Review: Sampling Distribution of the Sample Proportion, Binomial Distribution, Probability (7.5)

Binomial Distribution Example (N=3 Trials)
📌 Initial setup involved 200 green and 300 blue marbles, making the probability of success (drawing green, $P$) equal to $200/500 = 0.4$, and failure ($1-P$) equal to $0.6$.
🎲 The probability of drawing at least two green marbles in three draws with replacement is calculated as $P(\text{exactly 2 Green}) + P(\text{exactly 3 Green})$.
📊 The probability of exactly two green marbles was calculated by summing the probabilities of the three possible sequences (GGB, GBG, BGG), resulting in $0.288$.
✅ The final probability for at least two green marbles was $0.288 + 0.064 = \mathbf{0.352}$.

Binomial Distribution Application (N=5 Trials)
⚙️ When drawing five marbles, calculating $P(\text{at least 2 Green})$ requires summing probabilities for $k=2, 3, 4,$ and $5$ successes using the binomial formula.
🎯 Using the binomial formula for $k=2$ successes out of $n=5$ trials with $P=0.4$ yielded a probability of $\mathbf{0.3456}$.
📈 The total approximate probability for drawing at least two green marbles in five trials was calculated to be $\mathbf{0.6634}$.

Sampling Distribution and Central Limit Theorem (N=100 Trials)
📏 For $n=100$ trials and $P=0.4$, the Central Limit Theorem (CLT) conditions are met because $n \cdot P = 100 \cdot 0.4 = 40 \ge 10$ and $n \cdot (1-P) = 100 \cdot 0.6 = 60 \ge 10$.
📉 To find the approximate probability of drawing at least 35 green marbles ($\hat{p} \ge 0.35$), the $z$-score formula for proportions is applied: $z = \frac{\hat{p} - P}{\sqrt{P(1-P)/n}}$.
🧮 Plugging in values ($\hat{p}=0.35$, $P=0.4$, $n=100$), the resulting $z$-score is $\mathbf{-1.02}$.
✅ Since the $z$-score of $-1.02$ corresponds to an area of $0.1539$ to the left, the probability of drawing at least 35 (the area to the right) is $1 - 0.1539 = \mathbf{0.8461}$.

Key Points & Insights
➡️ For small numbers of trials ($n=3$ or $n=5$), exact probabilities are calculated using the binomial distribution formula or sample space enumeration.
➡️ For large sample sizes ($n=100$), the Central Limit Theorem is used to find an approximate probability via the sampling distribution of the sample proportion.
➡️ Always verify the CLT conditions ($n \cdot P \ge 10$ and $n \cdot (1-P) \ge 10$) before applying the normal approximation for proportions.

📸 Video summarized with SummaryTube.com on Dec 03, 2025, 17:41 UTC