Marble Problem Example

• Problem details: A jar contains:

  • 3 red marbles

  • 4 blue marbles

  • 5 green marbles

• Two marbles are drawn at random with replacement.

• Let defined variable y represent the number of red marbles drawn.

• Possible values for y:

  • 0, 1, or 2 red marbles.

  • Calculation of probability for exactly one red marble drawn.

Probability Calculation Steps

1. Case 1: First marble drawn is red.

   • Probability = $P(R) = \frac{3}{12}$.

   • Probability that the second marble is not red = $P(NR) = \frac{9}{11}$.

2. Case 2: First marble drawn is not red, second marble drawn is red.

   • Probability = $P(NR) = \frac{9}{12}$ and $P(R) = \frac{3}{11}$.

3. Total probability formula combines both cases:

   • Total Probability = $\frac{3 \times 9}{12 \times 11} + \frac{9 \times 3}{12 \times 11}$.

   • Result = $\frac{27}{66}$ ≈ 0.409 (40.9%).

Finding the Probability of Drawing One or More Red Marbles

• Two methods:

  1. Compute $P(y=1) + P(y=2)$.

  2. Use complement rule: P(y \geq 1) = 1 - P(y < 1) = 1 - P(y=0).

Individual Probabilities

• Previous result: $P(y=1) = 0.409$.

• Probability for both marbles being red:

  • $P(R<em>1) = \frac{3}{12}$, $P(R</em>2) = \frac{2}{11}$ → Total = $\frac{6}{132}$.

• Summarize findings:

  • Probability of at least one red marble = $P(y \geq 1)$ ≈ 0.454.

Introduction to Density Curves

• Density curves: Smooth curves modeling distribution shapes.

Key Properties of Density Curves

1. Always on or above the horizontal axis.

2. Total area under the curve = 1 (100%).

Normal Distribution

• A variable is normally distributed if it reflects a normal curve shape in a histogram, referred to as the bell curve.

• Empirical Rule:

  • 68% data lies within 1 standard deviation from the mean.

  • 95% data lies within 2 standard deviations from the mean.

  • 99.7% data lies within 3 standard deviations from the mean.

Types of Normal Distribution

1. Standard Normal Distribution:

   • Mean = 0, Standard Deviation = 1.

   • The area represents probability, where exactly half (0.5) lies below zero and the other half above.

2. General Normal Distribution Characteristics:

   • Symmetric about the mean.

   • Can be standardized using $Z = \frac{X - \mu}{\sigma}$.

Example: Cholesterol Levels in US Females

• Data regarding cholesterol levels follows a normal distribution:

  • Mean = 206 mg/dL;

  • Standard Deviation = 44.7 mg/dL.

• Calculating Z-Scores:

  • Example cholesterol level: 180 mg/dL.

  • Z-score = $Z = \frac{180 - 206}{44.7} ≈ -0.58$ interprets how below the mean this value falls.

Using Area Under Curve for Probability

• Area under the curve corresponds to the probability that a random sample falls within specified intervals.

Normal CDF Function Use

1. To find probabilities within a range, inputs required will include lower bound, upper bound, mean, and standard deviation.

   • Example of equilibrium between specific cholesterol levels.

Finding Percentiles with Reverse Norm

1. Find weights separating top 10% using the Inverse Norm function.

2. Example with salmon weight distributions:

   • Mean = 12 lbs, Standard Deviation = 2.5 lbs.

   • Area under curve for finding percentiles adjusts accordingly.

   • Resulting weight for top 10% = 15.2 lbs.

Battery Lifespan Example Analysis

• Mean battery life = 50 hours; Standard deviation = 6 hours.

• Bottom 25% separation calculated with Inverse Norm returning 45.95 hours.

Final Exam Review

• Practice using calculators for Normal CDF for a range of probabilities including:

  • Standardization and the interpretation of results.

• Class concludes with review on how percentiles are computed, ensuring clear understanding of the inverse norm application for determining weights in various contexts.