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:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

the answer is d

step-by-step explanation: