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Normal Distribution Basics Watch

1. I wonder if anyone can help me understand the basics here.

Course : Edexcel AS - S1 Module.
Lookup Table percenatge points : 0.3 = 0.5244

I want to understand each eventuality and how I should remember to do the right thing.

Assume X - N (mean, 4)

I am thinking about all the possibilities for a question that would involve the lookup, to find the mean, so these might be, e.g. :

1. P (X > 52) = 0.7
2. P (X < 52) = 0.7
3. P (X > 52) = 0.3
4. P (X < 52) = 0.3

... and the normalised bits are :

1. P (Z > (52-mean)/4) = 0.7
2. P (Z < (52-mean)/4) = 0.7
3. P (Z > (52-mean)/4) = 0.3
4. P (Z < (52-mean)/4) = 0.3

... it's all about changing the sign and knowing why I am doing it that I get a bit confused with, so and help in talking this through would be appreciated.

Cheers.
2. (Original post by Charries)
I wonder if anyone can help me understand the basics here.

Course : Edexcel AS - S1 Module.
Lookup Table percenatge points : 0.3 = 0.5244

I want to understand each eventuality and how I should remember to do the right thing.

Assume X - N (mean, 4)

I am thinking about all the possibilities for a question that would involve the lookup, to find the mean, so these might be, e.g. :

1. P (X > 52) = 0.7
2. P (X < 52) = 0.7
3. P (X > 52) = 0.3
4. P (X < 52) = 0.3

... and the normalised bits are :

1. P (Z > (52-mean)/4) = 0.7
2. P (Z < (52-mean)/4) = 0.7
3. P (Z > (52-mean)/4) = 0.3
4. P (Z < (52-mean)/4) = 0.3

... it's all about changing the sign and knowing why I am doing it that I get a bit confused with, so and help in talking this through would be appreciated.

Cheers.
I'll do 4 as an example and you can work your way through the rest.

P (X < 52) = 0.3 => P (Z < (52-mean)/4) = 0.3

but we know that P(Z< 0.5244) = 0.3

And this means that

(52-mean)/4 = 0.5244 => mean = 52 - 4*0.5244

Hope this has helped.
3. (Original post by Beansontoast yum)
I'll do 4 as an example and you can work your way through the rest.

P (X < 52) = 0.3 => P (Z < (52-mean)/4) = 0.3

but we know that P(Z< 0.5244) = 0.3

And this means that

(52-mean)/4 = 0.5244 => mean = 52 - 4*0.5244

Hope this has helped.
When standardising, shouldn't it be P (Z < (52-mean)/2) (2 not 4) Standard Deviation not Variance
4. Thanks - I understand now.

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