If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within LaTeX: \pm± 0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?

Answer:n=97Step-by-step explanation:The margin of error is the range of values below and above the sample statistic in a confidence interval.Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".Assuming the X follows a normal distribution[tex]X \sim N(\mu, \sigma=0.01)[/tex]We know that the margin of error for a confidence interval is given by:[tex]Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]   (1)The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.95=.05[/tex] and [tex]\alpha/2=0.025[/tex]Using the normal standard table, excel or a calculator we see that:[tex]z_{\alpha/2}=1.96[/tex]If we solve for n from formula (1) we got:[tex]\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}[/tex][tex]n=(\frac{z_{\alpha/2} \sigma}{Me})^2[/tex]And we have everything to replace into the formula:[tex]n=(\frac{1.96(0.05)}{0.01})^2 =96.04[/tex]And if we round up the answer we see that the value of n to ensure the margin of error required [tex]\pm=0.01 Liters[/tex] is n=97.