To find the probability of randomly selecting 30 values from a normally distributed population with a mean of 3.2 and a standard deviation of 0.7, and obtaining a sample mean greater than 3.6, we can use the Central Limit Theorem. According to this theorem, the sampling distribution of the sample mean will be normally distributed with the following parameters:
- Mean of the sample mean (\(\mu_{\bar{x}}\)) = Mean of the population = 3.2
- Standard deviation of the sample mean (\(\sigma_{\bar{x}}\)) = \(\frac{\sigma}{\sqrt{n}} = \frac{0.7}{\sqrt{30}}\)
First, calculate \(\sigma_{\bar{x}}\): \[ \sigma_{\bar{x}} = \frac{0.7}{\sqrt{30}} \approx \frac{0.7}{5.477} \approx 0.1270 \]
Next, calculate the z-score for a sample mean of 3.6 using the following formula: \[ z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \] Substituting the values: \[ z = \frac{3.6 - 3.2}{0.1270} \approx \frac{0.4}{0.1270} \approx 3.149 \]
Now, look up the z-score of 3.149 in the standard normal distribution table or use a calculator to find the probability.
The probability of a z-score greater than 3.149 is very small. Using the standard normal table or calculator, we find that: \[ P(Z > 3.149) \approx 0.0009 \]
Thus, the probability of randomly selecting 30 values and getting a mean greater than 3.6 is approximately:
\[ \boxed{0.0009} \]