1. **State the problem:** Meryl wants to estimate the population mean number of photos Instagram users have, based on a sample of 80 users with a sample mean of 120 photos and a known population standard deviation of 50. 2. **Point estimate:** The best single estimate (point estimate) of the population mean is the sample mean itself, which is $120$ photos. This is because the sample mean is an unbiased estimator of the population mean. 3. **Why the estimate may differ from the true mean:** The sample mean is unlikely to be exactly the same as the population mean due to sampling variability. Different samples may produce different means because of random chance. 4. **Calculate the standard error:** The standard error (SE) of the sample mean is given by $$ SE = \frac{\sigma}{\sqrt{n}} = \frac{50}{\sqrt{80}} \approx 5.590 \text{ (3 decimal places)} $$ (Note: The problem states 50/√120 = 4.564, but since the sample size is 80, we use 80.) 5. **Determine the 90% confidence interval:** For a 90% confidence level, the z-value (critical value) is approximately $z = 1.645$. 6. **Calculate margin of error:** $$ ME = z \times SE = 1.645 \times 5.590 \approx 9.196 $$ 7. **Calculate confidence interval:** $$ \text{Lower bound} = 120 - 9.196 = 110.804 $$ $$ \text{Upper bound} = 120 + 9.196 = 129.196 $$ 8. **Interpretation:** We are 90% confident that the true population mean number of photos lies between approximately 110.8 and 129.2 photos. **Final answer:** The 90% confidence interval for the population mean is $$ (110.8, 129.2) $$.