Standard deviation question
In a special IQ test, it is predicted that the mean score will be Score 100, the 90th percentile will fall at Score 110, the 99th percentile at Score 120, the 99.9th percentile at Score 130 and so on ad infinitum. What is the standard deviation of scores in the IQ test?
Original post by Big-Daddy
In a special IQ test, it is predicted that the mean score will be Score 100, the 90th percentile will fall at Score 110, the 99th percentile at Score 120, the 99.9th percentile at Score 130 and so on ad infinitum. What is the standard deviation of scores in the IQ test?
These values do not fit a normal distribution.
We X~N(100,60.88285) we would have P(X<110)=0.9 but P(X<120)=0.9948.
Given that it's not a normal distribution there isn't enough information to fully define the distribution and so the standard deviation cannot be found.
Where is this question from?
Original post by BabyMaths
These values do not fit a normal distribution.
We X~N(100,60.88285) we would have P(X<110)=0.9 but P(X<120)=0.9948.
Given that it's not a normal distribution there isn't enough information to fully define the distribution and so the standard deviation cannot be found.
We X~N(100,60.88285) we would have P(X<110)=0.9 but P(X<120)=0.9948.
Given that it's not a normal distribution there isn't enough information to fully define the distribution and so the standard deviation cannot be found.
Isn't it fully defined? Seems to me we need to find f(x) such that the integral dx from -infinity to 100 is 0.5, from -infinity to 110 is 0.9, -infinity to 120 is 0.99, -infinity to 130 is 0.999, etc. Recursively it also looks like the integral from -infinity to 90 would be 0.1, -infinity to 80 would be 0.01, -infinity to 70 would be 0.001, etc., i.e. the function is symmetrical around 100. Is it impossible to find such a function f(x)?
Original post by BabyMaths
Where is this question from?
My friend gave it to me.
Original post by BabyMaths
These values do not fit a normal distribution.
We X~N(100,60.88285) we would have P(X<110)=0.9 but P(X<120)=0.9948.
Given that it's not a normal distribution there isn't enough information to fully define the distribution and so the standard deviation cannot be found.
Where is this question from?
We X~N(100,60.88285) we would have P(X<110)=0.9 but P(X<120)=0.9948.
Given that it's not a normal distribution there isn't enough information to fully define the distribution and so the standard deviation cannot be found.
Where is this question from?
What happened? Do I need to rephrase my issue?
Original post by Big-Daddy
What happened? Do I need to rephrase my issue?
Nothing happened. I'd just forget about it if I were you.
If F(x) is P(X<x) then you know F(100), F(110), F(120) etc..
This does not fully define F.
Where did your friend get the question from?
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