Question

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 1

Given that:
mean,μ=35.6 min
std 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:-1.99
therefore:
z-score is given by:
(x-μ)/σ
hence:
-1.99=(x-35.6)/10.3
-20.497=x-35.6
x=35.6-20.497
x=15.103

Answer 2

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 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:

$Z-score=\frac{x-\mu}{\sigma}$

$-0.74=\frac{x-35.6}{10.3}$

$-0.74\times 10.3=x-35.6$

$-7.622=x-35.6$

$-7.622+35.6=x$

$x=27.978$

Approximately the value of x is 28.

The travel time is at least 28 minute.