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.

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 min

Standard deviation, σ=10.3 min

we 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:

Approximately the value of x is 28.

The travel time is at least 28 minute.

step-by-step explanation: you take each of the numbers and divide by the amount then you take the range