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In order for confidence interval\(\left( {\overline x - {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }},\overline x + {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }}} \right)\) to have width \(w\), the necessary sample size \(n\) is,\(n = {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\)

The average force required to break the binding to be within \(0.1\) means that the width of the interval is \(2(0.1) = 0.2\).

In order for confidence interval\(\left( {\overline x - {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }},\overline x + {Z_{\alpha /2}} \times \frac{\sigma }{{\sqrt n }}} \right)\) to have width \(w\), the necessary sample size \(n\) is,

\(n = {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\)

The smaller width \(w\), the larger \(n\) must be.

Using the formula for \(\sigma = 0.8\) and \({Z_{\alpha /2}} = 1.96\), the sample size is

\(\begin{aligned}{}n &= {\left( {2{Z_{\alpha /2}} \times \frac{\sigma }{w}} \right)^2}\\ &= {\left( {2(1.96) \times \frac{{0.8}}{{0.2}}} \right)^2}\\ &= 245.86\end{aligned}\)

But the sampler size needs to be an integer, therefore round it up to \(n = 246\).

Hence, The sampler size is\(n = 246\).