VCE Methods Question Thread!

[knightrider]

For this normal distribution curve attached.

isn't the max point meant to be

[knightrider]

For these  three curves in the image attached  normal distributions.

How would you find  the standard deviation for each curve.

in particular ii

[Cosec]

For these  three curves in the image attached  normal distributions.

How would you find  the standard deviation for each curve.

in particular ii

Although the bell curve never touches the x axis and seemingly goes for ever, you just break it up into 3 equal standard deviations per side. So say the mean/bell curve is centered around 0, and your bell curve 'finishes' at 9, the standard deviation will be 3.
So, part ii, the mean is 15, adn the curve "ends" at 20. So the difference is 5. So the standard deviation will be 5/3

[odeaa]

100 days to exam 1 fellow methdogs

shit just got real

[keltingmeith]

For these  three curves in the image attached  normal distributions.

How would you find  the standard deviation for each curve.

in particular ii

I'm going to go against Cosec - we have no sense of scale. We cannot find the standard deviation from this information alone, just relevant standard deviations (eg, sd (3)<sd (1)). VCAA will *never* ask this.

[knightrider]

Although the bell curve never touches the x axis and seemingly goes for ever, you just break it up into 3 equal standard deviations per side. So say the mean/bell curve is centered around 0, and your bell curve 'finishes' at 9, the standard deviation will be 3.
So, part ii, the mean is 15, adn the curve "ends" at 20. So the difference is 5. So the standard deviation will be 5/3

Thanks Cosec 

I'm going to go against Cosec - we have no sense of scale. We cannot find the standard deviation from this information alone, just relevant standard deviations (eg, sd (3)<sd (1)). VCAA will *never* ask this.

Thanks EulerFan101 

What other information would we need?

[knightrider]

For this normal distribution curve attached.

isn't the max point meant to be

Can anyone help with this?

Thanks 

[keltingmeith]

Thanks Cosec 


Thanks EulerFan101 

What other information would we need?

Anything, really. A point on the graph, a percentile. Essentially, you need a y-axis or area under the graph.

Can anyone help with this?

Thanks 

Depends - we don't know the mean/standard deviation.

[knightrider]

Anything, really. A point on the graph, a percentile. Essentially, you need a y-axis or area under the graph.

Depends - we don't know the mean/standard deviation.

Thanks EulerFan101 

The standard deviation was 1 ?

[knightrider]

How would you do this question?

A normally distributed variable has μ = 24 and σ = 3. Find the percentage of values between:
  30 and 33.

[silverpixeli]

How would you do this question?

A normally distributed variable has μ = 24 and σ = 3. Find the percentage of values between:
  30 and 33.

percentages, just like probabilities, will be given by areas (0.5 area under the pdf means 50% of the values are in that region, just like it would mean you have a 50% chance of getting a value in that region if you were talking about a random experiment)

This isn't one we can do by hand, you should have a function on your calculator for it. ti-nspire has a function to calculate the area between x=30 and x=33 on a normal curve with that mean and standard deviation. not sure about classpad/other calculators, but its definitely a CAS question

[Peanut Butter]

Hey guys! Just wondering if you could talk me through how to simplify the equations on the attached document.

Thanks! 

The first step you need to do is make all the exponentials have the lowest prime number base. So for this equation, the bases you are looking for are 2 and 3. So we simplify it like so:

[ 3^5n-4    x      (2 x 3 x 3)^n      x      (3^2)^3) ] divided by [ (2^2)^n+1     x     (3 x 3 x 2)^1-n     x     (2 x 3)^3-2n ]

We then expand.

[ 3^5n-4     x   2^n    x   2^n    x    3^n    x    3^6 ] divided by [ 2^2n+2   x   3^1-n   x   3^1-n   x    2^1-n   x  2^3-2n   x   3^3-2n  ]

And lastly, simplify.

[ 3^6n+2   x    2^2n ] divided by [2^-n+6    x    3^4-4n]

Which equals:
3^10n-2   x   2^3n-6

Hopefully that makes sense, if it doesn't just let me know

Sorry about the horrible formatting!!

P.S. I had to do this quickly, so hopefully their are no careless arithmetic errors!! 

[knightrider]

percentages, just like probabilities, will be given by areas (0.5 area under the pdf means 50% of the values are in that region, just like it would mean you have a 50% chance of getting a value in that region if you were talking about a random experiment)

This isn't one we can do by hand, you should have a function on your calculator for it. ti-nspire has a function to calculate the area between x=30 and x=33 on a normal curve with that mean and standard deviation. not sure about classpad/other calculators, but its definitely a CAS question

Thanks silverpixeli 

[knightrider]

For continuous random variables how come sometimes when they use probability notation these use the included signs and sometimes they use the not included signs.
e.g



OR SOMETIMES

what's the difference in these 2 for continuous random variables ?

[silverpixeli]

For continuous random variables how come sometimes when they use probability notation these use the included signs and sometimes they use the not included signs.
e.g



OR SOMETIMES

what's the difference in these 2 for continuous random variables ?

no difference numerically, you can think of it in terms of areas and hopefully see it with just a little geometric intuition:

what's the difference between the areas that those two probabilities describe, given a probability density function? you find the area between -2 and 5 including those points, or excluding them, it should make zero distance because whether you include the line on the end or not wont change the area because an infinitely thin line has an area of zero. So by taking away the edges and using < instead of <=, you're taking away zero area and therefore not changing the probability