\( \mu-1.96(\sigma / \sqrt{n})<\bar{x}<\mu+1.96(\sigma / \sqrt{n}) \)
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96 \left( \frac{\sigma}{\sqrt{n}} \right) < \bar{x} < \mu + 1.96 \left( \frac{\sigma}{\sqrt{n}} \right) \] This inequality represents the 95% confidence interval for the population mean \(\mu\) based on the sample mean \(\bar{x}\), the population standard Show more…
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