trial:

• A success is defined as a student receiving financial aid.

So we can define:

$x =$ a student receiving financial aid.

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Part b: What is the probability of success on a single trial?

The probability of success (receiving financial aid) on a single trial is given as 70%. Therefore:

$p = 0.70$.

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Part c: Make a table that shows the probability distribution

This is a binomial distribution where:

• $n = 4$ (number of trials),
• $p = 0.70$ (probability of success),
• $q = 1 - p = 0.30$ (probability of failure).

The probability mass function for a binomial distribution is given by:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where $k$ is the number of students receiving aid, $n = 4$, and $p = 0.70$.

Let's compute the probabilities for each value of $k$ (0, 1, 2, 3, 4):

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Let’s calculate each probability and create a table.Here is the probability distribution for the number of students receiving financial aid ($k = 0, 1, 2, 3, 4$):

\hline k & P(X = k) \\ \hline 0 & 0.0081 \\ 1 & 0.0756 \\ 2 & 0.2646 \\ 3 & 0.4116 \\ 4 & 0.2401 \\ \hline \end{array}$$ --- ### Part d: Find the probability that **AT MOST 2** of the students in the sample receive financial aid. This means we need to calculate $$P(X \leq 2)$$, which is the sum of the probabilities for $$X = 0, 1, 2$$. $$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$$The probability that **at most 2** of the students receive financial aid is: $$P(X \leq 2) = 0.3483$$ --- ### Part e: Find the probability that **AT LEAST 3** of the students in the sample receive financial aid. This means we need to calculate $$P(X \geq 3)$$, which is the sum of the probabilities for $$X = 3$$ and $$X = 4$$. $$P(X \geq 3) = P(X = 3) + P(X = 4)$$The probability that **at least 3** of the students receive financial aid is: $$P(X \geq 3) = 0.6517$$ --- ### Part f: Find the mean and the standard deviation using the shortcut formulas for the binomial distribution. - **Mean of the binomial distribution**: $$E(X) = n \cdot p$$ - **Standard deviation of the distribution**: $$\sigma = \sqrt{n \cdot p \cdot (1 - p)}$$ Let’s calculate both.It seems like I can’t do more advanced data analysis right now. Please try again later. However, you can easily calculate the mean and standard deviation using these formulas: - **Mean**: $$E(X) = n \cdot p = 4 \cdot 0.70 = 2.8$$ - **Standard Deviation**: $$\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{4 \cdot 0.70 \cdot 0.30} = \sqrt{0.84} \approx 0.9165$$ Would you like more details on any part? Here are some related questions for further understanding: 1. How do binomial distributions differ from normal distributions? 2. Can you show how to calculate the binomial coefficient? 3. What are real-world examples of binomial distributions? 4. How would the probabilities change if more students were selected? 5. How do we interpret variance in the binomial distribution? **Tip**: In binomial distributions, increasing the number of trials tends to make the distribution look more like a normal distribution!