Probability question that stumped me completely - reward

[mindless]

Asked of me in an interview today and I couldn't figure it out at all. Given a series that is normally distributed with mean 0 and standard deviation of 1, what is the probability that the next 100 elements will have an average greater than 1.5?

Now, the probability that the next element is greater than 1.5 is simple to find using CDF and a table to look up erf. And finding the probability that the next 100 are all greater than 1.5 is simply p(next > 1.5)^100. However, I'm stumped about looking at the average.

If I'm not missing something obvious, I'd like to email the interviewer and let him know that the question is really tricky, but I'd also like to show him the answer. If someone can help me get the answer, I'll transfer them $25 on pokerstars.

[thylacine]

[ QUOTE ]
Asked of me in an interview today and I couldn't figure it out at all. Given a series that is normally distributed with mean 0 and standard deviation of 1, what is the probability that the next 100 elements will have an average greater than 1.5?

Now, the probability that the next element is greater than 1.5 is simple to find using CDF and a table to look up erf. And finding the probability that the next 100 are all greater than 1.5 is simply p(next > 1.5)^100. However, I'm stumped about looking at the average.

If I'm not missing something obvious, I'd like to email the interviewer and let him know that the question is really tricky, but I'd also like to show him the answer. If someone can help me get the answer, I'll transfer them $25 on pokerstars.

[/ QUOTE ]

"I'd like to email the interviewer and let him know that the question is really tricky"

Don't do that.

"If I'm not missing something obvious ..."

You are.

The sum of 100 elements has mean 0, variance 100, SD 10.

The average of 100 elements has mean 0, SD 0.1.

So look up 15 on your table.

I'm not a statistician and I am only 90% sure that what I just said is right. [img]/images/graemlins/blush.gif[/img]

But I am 100% sure it is easy. [img]/images/graemlins/smile.gif[/img]

[felson]

thylacine is right.

[Utah]

{expected value X N} + {stddev x sqrt(N) x inverse of the standard normal cumulative distribution (Y) = 1.5

N = 100
expected value = 0
stddev = 1

Solve for Y, which I believe is .56

I think this is correct if I understand your problem correctly. I need to run out for a few hours so I dont have time to double check. I will check when I get back unless someone else verifies or discredits before I can double check.

[BruceZ]

[ QUOTE ]
Asked of me in an interview today and I couldn't figure it out at all. Given a series that is normally distributed with mean 0 and standard deviation of 1, what is the probability that the next 100 elements will have an average greater than 1.5?

[/ QUOTE ]

thylacine has it right. The average would have to be at least 15 standard deviations above the mean to be 1.5 or greater. This has a probability of zero for all practical purposes. I cannot even evaluate a probability this small in Excel.

This is a standard statistical problem. I would not make matters worse by telling your interviewer that it was "tricky". Even if you didn't know the distribution of the sample mean offhand, you should have been able to reason that an average of 1.5 would require that the sum of 100 elements be 150, and the standard deviation of the total is sqrt(100) = 10, so again we must be 15 standard deviations above the mean. This assumes you know that the distribution of the sample mean is normal if the original distribution is normal.

Why do you use the word "series"? I assume by this that you just mean a random variable, and the "elements" are just values of the random variable, or samples.

This also should have been posted in the probability forum.

[Lestat]

Bruce, I'm just curious what line of work you're in. I haven't seen a single question posted on here that you haven't thoroughly understood. I can't even imagine what it's like to be that good at math. You've got to be one of the best.

[bigpooch]

The probability is about 3.670966 x 10^(-51) using the
asymptotic series for the "error function".

Of course, the interviewer may have just wanted to know
that it was very small. It would be obvious to any
statistician.

[Utah]

[ QUOTE ]
{expected value X N} + {stddev x sqrt(N) x inverse of the standard normal cumulative distribution (Y) = 1.5

N = 100
expected value = 0
stddev = 1

Solve for Y, which I believe is .56

I think this is correct if I understand your problem correctly. I need to run out for a few hours so I dont have time to double check. I will check when I get back unless someone else verifies or discredits before I can double check.

[/ QUOTE ]

Ah. I misread the problem as I thought it was a cumulative total of 1.5. However, I believe the formula above would still work if you set it to 150 instead of 1.5. I would appreciate it if someone more versed in statistics than me could verify.