Probability Question - exponential distribution and normal distribution?

TheSetMiner

I'm stuck on two probability questions.
The first involves an exponential distribution: the expected value of an exponentially occurring event is 15 seconds, what is the probability that the value will exceed 60 seconds, given that the value has exceeded 25 seconds?
This one has thrown me as I am struggling to manipulate the usual e ^ lambda formula to suit this problem, can anyone at least push me in the right direction as I've been staring at it a while now ...

Also I think I have the answer to the following but would like confirmation as it seems suspiciously simple. mean of event = 167, standard dev = 10
sample size 7 selected, what is probability that their average height is less than 167. I have plugged the values into the (X − µ)/(σ/√n) but this gives me a z value of zero, could that be? so then the probability is simple .5?

Thanks to anyone who helps!

ray giraffe

You are correct in your answer for the last question, it is just 0.5

TheSetMiner

Thanks so that just leaves this dirty exponential problem. I'm gonna do some more research today and try to find some answers.

One last question I am having trouble with is this:
Suppose that male weight for 10 year olds is normally distributed
with mean 25kg and standard deviation 3kg, while the weight of 10 year old females
is normally distributed with mean 23.5kg and standard deviation
2.5kg. A male and female of this age are chosen at
random to play together on a seesaw. What is the probability that the
female is heavier than the male?

This one seems quite advanced. I am trying to approach it using normal distribution tables and z values but I can't see what I would do other than comparing their variances and means. In my head I'm thinking I need to somehow superimpose both their graphed distributions and find the point of intersection but I could be way off?

Yakuza

TheSetMiner wrote: »

Thanks
One last question I am having trouble with is this:
Suppose that male weight for 10 year olds is normally distributed
with mean 25kg and standard deviation 3kg, while the weight of 10 year old females
is normally distributed with mean 23.5kg and standard deviation
2.5kg. A male and female of this age are chosen at
random to play together on a seesaw. What is the probability that the
female is heavier than the male?

This one seems quite advanced. I am trying to approach it using normal distribution tables and z values but I can't see what I would do other than comparing their variances and means. In my head I'm thinking I need to somehow superimpose both their graphed distributions and find the point of intersection but I could be way off?

http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

The two variables are independent, so by subtracting the one distribution from the other, the you end up with a new normally-distributed variable whose mean is the difference of the two means, and whose variance is the sum of the two variances (note : variances, not standard deviations).

With the above in mind, you're looking for the probability of the above variable being less than zero (assuming you calculated it by taking the female weight from the male).

Yakuza

I just noticed the second part of your first question. The exponential distribution is memoryless, so the probability of there being no event before 60 seconds given there has been none up to 25 seconds is the same as the probability of no event occurring after 35 seconds, if measuring the time from now. Does that make it a bit easier?

TheSetMiner

Ok I am going to give the last question a go now with the assistance of that wiki page and see if I have any luck

Yakuza wrote: »

I just noticed the second part of your first question. The exponential distribution is memoryless, so the probability of there being no event before 60 seconds given there has been none up to 25 seconds is the same as the probability of no event occurring after 35 seconds, if measuring the time from now. Does that make it a bit easier?

I think I solved this question today using a cumulative density function, ie the integral of a probability density function. I got the odds of an event occuring after 25 seconds, then the odds after 60s and divided the latter by the former, think that makes sense?

Yakuza

If you replace 50 by 25 and 120 by 60 you'll be on the right track.
Or if you just work it out for 35 seconds, you'll get the same answer

TheSetMiner

Yakuza wrote: »

If you replace 50 by 25 and 120 by 60 you'll be on the right track.
Or if you just work it out for 35 seconds, you'll get the same answer

thanks man!