Using the z-distribution, the confidence interval is given by: (11.95, 12.65).
What is a z-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- [tex]\sigma[/tex] is the standard deviation for the population.
In this problem, we have a 90% confidence level, hence[tex]\alpha = 0.9[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.
The other parameters are given as follows:
[tex]\overline{x} = 12.3, \sigma = 1.5, n = 50[/tex]
Hence the bounds of the interval are:
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 12.3 - 1.645\frac{1.5}{\sqrt{50}} = 11.95[/tex]
[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 12.3 + 1.645\frac{1.5}{\sqrt{50}} = 12.65[/tex]
More can be learned about the z-distribution at https://brainly.com/question/25890103
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