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# Probabilities St. Normal Distribution

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1. Hi I want to check this out with everyone.

Assume the mean is 150 and the St. deviation is 5. I want to find the % probability that any 1 value will fall under 145 and over 160.

What I did was found the Z score for both values by the X - u / SD formula , then found the % probabilities of both reading it off the Z tables.

I then added the 2 % together and subtract the sum from 100% = the % probability they will fall o/s 145 - 160.

is this correct ?
2. (Original post by Zenomorph)
Hi I want to check this out with everyone.

Assume the mean is 150 and the St. deviation is 5. I want to find the % probability that any 1 value will fall under 145 and over 160.

What I did was found the Z score for both values by the X - u / SD formula , then found the % probabilities of both reading it off the Z tables.

I then added the 2 % together and subtract the sum from 100% = the % probability they will fall o/s 145 - 160.

is this correct ?
Doesn't sound right.

Your z values would be -1 and 2.

For the z value of -1, you'd need to do assuming you're using table (they normally only have +ve z values)

Then to find the probability of lying in the given range, you'd need

And outside that range you'd have

and then convert to a percentage.
3. (Original post by ghostwalker)
Doesn't sound right.

Your z values would be -1 and 2.

For the z value of -1, you'd need to do assuming you're using table (they normally only have +ve z values)

Then to find the probability of lying in the given range, you'd need

And outside that range you'd have

and then convert to a percentage.

Can I do it this way ? I take away the z value of -1 from z value of 2.
0.4772 - 0.3413 = 0.1359.

Then do 1 - 0.1359 = 0.8641 or 86.41 % of values o/s the range 145-160.

I think that should be the same as what you wrote, sorry I am not familiar with the notation of phi.

Thanks again.
4. (Original post by Zenomorph)

Can I do it this way ? I take away the z value of -1 from z value of 2.
0.4772 - 0.3413 = 0.1359.

Then do 1 - 0.1359 = 0.8641 or 86.41 % of values o/s the range 145-160.

I think that should be the same as what you wrote, sorry I am not familiar with the notation of phi.

Thanks again.
The method is correct in principle, HOWEVER your values for and are not.

= 0.9772

= 0.1587.

Where are you getting your values from? You seem to have the phi values of 1 and 2, and subtracted 0.5 from each.

Edit: Ironically, doing that would work with your original method.
5. (Original post by ghostwalker)
The method is correct in principle, HOWEVER your values for and are not.

= 0.9772

= 0.1587.

Where are you getting your values from? You seem to have the phi values of 1 and 2, and subtracted 0.5 from each.

Edit: Ironically, doing that would work with your original method.
I'm getting them from the regular Area under the normal curve values. I'm just reading it straight off, I think these are the correct tables, aren't they ?
6. (Original post by Zenomorph)
I'm getting them from the regular Area under the normal curve values. I'm just reading it straight off, I think these are the correct tables, aren't they ?
They're certainly not the usual ones.

phi(0) = 0.5, so any z value greater than 0, must give a phi(z) > 0.5.

Do you have a scan/link to them? Preferably with any header/footer information, and any preamble.
7. (Original post by ghostwalker)
They're certainly not the usual ones.

phi(0) = 0.5, so any z value greater than 0, must give a phi(z) > 0.5.

Do you have a scan/link to them? Preferably with any header/footer information, and any preamble.

exactly the same as this except mine goes on to 3.5 z.

http://www.mathsisfun.com/data/stand...ion-table.html

Anyway is the final answer 0.1815 ? cause I think get what you said
8. (Original post by Zenomorph)
exactly the same as this except mine goes on to 3.5 z.

http://www.mathsisfun.com/data/stand...ion-table.html

That's definitely non-standard - looks like they are doing something weird for the sake of a demo.

The table is giving the area under the curve between z=0 and the appropriate z value, whereas tables of the normal distribution usually give the total area to the left of the particular z value.

Anyway is the final answer 0.1815 ? cause I think get what you said
Yep.

Edit: I notice in the top left of the graph there is an options selection. Choosing "Up to z" gives you what would appear in standard tables (in the graph, albeit as a percentage rather than a probability). Their actual table below still doesn't reflect that though.
9. (Original post by ghostwalker)
That's definitely non-standard - looks like they are doing something weird for the sake of a demo.

The table is giving the area under the curve between z=0 and the appropriate z value, whereas tables of the normal distribution usually give the total area to the left of the particular z value.

Yep.

Edit: I notice in the top left of the graph there is an options selection. Choosing "Up to z" gives you what would appear in standard tables (in the graph, albeit as a percentage rather than a probability). Their actual table below still doesn't reflect that though.

Let me have another look at the tables and I'll come back to you in meantime my thanks to you.

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