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I'm having trouble understanding the difference between confidence interval and the width of confidence interval as given in the following question-enter image description here

I solved the first part of the question quite quickly but I really don't understand how to go about solving the second part. Is the difference between the upper limit and the lower limit of the interval supposed to be less than 0.5? Also, while trying to find a solution on the net, I came across the term margin of error which is apparently half of the width of confidence level. Is that relevant to this problem? Any help would be appreciated. Thanks!

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Is the difference between the upper limit and the lower limit of the interval supposed to be less than 0.5?

Yes, that is exactly what they are asking for. Or, if you read carefully, it should be only a little bit bigger than $0.5$, not necessarily smaller. Whatever they mean by that.

The confidence interval is an interval, with a lower and upper bound. The width of the confidence interval is a single real number, and the difference between the two bounds.

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The required confidence interval of $\mu$ is: $$\left(\bar{x}-z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}},\bar{x}+z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}\right)=\left(\bar{x}-1.96\cdot \frac{2}{\sqrt{n}},\bar{x}+1.96\cdot \frac{2}{\sqrt{n}}\right)$$ The width of the confidence interval is: $$\left(\bar{x}+1.96\cdot \frac{2}{\sqrt{n}}\right)-\left(\bar{x}-1.96\cdot \frac{2}{\sqrt{n}}\right)=\frac{7.84}{\sqrt{n}},$$ which must be slightly larger than $5$: $$\frac{7.84}{\sqrt{n}}>5 \Rightarrow \sqrt{n}<1.568 \Rightarrow n<2.5 \Rightarrow n=2$$ Note that $\frac{\sigma}{\sqrt{n}}$ is called Standard Error, while $z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{n}}$ is called Margin of Error (or Sampling Error). Indeed, the ME is half of the width of confidence interval.

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