Tough Discrete Probability Problem

James.wilson

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I'm struggling with this question, any help would be great. A news article says that the average person thinks about food 15 times a day or 0.625 times an hour, let Y equal the number of times a randomly selected person thinks about food in one day, find the probability the number of thoughts (discrete) are between 15 and 28 inclusive? P(15 ≤ Y ≤ 28)=? I've also found that our μ=1.6 by dividing 24 hours(in a day) by thinking about food 15 times per day.
 

tkhunny

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1) Why did you bother to scale to hourly if the original data are in days and the problem statement is in days?
2) Do you think the distribution to be Normal around the mean of 15? Did the problem statement say so?
3) What do you know of the Poisson Distribution?
4) What's μ? One should define a distribution and explain any parameters before putting them out there.
5) 0.625 = 1/1.6, so really, you have only one thing, not two things.
 

Jomo

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Lets see if what you say makes sense. You do something 15 times a day which is 1.6 times an hour. Does that sound right?

Everybody make arithmetic mistakes. The thing is that you have to think if your calculations makes sense! You just can't push buttons on a calculator and just accept the results. If you do something once per hour than you do this thing 24 times a day. If you do something more than once an hour (like 1.6 times an hour) than you do this thing more than 24 times a day. If you do something less than once per hour then you do this thing less than 24 times a day. You were told that you do this thing 15 times a day and that is less than once per hour NOT 1.6 times per hour
 

James.wilson

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1) Why did you bother to scale to hourly if the original data are in days and the problem statement is in days?
2) Do you think the distribution to be Normal around the mean of 15? Did the problem statement say so?
3) What do you know of the Poisson Distribution?
4) What's μ? One should define a distribution and explain any parameters before putting them out there.
5) 0.625 = 1/1.6, so really, you have only one thing, not two things.
For part a of my question it says "Let X be the amount of time, in hours that passes until a randomly selected person thinks about food. How many hours can you expect to pass between successive thoughts about food?" Which is where I got my μ which I can see is wrong now, and the problem does not say anything about the distribution being normal around the mean 15, I have learned about the poisson distribution and if I'm going about it correctly our μ is equal to .625
 

Jomo

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The problem says let Y equal the number of times a randomly selected person thinks about food in one day and asks you to compute
P(15 ≤ Y ≤ 28)=?

You go onto say that For part a of my question it says "Let X be the amount of time, in hours that passes until a randomly selected person thinks about food. How many hours can you expect to pass between successive thoughts about food? Can you please show me where you wrote this in your post? If you can't, then can you please post the entire problem as it was given to you?
 

James.wilson

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The problem says let Y equal the number of times a randomly selected person thinks about food in one day and asks you to compute
P(15 ≤ Y ≤ 28)=?

You go onto say that For part a of my question it says "Let X be the amount of time, in hours that passes until a randomly selected person thinks about food. How many hours can you expect to pass between successive thoughts about food? Can you please show me where you wrote this in your post? If you can't, then can you please post the entire problem as it was given to you?
IMG_2524.jpg
 

tkhunny

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Problem #1 -- The fact that it says "the average person" tells me the article is written poorly and not well thought out.
Problem #2 -- Are we to assume that the Daily Rate is perfectly scalable to an Hourly Rate? If we're talking about humans, that is preposterous. One may wish to include at least 6 hours of sleep.
Problem #3 -- Assuming that the "15 times a day" information is from a decent survey, the very idea that we should apply a population average to a single individual is a rookie error. We should not EVER think that way. You are selecting a sample of one (1)! What's the probability that the selected person acts anything like the population average? You want to talk about the average response of the population, NOT a single individual.

If you want to, you can write part C as P(X>3.1|X>1.5).

Problem #4 -- You could run the text past just one English Major. That would improve the parallelism problem quite a bit.

Having said that, lets see if you can follow the continuous part of the problem.

Define thinking of food as a Failure, we can talk about the Mean Time Between Failures. I'll ignore the sleeping, but you should consider it.

The failure rate, then is 15/24 - 15 failures in 24 hours. Because we're looking at a 24 hour period, let's use \(\displaystyle \omega = 24\), the maximum value we can get out of a day.

Check to see that we have a proper probability distribution: \(\displaystyle \int_{0}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)} = 1\). It's not REALLY unity, but it's pretty close.

The mean time, then, is given by the mean of this distribution. \(\displaystyle E[t] = \int_{0}^{24}\dfrac{15}{24}\cdot t\cdot e^{-t(15/24)} = 1.6, \;this\; is\;your\;\mu\).

