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Help With Statistics Problem by TisaBone: 9:25am On Mar 09, 2012 |
will someone help me with this math problem including step by step instructions of how you arrived at the answer? I would appreciate the help so much. A committee of 6 is selected from 7 male and 10 female. If at least 4 female must be on the committee, in how many ways can the committee be formed? ps: the only reason i put it in romance is because no one would answer it in the education section |
Re: Help With Statistics Problem by aletheia(m): 1:53pm On Mar 09, 2012 |
TisaBone:^ Here's an attempt: Since there must be at least 4 females on the committee of 6, the possible combinations are: 6F + 0M, 5F + 1M, 4F + 2M; and the question can be rephrased as what possible ways to select 6 Females, 5 Females + 1 Male and so on. The total number of ways = (number of ways to choose 6F + 0M) + (number of ways to choose 5F + 1M) + (number of ways to choose 4F + 2M). Order does not matter so. Number of ways to choose 6F + 0M = 10!/6!(10-6)! * 7!/0!(7-0)! = 210*1 = 210 Number of ways to choose 5F + 1M = 10!/5!(10-5)! * 7!/1!(7-1)! = 252*7 = 1764 Number of ways to choose 4F + 2M = 10!/4!(10-4)! * 7!/2!(7-2)! = 210*21 = 4410 So the total number of ways = 210 + 1764 + 4410 = 6384 I hope this helps. |
Re: Help With Statistics Problem by Mayflowa(m): 3:43am On Mar 26, 2012 |
^^^ You just refreshed my permutation and combination knowledge. I used to confuse the two. You made is so easy and palpable. Thanks so much! |
Re: Help With Statistics Problem by Jayjuice(m): 9:52pm On Jul 20, 2012 |
"in a study to determine iff there is an association between alcohol consumption and cigarette smokers a researcher collected the following data from 1090 respondent. Those who take alcohol and cigarette-580,those who take alcohol only-40, those who take cigarette only-200,and those who take neither 270. Is there any association between them".........im supose to usee the (o-e)2/e formula,but im kinda stuck...any help?? |
Re: Help With Statistics Problem by ishmael(m): 10:59pm On Jul 20, 2012 |
Jayjuice: "in a study to determine iffSee u 2mrw morning. |
Re: Help With Statistics Problem by Emmy3(m): 4:18am On Jul 24, 2012 |
Jayjuice: "in a study to determine iff dis is set. |
Re: Help With Statistics Problem by Emmy3(m): 5:06am On Jul 24, 2012 |
Let d uni set be U or £, Let dose dat take alcohol be AC, dose dat take cirgarette C, dose who take neither be N and X those dat take both. . . . . . So, U = 1090 AC = 40+580 = 620 C = 200+580 = 780 N = 270 n(AC) n n(C) = 580 <prove 1090=620+780+270-580> thus, there exist an association btw those dat takes cirgaratte and dose dat drinks; such that 580 consumes both. . . . |
Re: Help With Statistics Problem by ishmael(m): 11:28am On Jul 24, 2012 |
Em-my:Yes u got it there is an association btw smokers and those that drink. But that's not what they want u to find. U're required to perform a test of significance to know if there is actually a significant association btw smoking and drinking using the chi-squared test. Hope u know about "test of hypothesis?" |
Re: Help With Statistics Problem by Emmy3(m): 3:26am On Jul 29, 2012 |
ishmael: Yes u got it there is an association btw smokers and those that drink. But that's not what they want u to find. U're required to perform a test of significance to know if there is actually a significant association btw smoking and drinking using the chi-squared test. Hope u know about "test of hypothesis?"ya. . . Letme try |
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