Suppose Clay, a grocery store owner, is monitoring the rate at which customers enter his store. After watching customers enter for several weeks, he determines that the amount of time in between customer arrivals follows an exponential distribution with mean 25What is the 70th percentile for the amount of time between customers entering Clay's store? Round your answer to the nearest two decimal places. 70th percentile

Answer:70th percentile for the amount of time between customers entering is 30.10Step-by-step explanation:given data mean =  25to find out What is the 70th percentile for the amount of time between customers entering Clay's storesolutionwe know that here mean is mean = [tex]\frac{1}{\lambda}[/tex]   ..................1so here 25 =  [tex]\frac{1}{\lambda}[/tex] and we consider time value corresponding to 70th percentile = xso we can say P(X < x) = 1 - [tex]e^{- \lambda *x}[/tex]      P(X < x) = 70 % 1 - [tex]e^{\frac{-x}{25}}[/tex]  = 70 % 1 - [tex]e^{\frac{-x}{25} }[/tex]  = 0.70[tex]e^{\frac{- x}{25} }[/tex]  = 0.30take ln both side [tex]\frac{-x}{25}[/tex] = ln 0.30[tex]\frac{x}{25}[/tex]  = 1.203973x = 30.1070th percentile for the amount of time between customers entering is 30.10