Thread: Help with one short problem.

1.The driving time for an individual from his home to his work is distributed between 300 to 480 seconds.
So here are the two things I need to solve for this problem:

Compute the probability that the driving time will be less than or equal to 435 seconds.
Determine the expected driving time and its standard deviation.



How can I do this?

Originally Posted by Math123

1.The driving time for an individual from his home to his work is distributed between 300 to 480 seconds.
So here are the two things I need to solve for this problem:

Compute the probability that the driving time will be less than or equal to 435 seconds.
Determine the expected driving time and its standard deviation.



How can I do this?

First as you have typed the question there is no answer as you have not specified a distribution. But lets assume you mean uniformly distributed between 300 and 480 seconds.

Probability that travelling time is less than 435 is:

integral (p(x) dx, x=-inf,435) = integral (1/180) dx, x=300,435)

as p(x)=(1/180) for x in (300, 480), and 0 otherwise.

integral (p(x) dx, x=-inf, 435) = integral (1/180) dx, x=300, 435)
............................. ..=(1/180) (435-300)=135/180 = 0.75



By definition the mean:

m = integral (x p(x) dx, x=-inf, inf) = integral (x/180 dx, x=300, 480)
.............................. ........=480^2/360-300^2/360 = 390.



The definition of SD:

s = sqrt(V),

V = integral ((x-m)^2 p(x) dx, x=-inf, inf)

which I will leave for you to do.

RonL

I'm not getting it.

Originally Posted by Math123

I'm not getting it.

The uniform distribution between 300 and 480 has density:

p(x)=(1/180); x in (300, 480)
p(x)=0 otherwise.

Then you just use the text book definitions of the quatities you require
and compute the integrals.

p(x<a) = integral p(x) dx, x=-inf, a

mu= integral xp(x) dx, x=-inf, inf

var= integral (x-mu)^2 p(x) dx, x=-inf, inf

sigma=sqrt(var)

RonL