Statistics and Probability

[ak28it]

Registered events X in time interval t are distributed linearly n = n0 + bt. Find probability density function, then t = 10, n0 = 5 and b =2. Find average amount of registered events per day and Mean squared error. Find the probability to register an event per 5th and 6th days. What is the average registered amount of events per 5th and 6th days.

I got this problem with no idea how to begin, please help.

[jemidiah]

You're written the question very strangely. It's hard to reconstruct.

I imagine the number of events per day t is given by n = n0 + bt, where n0=5 and b=2. I have no idea why you wrote t=10--actually maybe that's supposed to be the last day data was collected? I don't know what PDF you want. The average number of events per day from this data is at least easy to calculate:

[(n0 + b*0) + (n0 + b*1) + ... + (n0 + b*10)]/11
= [11*n0 + b*(0+1+...+10)]/11
= [11*n0 + 10*11/2*b]/11
= n0 + 5b
= 5 + 10
= 15
= M

The mean squared error here is just the average of the squared difference between the mean M and each data point. The data points are n0 + b*0, ..., b0 + b*10 = 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, which have differences (magnitude only) of 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, which have squares 100, 64, 36, 16, 4, 0, 4, 16, 36, 64, 100, which have average 440/11 = 40.



By the way, I almost arbitrarily started this data at t=0. The given function would still give a positive number of events at t=-1.

[ak28it]

You're written the question very strangely. It's hard to reconstruct.

I imagine the number of events per day t is given by n = n0 + bt, where n0=5 and b=2. I have no idea why you wrote t=10--actually maybe that's supposed to be the last day data was collected? I don't know what PDF you want. The average number of events per day from this data is at least easy to calculate:

[(n0 + b*0) + (n0 + b*1) + ... + (n0 + b*10)]/11
= [11*n0 + b*(0+1+...+10)]/11
= [11*n0 + 10*11/2*b]/11
= n0 + 5b
= 5 + 10
= 15
= M

The mean squared error here is just the average of the squared difference between the mean M and each data point. The data points are n0 + b*0, ..., b0 + b*10 = 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, which have differences (magnitude only) of 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, which have squares 100, 64, 36, 16, 4, 0, 4, 16, 36, 64, 100, which have average 440/11 = 40.



By the way, I almost arbitrarily started this data at t=0. The given function would still give a positive number of events at t=-1.

Thanks Sir,
I was not able to initiate it .Even i have tried it solve but was unsuccessful.As u have explained it well So thanks for it .I hope you will try to solve my further questions.

[ak28it]

Hi jemidiah ,
Can you clear my doubt?
What do you mean by "linearly distributed"? I've never heard that term before.

Usually in linear regression statements, things will have a t-distribution or a normal distribution.

[jemidiah]

I meant that the data points are distributed evenly along a line. I later explicitly gave the data points as
5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

I hope I interpreted your question correctly. As I said, it was hard to understand.

[ak28it]

I meant that the data points are distributed evenly along a line. I later explicitly gave the data points as
5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25

I hope I interpreted your question correctly. As I said, it was hard to understand.

Thanks to clear all of my doubts step by step .Hope you will be there everytime to clear my doubts .
Thanks again .