2008 3(d)

[iGilly]

I'm completely lost and the answers aren't much help, so was hoping you could be haha

[Jestar]

The 65-95-99 rule states that 66% of the area lies withing the first standard dev, 95% within the second and 99% within the third.

So basically, to attain 99% you want to find what the deviation is when multiplied by three.

68.4-67 = 1.4 -> 1.4/3 = 0.47

It works the other way too:

67-65.6 = 1.4, same answer.

At least I think that's right, but it is approximate so I'm not really sure.

[cypriottiger]

do you have solutions?

[3Xamz]

The 65-95-99 rule states that 66% of the area lies withing the first standard dev, 95% within the second and 99% within the third.

So basically, to attain 99% you want to find what the deviation is when multiplied by three.

68.4-67 = 1.4 -> 1.4/3 = 0.47

It works the other way too:

67-65.6 = 1.4, same answer.

At least I think that's right, but it is approximate so I'm not really sure.

Question asks for .99 not .997 as you've suggested.
Also, why does the VCAA solutions use .995? Instead of 0.99 :S

[InitialDRulz]

3Xamz, i've been asking that question for the last 3 hours!!!!! thank god you just said it!!!

[Jestar]

The 65-95-99 rule states that 66% of the area lies withing the first standard dev, 95% within the second and 99% within the third.

So basically, to attain 99% you want to find what the deviation is when multiplied by three.

68.4-67 = 1.4 -> 1.4/3 = 0.47

It works the other way too:

67-65.6 = 1.4, same answer.

At least I think that's right, but it is approximate so I'm not really sure.

Question asks for .99 not .997 as you've suggested.
Also, why does the VCAA solutions use .995? Instead of 0.99 :S

Yeah, I realised that as I re-read the question. Ignore my answer :S

[3Xamz]

Okay, we know;

Pr(65.6 < X < 68.4) = 0.99 and the mean is 67.

How do we progress from here.. Urgh, this question stuffed me last year too and I still don't get it. =='

[m@tty]

Okay, so you know that the values are symetrical about the mean +-1.4 . That implies that Pr(X>68.4)=Pr(X<65.6)=0.005. From there you can use invnorm and solve for .

EDIT:

[3Xamz]

OMG! Urgh, so stupid. Here's my way of thinking about it now that I get it;

We know that the probability either side of the mean is 0.495.
Since continuous variable functions are symmetrical along the mean.

Hence let's convert this into a standard normal distribution.

We know, that from -infinity to Z, the probability must be 0.5+0.495 = 0.995.
Because till half way its 0.5, and from halfway to the required Z value is another 0.495.

So, find the value of Z that has a probability of 0.995 which is Z = 2.5758 ,
This Z value corresponds to the higher limit, hence 2.5758 = (X-mu)/S.D

Therefore; S.D = (X-mu)/2.5758.
X = 68.4 and mu = 67.
Solving this gives S.D = .54mm

EDIT: Or m@tty's much more convenient method, LOL T__T

[InitialDRulz]

according from iTute

[matt123]

OMG! Urgh, so stupid. Here's my way of thinking about it now that I get it;

We know that the probability either side of the mean is 0.495.
Since continuous variable functions are symmetrical along the mean.

Hence let's convert this into a standard normal distribution.

We know, that from -infinity to Z, the probability must be 0.5+0.495 = 0.995.
Because till half way its 0.5, and from halfway to the required Z value is another 0.495.

So, find the value of Z that has a probability of 0.995 which is Z = 2.5758 ,
This Z value corresponds to the higher limit, hence 2.5758 = (X-mu)/S.D

Therefore; S.D = (X-mu)/2.5758.
X = 68.4 and mu = 67.
Solving this gives S.D = .54mm

EDIT: Or m@tty's much more convenient method, LOL T__T

WAIT
lol wtf how did u find Z?

[m@tty]

invnorm function on calc.

[3Xamz]

OMG! Urgh, so stupid. Here's my way of thinking about it now that I get it;

We know that the probability either side of the mean is 0.495.
Since continuous variable functions are symmetrical along the mean.

Hence let's convert this into a standard normal distribution.

We know, that from -infinity to Z, the probability must be 0.5+0.495 = 0.995.
Because till half way its 0.5, and from halfway to the required Z value is another 0.495.

So, find the value of Z that has a probability of 0.995 which is Z = 2.5758 ,
This Z value corresponds to the higher limit, hence 2.5758 = (X-mu)/S.D

Therefore; S.D = (X-mu)/2.5758.
X = 68.4 and mu = 67.
Solving this gives S.D = .54mm

EDIT: Or m@tty's much more convenient method, LOL T__T

WAIT
lol wtf how did u find Z?

invNormCDf("L", 0.995,1,0) on Classpad 330.
You can get that Interactive -> Inv. Distribution -> invNormCDf

Or you can go old school like I do, and refer to the Standard Normal Cumulative Density Function table

[m@tty]

OMG! Urgh, so stupid. Here's my way of thinking about it now that I get it;

We know that the probability either side of the mean is 0.495.
Since continuous variable functions are symmetrical along the mean.

Hence let's convert this into a standard normal distribution.

We know, that from -infinity to Z, the probability must be 0.5+0.495 = 0.995.
Because till half way its 0.5, and from halfway to the required Z value is another 0.495.

So, find the value of Z that has a probability of 0.995 which is Z = 2.5758 ,
This Z value corresponds to the higher limit, hence 2.5758 = (X-mu)/S.D

Therefore; S.D = (X-mu)/2.5758.
X = 68.4 and mu = 67.
Solving this gives S.D = .54mm

EDIT: Or m@tty's much more convenient method, LOL T__T

Our methods are the same, just you explained it in more detail xD

[matt123]

OMG! Urgh, so stupid. Here's my way of thinking about it now that I get it;

We know that the probability either side of the mean is 0.495.
Since continuous variable functions are symmetrical along the mean.

Hence let's convert this into a standard normal distribution.

We know, that from -infinity to Z, the probability must be 0.5+0.495 = 0.995.
Because till half way its 0.5, and from halfway to the required Z value is another 0.495.

So, find the value of Z that has a probability of 0.995 which is Z = 2.5758 ,
This Z value corresponds to the higher limit, hence 2.5758 = (X-mu)/S.D

Therefore; S.D = (X-mu)/2.5758.
X = 68.4 and mu = 67.
Solving this gives S.D = .54mm

EDIT: Or m@tty's much more convenient method, LOL T__T

WAIT
lol wtf how did u find Z?

invNormCDf("L", 0.995,1,0) on Classpad 330.
You can get that Interactive -> Inv. Distribution -> invNormCDf

Or you can go old school like I do, and refer to the Standard Normal Cumulative Density Function table

I get 1.695 for Z when i do this.
But that is the wrong number.

Why is it 1 , 0
i always thought it was 0 , 0 ?

lol
wtf am i doing wrongggg