A professor would like to estimate the average number of hours his students spend studying for his class. He wants to estimate μwith 99% confidence and with a margin of error of at most 2 hours. From past experience he knows the standard deviation of number of hours spent studying to be 8 hours. How many students need to be surveyed to meet these requirements? A. 106B. 105C. 107D. 108

Answer:n=106.1≅106Option A which is 106 is correctStep-by-step explanation:In order to find the number of students need to be surveyed to meet the requirements we will use the following formula:[tex]n=\frac{Z^2*S^2}{E^2}[/tex]where:n is the number of students need o be surveyedZ is the distributionS is the standard DeviationE is the error MarginNow:S=8 ,E=2, CI= 99% or 0.99For Z we need to find Alpha/2Alpha=1-0.99Alpha=0.01Alpha/2=0.005 or 0.5%From the cumulative Standard Distribution Table:Z at alpha/2 =2.576n= (2.576^2* 8^2)/2^2n=106.1≅106Option A which is 106 is correct