Math Problem Statement

Sampling distribution from a normal population

Solution

A sampling distribution refers to the probability distribution of a statistic (e.g., mean, variance, proportion) based on a random sample from a population. When sampling from a normal population, the sampling distribution has specific properties depending on the statistic being considered and the sample size. Here's a breakdown:


1. Sampling Distribution of the Sample Mean

  • If the population is normally distributed with mean μ\mu and standard deviation σ\sigma, then:
    • The sample mean xˉ\bar{x} is also normally distributed.
    • The mean of the sampling distribution is μxˉ=μ\mu_{\bar{x}} = \mu.
    • The standard deviation of the sampling distribution (known as the standard error) is: σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} where nn is the sample size.
  • Formula: XˉN(μ,σ2n)\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)

2. Central Limit Theorem (CLT)

  • Even if the population is not normal, for a sufficiently large sample size (n30n \geq 30), the sampling distribution of the sample mean approaches a normal distribution.
  • This allows us to apply normal distribution properties even to non-normally distributed populations when nn is large.

3. Sampling Distribution of the Sample Variance

  • When sampling from a normal population, the sample variance (s2)(s^2) has a distribution related to the chi-square distribution: (n1)s2σ2χn12\frac{(n-1)s^2}{\sigma^2} \sim \chi^2_{n-1} where n1n-1 represents the degrees of freedom.

4. Sampling Distribution of the Sample Proportion

  • If sampling involves proportions (e.g., number of successes in a sample of size nn), and the population is normal or approximated as normal, the sampling distribution of the sample proportion is: p^N(p,p(1p)n)\hat{p} \sim N\left(p, \sqrt{\frac{p(1-p)}{n}}\right) where pp is the population proportion.

Example Use:

If a population has μ=50\mu = 50 and σ=10\sigma = 10, and you take a sample of size n=25n = 25, then:

  • The sample mean has μxˉ=50\mu_{\bar{x}} = 50 and σxˉ=1025=2\sigma_{\bar{x}} = \frac{10}{\sqrt{25}} = 2.

Would you like more detailed examples, visualizations, or an application of these principles?


Related Questions:

  1. How does the Central Limit Theorem affect the sampling distribution for non-normal populations?
  2. What is the role of standard error in hypothesis testing?
  3. How do sample size and variability affect the shape of the sampling distribution?
  4. What happens to the sampling distribution as the sample size increases?
  5. How do we calculate probabilities using the sampling distribution of the sample mean?

Tip:

Always verify whether the sample size is large enough for approximations (like CLT) when dealing with non-normal populations.

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Math Problem Analysis

Mathematical Concepts

Probability Distribution
Sampling Distribution
Central Limit Theorem
Chi-Square Distribution

Formulas

Standard error: σ_x̄ = σ / √n
Sample variance: (n-1)s^2 / σ^2 ~ χ²_(n-1)
Sample proportion: p̂ ~ N(p, √(p(1-p)/n))

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Level (Statistics or Probability)