Social Networking Sites A recent survey of 10 social networking sites has a mean of 12.67 million visitors for a specific month. The standard deviation was 3.3 miltion. Find the $95 \%$ confidence interval of the true mean. Assume the variable is normally distributed. Use a graphing calculator and round the answers to at least one decimal place.
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Step 1 ：We are given that the mean number of visitors for 10 social networking sites is 12.67 million, the standard deviation is 3.3 million, and we are asked to find the 95% confidence interval of the true mean. We assume the variable is normally distributed.

Step 2 ：The formula for the confidence interval of the mean for a normally distributed variable is given by: \(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(\sigma\) is the standard deviation of the population, and \(n\) is the size of the sample.

Step 3 ：Substituting the given values into the formula, we have: \(\bar{x} = 12.67\) million, \(z = 1.96\) (for a 95% confidence level), \(\sigma = 3.3\) million, and \(n = 10\).

Step 4 ：Calculating the margin of error, we get: \(1.96 \times \frac{3.3}{\sqrt{10}} = 2.0453611905969074\).

Step 5 ：Subtracting the margin of error from the mean, we get the lower bound of the confidence interval: \(12.67 - 2.0453611905969074 = 10.624638809403093\).

Step 6 ：Adding the margin of error to the mean, we get the upper bound of the confidence interval: \(12.67 + 2.0453611905969074 = 14.715361190596907\).

Step 7 ：Rounding the lower and upper bounds to at least one decimal place, we get: \(10.6\) and \(14.7\) respectively.

Step 8 ：Final Answer: The 95% confidence interval of the true mean is \(\boxed{10.6<\mu<14.7}\) million visitors.