# Uniform Distribution Probability

## Homework Statement

The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent.

Part a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between 536 and 637?

Part b) What is the approximate probability (to 2 decimal places) that the average of the 95 wait times exceeds 6 minutes?

Part c) Find the probability (to 2 decimal places) that 92 or more of the 95 wait times exceed 1 minute. Please carry answers to at least 6 decimal places in intermediate steps.

Part d) Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 95 wait times recorded exceed 5 minutes.

## The Attempt at a Solution

I know that for part c and d I need to use binomial distribution, however I am unsure of what to do for part a and b

What I tried for a was ((637/95)-(536/95))/12 and for b I thought I could just do 7/12 since it is uniformly distributed, but neither are correct. Can anyone tell me if there is a certain formula to follow? Or what I am doing wrong? Thanks[/B]

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

The wait time (after a scheduled arrival time) in minutes for a train to arrive is Uniformly distributed over the interval [0,12]. You observe the wait time for the next 95 trains to arrive. Assume wait times are independent.

Part a) What is the approximate probability (to 2 decimal places) that the sum of the 95 wait times you observed is between 536 and 637?

Part b) What is the approximate probability (to 2 decimal places) that the average of the 95 wait times exceeds 6 minutes?

Part c) Find the probability (to 2 decimal places) that 92 or more of the 95 wait times exceed 1 minute. Please carry answers to at least 6 decimal places in intermediate steps.

Part d) Use the Normal approximation to the Binomial distribution (with continuity correction) to find the probability (to 2 decimal places) that 56 or more of the 95 wait times recorded exceed 5 minutes.

## The Attempt at a Solution

I know that for part c and d I need to use binomial distribution, however I am unsure of what to do for part a and b

What I tried for a was ((637/95)-(536/95))/12 and for b I thought I could just do 7/12 since it is uniformly distributed, but neither are correct. Can anyone tell me if there is a certain formula to follow? Or what I am doing wrong? Thanks[/B]

In part (a) you are dealing with the random sum ##S = \sum_{i=1}^{95} W_i##, where the individual wait times ##W_1, W_2, \dots, W_{95}## are an independent sample from the distribution ##\text{Unif}(0,12)##. You are being asked to find ##P(536 \leq S \leq 637)##, or maybe ##P(536 < S < 637)##---it is not clear which.

In principle you can locate formulas for the probability density of a sum of 95 independent uniform random variables, but the formula is nasty and difficult to use, unless you compute intermediate results to hundreds of decimal places of accuracy. So, the only sensible approach is to use a good approximation to the density of ##S##, and I would bet that your textbook and/or course notes have useful suggestions about that.

In part (b) you are dealing with ##S/95##, and want ##P(S/95 > 6)##. Again, you will need to use an approximation to the exact distribution.

RUber
Homework Helper
With a reasonably large sample, your uniform distribution will start normalizing, so the average should tend to 6 minutes. Thus, you would expect a fairly high probability that the total time would be close to 6x95=570 mins.
As Ray said, check for a formula in your text that might help you approximate the cumulative probability in that range.

6 minutes is your midpoint. The probability is equal on both sides -- regardless of how many observations.

With the binomial distribution, you have p and 1-p raised to powers based on how many observations you see of each.
92 or more exceed 1 min means that p=prob(x>1) , and 1-p = prob(x<1).

Remember when you take the normal approximation, you will need to adjust for the fact that you have while integer data for frequency. Usually that means you include .5 below the cutoff for > problems and .5 above for < problems.