# Probability Problem

• Dec 9th 2006, 06:21 PM
ayato
Probability Problem
Ok, here's the problem i've been having some trouble with:

Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

If the longest each will wait is 10 minutes, find the probability that they'll meet?

I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.
• Dec 9th 2006, 08:59 PM
CaptainBlack
Quote:

Originally Posted by ayato
Ok, here's the problem i've been having some trouble with:

Ann and Bob have agreed to meet for dinner "around four". The number of minutes after 4:00 that Ann will arrive is approximately normally distributed with mean 10 and standard deviation 3.1, and the number of minutes after 4:00 that Bob will arrive is approximately normally distributed with mean 20 and standard deviation 4.2.

If the longest each will wait is 10 minutes, find the probability that they'll meet?

I know the answer will be around the 1 range but i don't really know how to show the work for it, at least for the work that i showed for it, i'm not confident about it. I'll be very appreciated if someone can help me out. Thank You in advance.

Consider the difference D in arrival times, this will be normally distributed with
mean 10 minutes, and sd sqrt(3.1^2+4.2^2). Then you want: p(-10<D<10).

RonL
• Dec 10th 2006, 10:46 AM
ayato
Quote:

Originally Posted by CaptainBlack
Consider the difference D in arrival times, this will be normally distributed with
mean 10 minutes, and sd sqrt(3.1^2+4.2^2). Then you want: p(-10<D<10).

RonL

Thanks alot, so let me get this straight:

sqrt(3.1^2+4.2^2)
p(-10<D<10)

z> (-10-10)/sqrt(3.1^2+4.2^2)
= -3.83
P= practically 1

Z<(10-10)/sqrt(3.1^2+4.2^2)
=0
P= .5000

1-.5000
=.5000
Is this right?

• Dec 10th 2006, 10:50 AM
CaptainBlack
Quote:

Originally Posted by ayato
Thanks alot, so let me get this straight:

sqrt(3.1^2+4.2^2)
p(-10<D<10)

z> (-10-10)/sqrt(3.1^2+4.2^2)
= -3.83
P= practically 1

Z<(10-10)/sqrt(3.1^2+4.2^2)
=0
P= .5000

1-.5000
=.5000
Is this right?