broodjepanda Posted January 10, 2017 Posted January 10, 2017 I have two questions for my homework, it would be very nice if someone can help me with this.Question 1: Exam candidates for the driving test have success rate of 60%, how likely is it that 60 students or more pass from the 100 students. Question 2: a machine fills the bottles with cola, it is distributed normally with an average of u = 300 ml and a standard deviation of 3mlHow likely is it that a Random bottle contains less than 299 milliliters, give an explanationHow likely is it that the average content of any 6 bottles is less than 299ml? Please specify Sorry for the bad english
Solution Posted January 10, 2017 Posted January 10, 2017 I have two questions for my homework, it would be very nice if someone can help me with this. Question 1: Exam candidates for the driving test have success rate of 60%, how likely is it that 60 students or more pass from the 100 students. Question 2: a machine fills the bottles with cola, it is distributed normally with an average of u = 300 ml and a standard deviation of 3ml How likely is it that a Random bottle contains less than 299 milliliters, give an explanation How likely is it that the average content of any 6 bottles is less than 299ml? Please specify Sorry for the bad english Question 1: 100% statistically 60% should pass, which means it's 100% LIKELY that 60 will pass. Question 2: with a standard deviation of 3 it'll most likely be between -3 and +3 deviation from 300ml. With 6 bottles you ought to end up with -1, +1, -2, +2, -3, +3 2 of them will end up being lower than 299 (298 and 297ml) your chances of hitting one of these 2 bottles when going through them is 2/6 = 1/3. so 33.33% riiiiiiight?
Team Cape Posted January 10, 2017 Posted January 10, 2017 (edited) 1. I assumed the first one was binomial, so i used binomcdf(100, 0.6, 60) on my caluclator and got a 53.79% chance. Not totally sure on this one, but this is what I got. 2. It's given that it's normal, so use normalcdf(-infinity, 299, 300, 3) (or you could use a table but thats very tedious). Comes out to 36.944% - This means that 36.944% of bottles will contain less than 299mL. This one is definitely right. 3. I multiplied the distribution by 6 (so it would be ~ Nor(1800, 18)), and I decided to look for 1794. I believe this should be the same probability (someone can correct me if I'm wrong). I plugged in normalcdf(-infinity, 1794, 1800, 18), and I got 36.944% Edit: I did all this in AP Stats like a couple months ago, so my memory is hazy. This is what I got though. I don't think #3 is right. Edited January 10, 2017 by Imateamcape 1
broodjepanda Posted January 10, 2017 Author Posted January 10, 2017 1. I assumed the first one was binomial, so i used binomcdf(100, 0.6, 60) on my caluclator and got a 53.79% chance. Not totally sure on this one, but this is what I got. 2. It's given that it's normal, so use normalcdf(-infinity, 299, 300, 3) (or you could use a table but thats very tedious). Comes out to 36.944% - This means that 36.944% of bottles will contain less than 299mL. This one is definitely right. 3. I multiplied the distribution by 6 (so it would be ~ Nor(1800, 18)), and I decided to look for 1794. I believe this should be the same probability (someone can correct me if I'm wrong). I plugged in normalcdf(-infinity, 1794, 1800, 18), and I got 36.944% Edit: I did all this in AP Stats like a couple months ago, so my memory is hazy. This is what I got though. I don't think #3 is right. Well you got me in the right direction, thanks! 1
Solution Posted January 10, 2017 Posted January 10, 2017 1. I assumed the first one was binomial, so i used binomcdf(100, 0.6, 60) on my caluclator and got a 53.79% chance. Not totally sure on this one, but this is what I got. 2. It's given that it's normal, so use normalcdf(-infinity, 299, 300, 3) (or you could use a table but thats very tedious). Comes out to 36.944% - This means that 36.944% of bottles will contain less than 299mL. This one is definitely right. 3. I multiplied the distribution by 6 (so it would be ~ Nor(1800, 18)), and I decided to look for 1794. I believe this should be the same probability (someone can correct me if I'm wrong). I plugged in normalcdf(-infinity, 1794, 1800, 18), and I got 36.944% Edit: I did all this in AP Stats like a couple months ago, so my memory is hazy. This is what I got though. I don't think #3 is right. Wew dis makes more sense 2
Team Cape Posted January 10, 2017 Posted January 10, 2017 Well you got me in the right direction, thanks! Ahhh here we go. For #3: u is an unbiased estimator so we can assume the mean is the same. Since the parent distribution is x also normal, we also know that it's normal. This should follow the 10% rule as well. So the distribution we use is ~ Nor(300, 3/sqrt(6)) which is ~ Nor(300, 1.225). So now we can plug this into the calculator. normalcdf(-infinity, 299, 300, 1.225) = 20.711% chance. I'm fairly sure this is correct.
ez11 Posted January 10, 2017 Posted January 10, 2017 Question 1: 100% statistically 60% should pass, which means it's 100% LIKELY that 60 will pass. Question 2: with a standard deviation of 3 it'll most likely be between -3 and +3 deviation from 300ml. With 6 bottles you ought to end up with -1, +1, -2, +2, -3, +3 2 of them will end up being lower than 299 (298 and 297ml) your chances of hitting one of these 2 bottles when going through them is 2/6 = 1/3. so 33.33% riiiiiiight? that wasnt a very good @Solution 1
Solution Posted January 10, 2017 Posted January 10, 2017 that wasnt a very good @Solution 2shit at statistics lul this is why I went with trigonometry n shit, much easier
Team Cape Posted January 10, 2017 Posted January 10, 2017 2shit at statistics lul this is why I went with trigonometry n shit, much easier :P
