Probability Distribution Problem

In summary: XZlLCB0aGUgbm90aWZpY2F0aW9uIG9mIFQxID0gdGltZSB0byBmaXJzdCBsb2ctb24oYWZ0ZXJpb3JpbmcgdGltZXMpLCBUMiA9IHRpbWUgYmV0dGVybmVzcyBmaXJzdCBhbmQgUj0gYWZ0ZXJpb3JpbmcgVGk+VGQsIFQzID0gdGltZSBiZXR0ZXIgZm9yIFQ0LCBUNSA9I
  • #1
ashwinnarayan
18
0

Homework Statement



Log-ons to a certain computer website occur randomly at a uniform average rate of 2.4 per minute. State the distribution of the number N of log-ons that occur during a period of t minutes.

Obtain the probablity that at least one log on occurs during a period of t minutes.

Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons.

Identify the distribution of T and state its mean and variance.

Homework Equations


The Attempt at a Solution



This one's a bit embarrassing really. What I don't get is the "Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons." part. It's only 3 marks for this section of the question so it's probably something simple. I must be missing some key fact or something.

Here's what I did for the first bit.

The distribution of N is Poisson with a mean of 2.4t so X~Po(2.4t)

The probability of at least one log on is P(X≥1) = 1 - P(X=0) = 1 - e-2.4t

And now I am stuck. This is from a really old A-Level Further Mathematics exam so I only had an examiner's report which gave the answers with a short and confusing explanation. It said that "The probability density function is obtained by taking the derivative of their answer 1 - e-2.4t. Which gives a negative exponential distribution 2.4e-2.4t. The mean and variance is easily 1/2.4 and (1/2.4)2 respectively."

What I don't understand is why the probablity density function is the derivative of that expression.
 
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  • #2
ashwinnarayan said:

Homework Statement



Log-ons to a certain computer website occur randomly at a uniform average rate of 2.4 per minute. State the distribution of the number N of log-ons that occur during a period of t minutes.

Obtain the probablity that at least one log on occurs during a period of t minutes.

Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons.

Identify the distribution of T and state its mean and variance.

Homework Equations





The Attempt at a Solution



This one's a bit embarrassing really. What I don't get is the "Hence obtain the probability density function of T, where T minutes is the interval between successive log-ons." part. It's only 3 marks for this section of the question so it's probably something simple. I must be missing some key fact or something.

Here's what I did for the first bit.

The distribution of N is Poisson with a mean of 2.4t so X~Po(2.4t)

The probability of at least one log on is P(X≥1) = 1 - P(X=0) = 1 - e-2.4t

And now I am stuck. This is from a really old A-Level Further Mathematics exam so I only had an examiner's report which gave the answers with a short and confusing explanation. It said that "The probability density function is obtained by taking the derivative of their answer 1 - e-2.4t. Which gives a negative exponential distribution 2.4e-2.4t. The mean and variance is easily 1/2.4 and (1/2.4)2 respectively."

What I don't understand is why the probablity density function is the derivative of that expression.


Let T1 = time to first log-on (after you start measuring times), T2 = time between first and second log-on, etc. The T1, T2, ... are the inter-arrival times of a Poisson process. Assuming the counting probabilities are independent between non-overlapping intervals, and that N = no. arrivals in (t1,t2) has the Poisson distribution with mean a*(t2-t1) [with a = your 2.4], we can show--by a lengthy but not particularly difficult argument--that T1, T2, ... are independent and identically distributed. Furthermore, the number arriving by time t is N(t), where Pr{N(t) = n} = Pr{T1+T2+...+Tn < t, T1+T2+...+Tn+T(n+1) > t} = (a*t)^n *exp(-a*t)/n!, where a = 2.4. In particular, Pr{T1 > t} = Pr{N(t)=0} = exp(-a*t), so the density f(t) of T1 satisfies f(t) = (d/dt) Pr{T1 > t} = a*exp(-a*t), because Pr{T1 > t} = integral_{x=t..infinity} f(x) dx. Also, T2, T3, T4,... all have the same density.

RGV
 

What is a probability distribution?

A probability distribution is a mathematical function that describes the likelihood of a random variable taking on different values. It shows the possible outcomes of an event and their corresponding probabilities.

What is the difference between discrete and continuous probability distributions?

A discrete probability distribution deals with outcomes that can only take on specific values, while a continuous probability distribution deals with outcomes that can take on any value within a specified range.

How do you calculate the mean of a probability distribution?

The mean of a probability distribution is calculated by multiplying each outcome by its corresponding probability and then adding all of these products together. This is also known as the expected value or the average value.

What is the purpose of a probability distribution in statistics?

A probability distribution is used in statistics to model real-world scenarios and make predictions about the likelihood of certain outcomes. It helps to understand the probability of an event occurring and its potential outcomes.

Can a probability distribution be skewed?

Yes, a probability distribution can be skewed, meaning that the data is not evenly distributed. This can happen in both discrete and continuous distributions and can affect the shape and interpretation of the data.

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