To calculate the probability that the mean length of 35 randomly chosen items is greater than 15.8 inches, we will use the Central Limit Theorem (CLT). According to the CLT, the sampling distribution of the sample means will be approximately normally distributed if the sample size is large enough.
Step 1: Calculate the mean and standard deviation of the sample mean.
Given:
- Population mean \( \mu = 16.2 \) inches
- Population standard deviation \( \sigma = 4.6 \) inches
- Sample size \( n = 35 \)
The mean of the sampling distribution \( \mu_{\bar{x}} \) is the same as the population mean: \[ \mu_{\bar{x}} = \mu = 16.2 \]
The standard deviation of the sampling distribution \( \sigma_{\bar{x}} \), also known as the standard error of the mean, is calculated using the formula: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{4.6}{\sqrt{35}} \approx \frac{4.6}{5.916} \approx 0.778 \]
Step 2: Calculate the z-score for a sample mean of 15.8 inches.
We can find the z-score using the formula: \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \] Where \( \bar{x} = 15.8 \) inches: \[ z = \frac{15.8 - 16.2}{0.778} = \frac{-0.4}{0.778} \approx -0.514 \]
Step 3: Find the probability that the sample mean is greater than 15.8 inches.
We need to find \( P(\bar{x} > 15.8) \), which is equivalent to finding \( P(z > -0.514) \).
Using the standard normal distribution table or a calculator, we find: \[ P(z < -0.514) \approx 0.3034 \] Thus: \[ P(z > -0.514) = 1 - P(z < -0.514) \approx 1 - 0.3034 = 0.6966 \]
Conclusion
The probability that the mean length of 35 randomly chosen items is greater than 15.8 inches is approximately: \[ \boxed{0.6966} \]