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Papessa [141]
2 years ago
12

A random sample of n D 225 and xN D 21 was drawn from a normal population with a known

Mathematics
1 answer:
Rainbow [258]2 years ago
6 0

Using the z-distribution, the 95% confidence interval of the population mean is: (17.5, 24.5).

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

\overline{x} \pm z\frac{\sigma}{\sqrt{n}}

In which:

  • \overline{x} is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • \sigma is the standard deviation for the population.

In this problem, we have a 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so the critical value is z = 1.96.

For this problem, the other parameters are:

\overline{x} = 21, \sigma = 26.8, n = 225

Hence the bounds of the interval are:

\overline{x} - z\frac{\sigma}{\sqrt{n}} = 21 - 1.96\frac{26.8}{\sqrt{225}} = 17.5

\overline{x} + z\frac{\sigma}{\sqrt{n}} = 21 + 1.96\frac{26.8}{\sqrt{225}} = 24.5

The interval is: (17.5, 24.5).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

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