Central Limit Theorem (CLT) EX

Central Limit Theorem (CLT)

  • The CLT states that for a random sample of size n from a population with mean (μ) and variance (σ²):

    • As n becomes large enough (typically at least n ≥ 30), the sampling distribution of the sample mean (x̄) approaches a normal distribution.

    • The mean of the sampling distribution of x̄ is μ, and the variance is σ²/n.

  • This theorem allows us to perform probability calculations involving the sample mean in a similar fashion to those we've done for the normal distribution.

  • The standardized version of x̄ is calculated as:

    • z = (x̄ - μ) / (σ / √n)

    • where z is approximately standard normal.

Example Problem Setup

  • Scenario: Testing a precision thermometer.

    • 36 temperature measurements (x₁, x₂, …, x₃₆) taken from a block of ice expected to be 0°C.

    • Each measurement is normally distributed with:

      • Mean (μ) = 0°C

      • Standard deviation (σ) = 0.2°C

      • Measurements are independent.

  • Objective: Find probabilities for:a) A single measurement differing by less than 0.1°C from 0°C.b) The sample mean of the 36 measurements differing by less than 0.1°C from 0°C.

Part A: Single Measurement Probability

  • Calculation:

    • Need to find P(-0.1 < x₁ < 0.1).

    • Standardize:

      • z = (x - μ) / σ

      • For -0.1: z = (-0.1 - 0) / 0.2 = -0.5

      • For 0.1: z = (0.1 - 0) / 0.2 = 0.5

    • Thus, find P(-0.5 < z < 0.5).

    • Using standard normal distribution table:

      • P(z < 0.5) = 0.6915

      • P(z < -0.5) = 0.3085

    • Final Calculation:

      • P(-0.1 < x₁ < 0.1) = 0.6915 - 0.3085 = 0.3830 (approximately).

Part B: Sample Mean Probability

  • Setup:

    • x̄ is approximately normal with:

      • Mean (μ) = 0

      • Standard deviation (σ) = 0.2 / √36 = 0.2 / 6 ≈ 0.0333

  • Calculation:

    • Need to find P(-0.1 < x̄ < 0.1).

    • Standardize:

      • For -0.1: z = (-0.1 - 0) / (0.2/6) = -3

      • For 0.1: z = (0.1 - 0) / (0.2/6) = 3

    • Thus, find P(-3 < z < 3).

    • Using standard normal distribution table:

      • P(z < 3) = 0.9987

      • P(z < -3) = 0.0013

    • Final Calculation:

      • P(-0.1 < x̄ < 0.1) = 0.9987 - 0.0013 = 0.9974 (approximately).

Analysis of Results

  • Comparative Results:

    • Part A: Probability for single measurement = 0.3830.

    • Part B: Probability for sample mean = 0.9974.

  • Explanation:

    • The probability for a sample mean is significantly higher because:

      • The standard deviation for x̄ is smaller (σ/√n), indicating less variability in the sample mean.

      • The area under the normal curve between -0.1 and 0.1 for the sample mean is much larger compared to that for a single measurement.

Conclusion

  • This video illustrates the application of the Central Limit Theorem in calculating probabilities related to the sample mean and highlights the differences in variability between single measurements and the mean of a sample.