Suppose abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min. determine the travel time x such that 22.96% of the 60 days have a travel time that is at least x.

Answer:The travel time is at least 28 minute.Step-by-step explanation:Given : Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min. To determine : The travel time x such that 22.96% of the 60 days have a travel time that is at least x.Solution : Mean, μ=35.6 minStandard deviation, σ=10.3 minwe are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.Using z-table, the z-score that will give us 0.2296 is -0.74 (Shown in attached graph)The formula of z-score is given by: [tex]Z-score=\frac{x-\mu}{\sigma}[/tex][tex]-0.74=\frac{x-35.6}{10.3}[/tex][tex]-0.74\times 10.3=x-35.6[/tex][tex]-7.622=x-35.6[/tex][tex]-7.622+35.6=x[/tex][tex]x=27.978[/tex]Approximately the value of x is 28.The travel time is at least 28 minute.