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January 30th, 2016, 09:44 AM   #1
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Normal Distribution: Calculating Mean and Standard Deviation

Hi All,

As usual have been driving myself mad with this question;

The heights of seedlings are normally distributed. Given that 10% of the seedlings are taller than 15cm and 5% are shorter than 4cm, find the mean and standard deviation of the heights.

Have been playing around getting lots of different answers.

Any help would be much appreciated, Thanks.
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January 30th, 2016, 03:58 PM   #2
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It looks like you should start with a normal distribution table. Find the points corresponding to 10% and above (above mean) and 4% and below (below mean). Linear transform to your coordinates. Finally use same linear transformation on 0 and 1 of the table to get your mean and standard deviation (subtracting the mean).
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February 4th, 2016, 05:09 AM   #3
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workings

so would I be right in saying?;

1.2816 = (15-mu) / sigma

and

-0.6449 = (4-mu) / sigma

Then use simultaneous equations to solve for both?
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February 4th, 2016, 04:42 PM   #4
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Assuming you constants are correct, this is the right idea.
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February 5th, 2016, 07:07 AM   #5
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I'm happy with the 1.2816 but I'm not convinced that -0.6449 is the correct value
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February 5th, 2016, 02:14 PM   #6
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Quote:
Originally Posted by wahhdoe View Post
I'm happy with the 1.2816 but I'm not convinced that -0.6449 is the correct value
These should have been obtained from a normal distribution table. I looked it up. Your upper number is correct, but I got ~ -1.645 for the lower number. It looks like you left out the 1. Note: the 5% or lower must be larger in magnitude than the 10% or above.
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February 10th, 2016, 02:37 AM   #7
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got there! wicked!!
thanks for your help!
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