2Far ★ Posted August 14, 2020 Share #1 Posted August 14, 2020 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? Link to comment Share on other sites More sharing options...
maddmaxx ★ Posted August 14, 2020 Share #2 Posted August 14, 2020 42 1 1 Link to comment Share on other sites More sharing options...
jsharr ★ Posted August 14, 2020 Share #3 Posted August 14, 2020 Look I couldn’t even get past the questions were the train leaves the station in Chicago at 4 PM traveling 80 miles an hour. 1 Link to comment Share on other sites More sharing options...
Razors Edge ★ Posted August 14, 2020 Share #4 Posted August 14, 2020 1 minute ago, jsharr said: Look I couldn’t even get past the questions were the train leaves the station in Chicago at 4 PM traveling 80 miles an hour. Statistically speaking, 110% of students struggle with that problem (at first). It gets better. 2 1 Link to comment Share on other sites More sharing options...
JerrySTL ★ Posted August 14, 2020 Share #5 Posted August 14, 2020 Mean, median, or mode averages. That's all I remember from statistics. Link to comment Share on other sites More sharing options...
bikeman564™ Posted August 14, 2020 Share #6 Posted August 14, 2020 2 Link to comment Share on other sites More sharing options...
2Far ★ Posted August 14, 2020 Author Share #7 Posted August 14, 2020 So far, whilst mildly entertaining, you folks aren’t much help. Link to comment Share on other sites More sharing options...
maddmaxx ★ Posted August 14, 2020 Share #8 Posted August 14, 2020 37 minutes ago, jsharr said: Look I couldn’t even get past the questions were the train leaves the station in Chicago at 4 PM traveling 80 miles an hour. So you're saying that you fall outside the 3 sigma probability of being able to deal with this question. Link to comment Share on other sites More sharing options...
maddmaxx ★ Posted August 14, 2020 Share #9 Posted August 14, 2020 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. Link to comment Share on other sites More sharing options...
Ralphie ★ Posted August 14, 2020 Share #10 Posted August 14, 2020 Wouldn't it just be 1/4 person? Link to comment Share on other sites More sharing options...
Kzoo Posted August 14, 2020 Share #11 Posted August 14, 2020 Just another example if "Why do I need to learn that stuff. I'll never need it in the real world." And that's my final answer. 1 1 Link to comment Share on other sites More sharing options...
Kzoo Posted August 14, 2020 Share #12 Posted August 14, 2020 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? Link to comment Share on other sites More sharing options...
Mr. Silly Posted August 14, 2020 Share #13 Posted August 14, 2020 I think it would be 1 in 4 chance of having a mixed eye person in you group of 100. 2 Link to comment Share on other sites More sharing options...
MickinMD ★ Posted August 14, 2020 Share #14 Posted August 14, 2020 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! Link to comment Share on other sites More sharing options...
Zephyr Posted August 14, 2020 Share #15 Posted August 14, 2020 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 Link to comment Share on other sites More sharing options...
Razors Edge ★ Posted August 14, 2020 Share #16 Posted August 14, 2020 51 minutes ago, Kzoo said: Just another example if "Why do I need to learn that stuff. I'll never need it in the real world." And that's my final answer. Are you claiming THIS place - the SWCF aka Badfinger - is the "real world"????? Link to comment Share on other sites More sharing options...
Kzoo Posted August 14, 2020 Share #17 Posted August 14, 2020 Just now, Razors Edge said: Are you claiming THIS place - the SWCF aka Badfinger - is the "real world"????? no ummmmm.............. maybe Link to comment Share on other sites More sharing options...
maddmaxx ★ Posted August 14, 2020 Share #18 Posted August 14, 2020 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! Does this allow for the concept that the first eye checked can be either color but the second has to be specific? Link to comment Share on other sites More sharing options...
2Far ★ Posted August 14, 2020 Author Share #19 Posted August 14, 2020 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. Link to comment Share on other sites More sharing options...
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