Calculating the Probability of Sample Mean Using Normal Distribution and Central Limit Theorem

Understanding how to calculate the probability of a sample mean can be crucial in various fields, including manufacturing, quality control, and statistical analysis. This article will guide you through the process using statistical methods, specifically focusing on the Central Limit Theorem and normal distribution principles.

### Understanding the Problem

A manufacturer has noticed that the lengths of their items are normally distributed with a mean (μ) of 16.4 inches and a standard deviation (σ) of 5.4 inches. We need to determine the probability that the mean length of 9 randomly chosen items will be less than 13.4 inches.

### Step 1: Identify the Parameters Population Mean ((mu)): 16.4 inches Population Standard Deviation ((sigma)): 5.4 inches Sample Size ((n)): 9 ### Step 2: Calculate the Standard Error of the Mean (SEM)

The standard error of the mean (SEM) is a measure of the variability of the sample mean. It is calculated using the formula:

SEM (frac{sigma}{sqrt{n}})

Substituting the given values:

SEM (frac{5.4}{sqrt{9}} frac{5.4}{3} 1.8) inches

### Step 3: Standardize the Sample Mean

We need to convert the sample mean to a z-score to find the probability using the standard normal distribution. The z-score is calculated using the formula:

z (frac{bar{x} - mu}{SEM})

For a sample mean ((bar{x})) of 13.4 inches:

z (frac{13.4 - 16.4}{1.8} frac{-3.0}{1.8} approx -1.67)

### Step 4: Find the Probability Corresponding to the z-Score

Using a standard normal distribution table or calculator, we find the cumulative probability for the z-score of -1.67:

P(Z ### Conclusion

The probability that the mean length of 9 randomly chosen items is less than 13.4 inches is approximately 0.0478, or 4.78%.

### Understanding the Sample Size and Central Limit Theorem

For large sample sizes (n > 30), the Central Limit Theorem (CLT) applies, stating that the sample mean of a sufficiently large sample from any population has a normal distribution. However, in this particular example, the sample size is 9, which is relatively small.

### Calculation for a Larger Sample Size

For the sake of completeness, let's consider a larger sample size of 23 items.

#### Step 1: Calculate the Mean and Variance of the Sample Mean Sample Mean ((bar{x})) 13.4 inches Sample Standard Deviation (std(x)) 1.1 inches Sample Size (n) 23

The mean of the sampling distribution of the sample mean ((mu_{bar{x}})) is equal to the population mean:

(mu_{bar{x}} 13.4) inches

The standard deviation of the sampling distribution of the sample mean (also known as the standard error, SEM) is:

SEM (frac{sigma}{sqrt{n}} frac{1.1}{sqrt{23}} approx 0.229) inches

#### Step 2: Standardize and Find the Probability

Using the z-score formula:

z (frac{bar{x} - mu}{SEM})

Substituting the values:

z (frac{13 - 13.4}{0.229} frac{-0.4}{0.229} approx -1.75)

Using a standard normal distribution table or calculator, we find the probability of a z-score less than -1.75:

P(Z ### Conclusion for the Larger Sample Size

The probability that the mean length of a sample of 23 items is less than 13 inches is approximately 0.0401, or 4.01%.

### Final Thoughts

The Central Limit Theorem and normal distribution principles are powerful tools for manufacturing and statistical analysis. Understanding these concepts can help in making informed decisions about product quality and production processes.