2. Hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes. Between how many minutes must be the lengths of at least 93.75% of these surgical procedures? Answer:

Answer:93.75% of these surgical procedures takes at-least 80 to 160 minutes.Step-by-step explanation:Given : Hospital records show that a certain surgical procedure takes on the average of 120 minutes with a standard deviation of 10 minutes.To find : Between how many minutes must be the lengths of at least 93.75% of these surgical procedures?Solution : The average mean is [tex]\mu=120[/tex] minutes.The standard deviation is [tex]\sigma=10[/tex] minutes.At least 93.75% of these surgical procedures,Applying Chebyshev's theorem,For any constant k>1, no more than [tex]\frac{1}{k^2}[/tex] of teh data set lie outside the k standard deviations away from the mean i.e. at-least [tex](1-\frac{1}{k^2})[/tex] of the distribution value fall within 'k' standard deviations.So, [tex](1-\frac{1}{k^2})=0.9375[/tex][tex]\frac{1}{k^2}=0.0625[/tex][tex]k^2=\frac{1}{0.0625}[/tex][tex]k^2=\frac{10000}{625}[/tex][tex]k^2=16[/tex][tex]k=4[/tex]Length must be given by,[tex]L=(\mu-k\times \sigma,\mu+k\times \sigma)[/tex][tex]L=(120-4\times 10,120+4\times 10)[/tex][tex]L=(120-40,120+40)[/tex][tex]L=(80,160)[/tex]Therefore, 93.75% of these surgical procedures takes at-least 80 to 160 minutes.