Proportion/probability?

Question

The travel time for a college student to get between home and school is uniformly distributed between 40 and 70 minutes. This student leaves home each school morning to go to school exactly 1 hour before her class time starts. On what proportion of school days will she be late to class?

Answer 1

1/3 is the proportion of school days she will be late to class.

She will be on time if her travel time is between 40 and 60.
She will be late if her travel time is between 60 and 70.

60-40=20
70-60=10

She will be late 10/(10+20) = 1/3 of the time.

Answer 2

She'll be late on the days when it takes more than 60 minutes.

The total range is 30. The bad range is 60 to 70 = 10.

The proportion is 10/30 or 1/3.

Answer 3

Let X have the uniform distribution between α and β.

the probability density function is:

f(x) = 0 x < α
f(x) = 1/(β - α) α ≤ x ≤ β
f(x) = 0 x > β

the mean is: (β - α) / 2
the variance is: (β - α)² / 12

the student will be late if her walk takes between 60 and 70 minutes

P( 60 < X < 70) =

70
∫ f(x) dx
60

= 1/3

this student will be late 1/3 of the time.

Answer 4

70-40=30 minute range.
between 40 and 60 minutes she won't be late; 20 min / 30 = 2/3 = 66.7%
between 60 and 70 minutes she will. 10 /30 =1/3=33.3%.

Answer 5

= (60 - 40)/60
= 20/60
= 1/3

Proportion is 1/3 or 1:3 chances that he/she would be late.