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# Find the interval $\left[\mu-z \frac{\sigma}{\sqrt{n}}, \mu+z \frac{\sigma}{\sqrt{n}}\right]$ within which $95$ percent of the sample means would be expected to fall, assuming that each sample is from a normal population.$\mu=1,000, \sigma=15, n=9$

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To determine the interval with $95\%$ of the sample mean and known standard deviation. Let's start by defining a confidence interval for the mean of the population. But first and foremost, we must select the most acceptable formula for our objectives. For situations where the standard deviation is known, we can use the following formula.

\begin{aligned} \mu \pm z_\frac{\alpha}{2} \frac{\sigma}{\sqrt{n}} \end{aligned}

where,

• $n$ - is the sample size,
• $z_{\frac{\alpha}{2}}$ - is the appropriate $z$-value from the standard normal distribution table corresponding to level of significance,
• $\mu$ - is the population mean,
• $\sigma$ - is the population standard deviation.

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