The margin of error of the random selection is 0.29
The given parameters are:
[tex]n = 320[/tex] --- the sample size
[tex]\sigma = 2.9[/tex] --- the standard deviation
[tex]\bar x = 24[/tex] --- the mean
[tex]\alpha = 90\%[/tex] --- the confidence level.
The margin of error (E) is calculated as follows:
[tex]E = z \times \sqrt{\frac{\sigma^2}{n}}[/tex]
So, we have:
[tex]E = z \times \sqrt{\frac{3.2^2}{320}}[/tex]
[tex]E = z \times \sqrt{\frac{10.24}{320}}[/tex]
The z-value for 90% confidence level is 1.645.
Substitute 1.645 for z
[tex]E = 1.645 \times \sqrt{\frac{10.24}{320}}[/tex]
[tex]E = 1.645 \times \sqrt{0.032}[/tex]
Take square roots
[tex]E = 1.645 \times 0.1789[/tex]
Multiply
[tex]E = 0.2943[/tex]
Approximate
[tex]E = 0.29[/tex]
Hence, the margin of error is 0.29
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