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So, my last statistics class was looong ago, so I'm askin'


2Far

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Let's say I have a total sample size of 5,000,000

And out of that 5 x 10^6 group, 13,000 have, say, one green eye and one brown eye. So, that's ~0.26% of the group, or about 1 in 400.

So if I have a subgroup of, say 100 people from my main group, what are the odds of having a green/brown person in the group? 1 in 4 seems awful high?? Dunno.

Whatcha got?

 

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1 hour ago, 2Far said:

Let's say I have a total sample size of 5,000,000

And out of that 5 x 10^6 group, 13,000 have, say, one green eye and one brown eye. So, that's ~0.26% of the group, or about 1 in 400.

So if I have a subgroup of, say 100 people from my main group, what are the odds of having a green/brown person in the group? 1 in 4 seems awful high?? Dunno.

Whatcha got?

 

1/100 + 1/99+ 1/98 +1/97..................carry on.  If I remember properly this might be called something to do with the word factorial.

Perhaps it's not a sum, but rather some multiple.....I don't know.

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1 hour ago, 2Far said:

So if I have a subgroup of, say 100 people from my main group, what are the odds of having a green/brown person in the group?

So were these 100 samples plucked from the left side of the bell curve or the right side?  And what color eyes did their parents have?  Are some of them wearing colored contacts?

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13000/5000000 x 100 = 0.26.  1/0.26 = 3.85.   There's a 26% probability one of those 100 people will have 1 green, 1 brown eye, or 1 chance in 3.85,

This assumes the 13000 and 5000000 are exact numbers.  If it's rounded to the nearest million, the significant figure rules require a one sig.fig. answer, so 1 in 4.

Mila Kunis is the most attractive person I know of with one green eye and one brown eye, though her right eye is a sort of light brown.

When she tried out for That 70's Show, she was told they were only auditioning people 18 and older.  She replied, "Well, I'll be 18 on my birthday," and was allowed to audition.

She didn't say which birthday!  She was 14!

mila-kunis-with-brown-hair-and-natural-light-makeup | Brown hair ... image.png.5de850a2b242198414ba95b4e16c1103.png

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22 minutes ago, MickinMD said:

13000/5000000 x 100 = 0.26.  1/0.26 = 3.85.   There's a 26% probability one of those 100 people

I hate to ask this cause you are way smarter than me, but are you calculations right?  You multiplied by 100 twice.  13000/5000000 = 0.0026  X 100 = .26 which means a 0.26% chance, not 26%.

Think about it, 13000 is not 1/4 of 5000000

But what do I know..., mathing makes my head hurt and I have not had coffee yet

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38 minutes ago, MickinMD said:

13000/5000000 x 100 = 0.26.  1/0.26 = 3.85.   There's a 26% probability one of those 100 people will have 1 green, 1 brown eye, or 1 chance in 3.85,

This assumes the 13000 and 5000000 are exact numbers.  If it's rounded to the nearest million, the significant figure rules require a one sig.fig. answer, so 1 in 4.

Mila Kunis is the most attractive person I know of with one green eye and one brown eye, though her right eye is a sort of light brown.

When she tried out for That 70's Show, she was told they were only auditioning people 18 and older.  She replied, "Well, I'll be 18 on my birthday," and was allowed to audition.

She didn't say which birthday!  She was 14!

mila-kunis-with-brown-hair-and-natural-light-makeup | Brown hair ... image.png.5de850a2b242198414ba95b4e16c1103.png

Does this allow for the concept that the first eye checked can be either color but the second has to be specific?

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22 minutes ago, Zephyr said:

I hate to ask this cause you are way smarter than me, but are you calculations right?  You multiplied by 100 twice.  13000/5000000 = 0.0026  X 100 = .26 which means a 0.26% chance, not 26%.

Think about it, 13000 is not 1/4 of 5000000

But what do I know..., mathing makes my head hurt and I have not had coffee yet

I read it as the first 100 was to get percent and the second 100 was for the 100 people.

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