Answer :
To find the 99.9% confidence interval for the population mean [tex]\mu[/tex] given a sample, we use the following formula for the confidence interval:\n\n[tex]\text{CI} = \left( \bar{x} - z \cdot \frac{s}{\sqrt{n}}, \bar{x} + z \cdot \frac{s}{\sqrt{n}} \right)[/tex]\n\nWhere:\n\n- [tex]\bar{x}[/tex] is the sample mean. In this case, [tex]\bar{x} = 51.7[/tex].\n- [tex]s[/tex] is the sample standard deviation. Here, [tex]s = 21.8[/tex].\n- [tex]n[/tex] is the sample size. We have [tex]n = 44[/tex].\n- [tex]z[/tex] is the z-score corresponding to the desired confidence level, which is 99.9% in this case.\n\nSince we need the z-score for a 99.9% confidence level, we look up the z-score such that the area in each tail is 0.05% (since 100% - 99.9% = 0.1%, and this is split between two tails in the normal distribution). The critical z-score is approximately [tex]3.291[/tex].\n\nLet's calculate the margin of error (MOE):\n\n[tex]\text{MOE} = z \cdot \frac{s}{\sqrt{n}} = 3.291 \cdot \frac{21.8}{\sqrt{44}} \approx 10.822[/tex]\n\nNow, we can find the confidence interval:\n\n[tex]\text{Lower bound} = 51.7 - 10.822 = 40.878[/tex]\n[tex]\text{Upper bound} = 51.7 + 10.822 = 62.522[/tex]\n\nThus, the 99.9% confidence interval for the population mean [tex]\mu[/tex] is approximately [tex](40.9, 62.5)[/tex] when rounded to one decimal place.