P(X<1.7) is pretty simple, now: \(\displaystyle \int_{0}^{1.7}\dfrac{15}{24}\cdot e^{-t(15/24)} = 0.654409\), thereabouts.

The next one is a little more fun: P(X>3.1|X>1.5) = \(\displaystyle \dfrac{\int_{3.1}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)}}{\int_{1.5}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)}} = \dfrac{0.144063}{0.391605} = 0.3679\)

That's enough. I'll leave the last one for you and let you add up a bunch of Poisson probabilities. I gave you as much as I did because I thought maybe I owed it to you after telling how how bad is the whole conception of the problem statement.

Note: There is only one variable, t, thus for the sake of visual clarity, I omitted all the little "dt" symbols. As there is no ambiguity, this is not a problem. If you had a calculus professor who insisted that there is or has been only one authoritative notation for such things, you have not been told the truth. On the other hand, if your teacher insists or you'll get a test problem wrong, you should probably put them in. And, yes, it would have been easier to put them in than it was to write this disclaimer. :)
 

tkhunny

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BTW -- Welcome to this lovely forum. Show your work and start with the WHOLE problem statement in the very first post. This ALWAYS makes the conversation go more smoothly. Thanks for your patients.
 

James.wilson

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Problem #1 -- The fact that it says "the average person" tells me the article is written poorly and not well thought out.
Problem #2 -- Are we to assume that the Daily Rate is perfectly scalable to an Hourly Rate? If we're talking about humans, that is preposterous. One may wish to include at least 6 hours of sleep.
Problem #3 -- Assuming that the "15 times a day" information is from a decent survey, the very idea that we should apply a population average to a single individual is a rookie error. We should not EVER think that way. You are selecting a sample of one (1)! What's the probability that the selected person acts anything like the population average? You want to talk about the average response of the population, NOT a single individual.

If you want to, you can write part C as P(X>3.1|X>1.5).

Problem #4 -- You could run the text past just one English Major. That would improve the parallelism problem quite a bit.

Having said that, lets see if you can follow the continuous part of the problem.

Define thinking of food as a Failure, we can talk about the Mean Time Between Failures. I'll ignore the sleeping, but you should consider it.

The failure rate, then is 15/24 - 15 failures in 24 hours. Because we're looking at a 24 hour period, let's use \(\displaystyle \omega = 24\), the maximum value we can get out of a day.

Check to see that we have a proper probability distribution: \(\displaystyle \int_{0}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)} = 1\). It's not REALLY unity, but it's pretty close.

The mean time, then, is given by the mean of this distribution. \(\displaystyle E[t] = \int_{0}^{24}\dfrac{15}{24}\cdot t\cdot e^{-t(15/24)} = 1.6, \;this\; is\;your\;\mu\).

P(X<1.7) is pretty simple, now: \(\displaystyle \int_{0}^{1.7}\dfrac{15}{24}\cdot e^{-t(15/24)} = 0.654409\), thereabouts.

The next one is a little more fun: P(X>3.1|X>1.5) = \(\displaystyle \dfrac{\int_{3.1}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)}}{\int_{1.5}^{24}\dfrac{15}{24}\cdot e^{-t(15/24)}} = \dfrac{0.144063}{0.391605} = 0.3679\)

That's enough. I'll leave the last one for you and let you add up a bunch of Poisson probabilities. I gave you as much as I did because I thought maybe I owed it to you after telling how how bad is the whole conception of the problem statement.

Note: There is only one variable, t, thus for the sake of visual clarity, I omitted all the little "dt" symbols. As there is no ambiguity, this is not a problem. If you had a calculus professor who insisted that there is or has been only one authoritative notation for such things, you have not been told the truth. On the other hand, if your teacher insists or you'll get a test problem wrong, you should probably put them in. And, yes, it would have been easier to put them in than it was to write this disclaimer. :)
Thank you for all your effort, I took the time to calculate the last one out and got 0.5334856
Also, earlier in the forum Jomo said that the mu cannot be 1.6 because that wouldnt make any sense
 
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tkhunny

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Jomo

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Otis, I am by far no expert with statistics. In fact I refused to teach it my whole teaching career. Now I see how you calculated E(t) = 1.6 and your work looks good to me. I just can't believe that E(t) = 15 is wrong? How is that? They said on average 15 times per day. What am I missing!?
 

tkhunny

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E[t] is the mean time between failures, not the number of failures during the day.

24 / 1.6 = 15
 

Otis

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