Help with statistics! (Probability)
- Thread starter Mattydv
- Start date
Hey mate, for this you're gonna need access to a T Distribution table. I'm no expert at Statistics, but I had a workshop for pretty much the exact same stuff yesterday arvo - so I'll try and give you a hand.
Remember Degrees of Freedom - 1. (Not sure why, but that's just what happens).
So for the first one you want to find the Z score of the probability.
So the formula is: z = (x/n) -μ / σ
(Where x is the number of days and n is the amount of riders)
this is where I get unclear because yesterday we had to fiddle with the SD, because the one given was the SD of the sample - but I don't think you need to do it because your sample is only 1?
Yours gives me: z = ((380/1) - 400)/ 30 = -0.6667
with a corresponding probability of: 0.476
[One thing is, that might be the probability of 380 days or more - since I'm following the formula for finding the probability of the weight of a bag being at least x grams or over...]
Not quite sure if that's right or not haha but that's what I came up with. Hopefully someone on here is better at stats than I am (I failed it last semester
but I'm doing better so far this semester)
Remember Degrees of Freedom - 1. (Not sure why, but that's just what happens).
So for the first one you want to find the Z score of the probability.
So the formula is: z = (x/n) -μ / σ
(Where x is the number of days and n is the amount of riders)
this is where I get unclear because yesterday we had to fiddle with the SD, because the one given was the SD of the sample - but I don't think you need to do it because your sample is only 1?
Yours gives me: z = ((380/1) - 400)/ 30 = -0.6667
with a corresponding probability of: 0.476
[One thing is, that might be the probability of 380 days or more - since I'm following the formula for finding the probability of the weight of a bag being at least x grams or over...]
Not quite sure if that's right or not haha but that's what I came up with. Hopefully someone on here is better at stats than I am (I failed it last semester
but I'm doing better so far this semester)
PINT of Stella. mate!
Many, many Scotches
If we had a pool room it would probably need a good clean. This thread would be in there though.
pharmaboy
Eats Squid
Rotorburners, I need your help! Ordinarily I'd go see a tutor about this sort of thing, but drop-in sessions are booked up for almost a fortnight, and I really don't want to fall that far behind - so I'm appealing to RB's academic collective to give me some support!
I'm not necessarily after answer's, but some sort of explanation of how I'd approach and work through the multiple associated questions would be helpful. And because you'll be doing the job that someone else would ordinarily be paid to do, I'll throw in a reward
The time the average punter spends mountain biking over 3 years varies according to a distribution that is approximately normal with a mean of 400 days, and a standard deviation of 30 days.
What is the probability that a randomly selected punter will ride less than 380 days?
What percentage of riders get out on the trails between 380 and 420 days?
Between what values do the amount of days of the middle 85% fall?
In the above question, a range of 85% has been determined. What is the probability that 15 randomly chosen riders will all get out on the trails in in that 85% range?
geez matty, havent done any stats for 25 years, but the first thing that strikes me is that 380/400 happens to be 95% - which is the std point for statistical relevance and also happens to be near as dammit 2 standard deviations (well, it doesnt happen to be, it is) - look up SD on wikipedia, work through it, and at the end you'll understand it, then you can figure out the answers for yourself and have learnt summfin.
Old C
Likes Bikes
You should use a normal distribution table. Everything is in terms of the number of standard deviations away from the mean:
The time the average punter spends mountain biking over 3 years varies according to a distribution that is approximately normal with a mean of 400 days, and a standard deviation of 30 days.
What is the probability that a randomly selected punter will ride less than 380 days? 400 - 380 = 20. 20/30 = 0.66 Standard Deviations away from the mean, so the chance is taken from the table at that SD. Z = 0.66, alpha = 0.7454. So the chance is 1 - 0.7454 = 25.14%
What percentage of riders get out on the trails between 380 and 420 days? +/- 0.66 standard deviations. Keep in mind that +/- 1 standard deviation is roughly 66% of the population; +/- 0.66 is less, so from the table = 2 * 0.2514 = 50.28%
Between what values do the amount of days of the middle 85% fall? +/- 85/2 = +/- 42.5% From the table 0.5 - 0.425 = 0.075 = 1.44 Standard Deviations = 1.44 * 30 = 43.2 days +/- from 400 = 356.8 to 443.2
In the above question, a range of 85% has been determined. What is the probability that 15 randomly chosen riders will all get out on the trails in in that 85% range? . 0.85^15 = 8.735%
I hope that helps, I am a little rusty with the Z table.
The time the average punter spends mountain biking over 3 years varies according to a distribution that is approximately normal with a mean of 400 days, and a standard deviation of 30 days.
What is the probability that a randomly selected punter will ride less than 380 days? 400 - 380 = 20. 20/30 = 0.66 Standard Deviations away from the mean, so the chance is taken from the table at that SD. Z = 0.66, alpha = 0.7454. So the chance is 1 - 0.7454 = 25.14%
What percentage of riders get out on the trails between 380 and 420 days? +/- 0.66 standard deviations. Keep in mind that +/- 1 standard deviation is roughly 66% of the population; +/- 0.66 is less, so from the table = 2 * 0.2514 = 50.28%
Between what values do the amount of days of the middle 85% fall? +/- 85/2 = +/- 42.5% From the table 0.5 - 0.425 = 0.075 = 1.44 Standard Deviations = 1.44 * 30 = 43.2 days +/- from 400 = 356.8 to 443.2
In the above question, a range of 85% has been determined. What is the probability that 15 randomly chosen riders will all get out on the trails in in that 85% range? . 0.85^15 = 8.735%
I hope that helps, I am a little rusty with the Z table.
Mattydv
Likes Bikes and Dirt
Pharmaboy, the 380/400 were randomly chosen numbers by me. It could have been 374 and 403 for all it mattered, I was just hoping for someone to work through it so I could see the process on a question of my choosing.
I've been through all my notes (and Wikipedia, and Khan Academy, and other YouTube lessons), but I think it's different to see poorly worked examples and questions I'm likely to get minor variations of. Rotorburn has a myriad of people much smarter than myself who are usually all too willing to help answer any (stupid) questions. Why not make the most of it?
Thankyou for your response Old C, my only query is why does 0.075 = 1.44 SD's? Why is the population SD not 0.66, thus making 0.075, 0.11 of a SD?
I've been through all my notes (and Wikipedia, and Khan Academy, and other YouTube lessons), but I think it's different to see poorly worked examples and questions I'm likely to get minor variations of. Rotorburn has a myriad of people much smarter than myself who are usually all too willing to help answer any (stupid) questions. Why not make the most of it?
Thankyou for your response Old C, my only query is why does 0.075 = 1.44 SD's? Why is the population SD not 0.66, thus making 0.075, 0.11 of a SD?
Matt H
Eats Squid
Wow, booked up for almost a fortnight? Most of my subjects have near-daily consultation sessions with tutors/lecturers. I guess I really take that for granted eh...
Mattydv
Likes Bikes and Dirt
It's not as bad for other programs, but the classes for the degree I'm doing don't really coincide with drop in sessions, hence everybody in the class wants time.