A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate with 98% confidence to within 0.1 lb, the average force required to break the binding? Assume that σ is known to be 0.6 lb. (Use the value of z rounded to two decimal places.)

Answer:538 books should be tested.Step-by-step explanation:We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]Now, find the margin of error M as such[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?n books should be tested.n is found when [tex]M = 0.08[/tex]We have that [tex]\sigma = 0.72[/tex][tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex][tex]0.08 = 2.575*\frac{0.72}{\sqrt{n}}[/tex][tex]0.08\sqrt{n} = 2.575*0.72[/tex][tex]\sqrt{n} = \frac{2.575*0.72}{0.08}[/tex][tex](\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}[/tex][tex]n = 537.1[/tex]Rounding up538 books should be tested.