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

Dis made chatbox lit fam come 

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?

 

 

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, 20179 yr by Imateamcape

 

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. 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

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.

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 

that wasnt a very good @Solution 

2shit at statistics lul this is why I went with trigonometry n shit, much easier

2shit at statistics lul this is why I went with trigonometry n shit, much easier

 

:P