The standard deviation of the sampling distribution is called the standard error and is calculated as: \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\)

Standardization formula for sampling distribution: \(z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}\)

Sampling distribution is used because it is more efficient and allows us to estimate population parameters without having to measure the entire population, as well as calculate probabilities based on sample size.

Example 1:

Height distribution of Canadians: 𝜇 = 160 , 𝜎 = 7 , 𝑛 = 10 .

Standard error:

\(\frac{7}{\sqrt{10}}\) = 2.21

Probability \({\bar{X}}\) < 157

z = \(\frac{157 - 160}{2.21}\) = -1.36

From the z table, the probability = 0.0869 (8.69%).

Example question 2:

Proportion of people taller than 170 cm:

Standardization: z = \(\frac{170 - 160}{7}\) = 1.43

Right area 1 - 0.9236 = 0.0764

So the proportion is 7.64% .

A general rule of thumb:

• The CLT can be safely applied when 𝑛 ≥ 30 .

• If the population itself is already normally distributed, then the sampling distribution will be normal even for small samples.

When evaluating situations to determine whether the sampling distribution will be approximately normal:

• Cases where 𝑛 < 30 and the population is not normal → not normal.

• Cases where 𝑛 ≥ 30 , or the population is normal → approximately normal.

The sample proportion and population proportion are calculated using the same idea:

Sample Proportion \(\hat{p} = \frac{\text{number of successes}}{\text{sample size}}\)

Population Proportion \(p = \frac{\text{number of successes in population}}{\text{population size}}\)

Since each sample contains different individuals, repeated samples can produce different values of \(\hat{p}\).

Plotting all these \(\hat{p}\) values forms the sampling distribution of the sample proportion.

This distribution has a mean and a standard deviation:

Mean of the sampling distribution

\(\mu_{\hat{p}} = p\)

Standard deviation (Standard Error) \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\)

Z-score formula for proportions \(Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}\)

To apply the Central Limit Theorem (CLT) for sample proportions, two conditions must be satisfied:

Conditions for CLT to apply \(np \ge 10\)

\(n(1 - p) \ge 10\)

When both conditions are met, the sampling distribution of \(\hat{p}\) is approximately normal, allowing us to use Z-tables and normal distribution methods. Sampling distributions for proportions are also related to the binomial distribution, since proportions are essentially standardized binomial outcomes.

1. Probability of Green & Blue

• There are 200 green and 300 blue marbles → Probability of green:

\(p = \frac{200}{500} = 0.4\)

• Probability of blue: \(q = 1 - p = 0.6\)

• When drawing three times with replacement, the probability of a specific sequence (e.g., GBB) is: \(0.4 \times 0.6 \times 0.6\)

2. Probability of “At Least Two Green” (Manual Sample Space Method)

Enumerate all sequences with ≥2 greens and add their probabilities.

Example: Probability(G G B), Probability(G B G), Probability(B G G), Probability(G G G)

Final combined probability from the example: 0.352

3. Using Binomial Formula for Larger n

Instead of listing sample space, use the binomial formula: \(P(X = k) = \binom{n}{k} p^{k} (1-p)^{\,n-k}\)

To find at least 2 successes: \(P(X \ge 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)\)

Example calculation for \(n = 5\), \(p = 0.4\) gives total probability:

\[ P(X \ge 2) = 0.66304 \]

4. Important Note

Using CLT gives an approximation, not the exact binomial probability.

Exact results require the full binomial sum or sample space enumeration, which becomes impractical for large 𝑛