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Set A consists of consecutive integers. What is the median of all the
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28 Oct 2017, 23:09
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Set A consists of consecutive integers. What is the median of all the numbers in set A? (1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \(\sqrt{2}\)
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Set A consists of consecutive integers. What is the median of all the
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28 Oct 2017, 23:19
souvonik2k wrote: Set A consists of consecutive integers. What is the median of all the numbers in set A? (1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \((2)^½\)
Please give kudos, if u liked my post! hi.. (1) The smallest number in set A is 4. we just know the smallest number . we require to know the number of items or the largest number to know the median insuff (2) The standard deviation of all the numbers in set A is \((2)^½\) since theterms are consecutive, we can know the number of elements in set.. \(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements but where are these 3 elements... insuff combined 3 elements starting with 4.. 4,5,6 median 5 suff c souvonik2k, in response to your query mentioned in post below the consecutive numbers have a difference of 1 ... so two cases .. 1) ODD numbers in the list.. MEDIAN is also the MEAN.. SD depends on how each element is away from median.. If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\) If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\) and so on, it will keep increasing with more elements added 2) EVEN numbers in the list median=mean = average of centre two numbers if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)hope it helps
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Re: Set A consists of consecutive integers. What is the median of all the
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28 Oct 2017, 23:27
chetan2u wrote: souvonik2k wrote: Set A consists of consecutive integers. What is the median of all the numbers in set A? (1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \((2)^½\)
Please give kudos, if u liked my post! hi.. (1) The smallest number in set A is 4. we just know the smallest number . we require to know the number of items or the largest number to know the median insuff (2) The standard deviation of all the numbers in set A is \((2)^½\) since theterms are consecutive, we can know the number of elements in set.. \(\sqrt{2}=\sqrt{ 1^2+0+1^2}\), so 3 elementsbut where are these 3 elements... insuff combined 3 elements starting with 4.. 4,5,6 median 5 suff c Hi chetan2uI couldn't understand this highlighted part. Cannot there be any other possibility than 3 numbers. Please explain.



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Re: Set A consists of consecutive integers. What is the median of all the
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09 Nov 2017, 00:12
niks18 ! some feedback on this one please !



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Set A consists of consecutive integers. What is the median of all the
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09 Nov 2017, 02:14
souvonik2k wrote: Set A consists of consecutive integers. What is the median of all the numbers in set A?
(1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \(\sqrt{2}\) Hi kunalsinghNSto find the median we need to know the number of elements in the set and as the set has consecutive elements, we need any one number from the set. Statement 1: provides one number from the set but we don't know the number of elements in the set. InsufficientStatement 2: let the number of elements in the set we \(n\) and \(a\) be the first number so our set will be {\(a, a+1, a+2.....a+(n1)\)} Use simple mathematical process to calculate st. deviation of this set. So average of the set will be \(= \frac{{a+a+1......a+(n1)}}{n}\) \(=\frac{{an+1+2...n1}}{n}\) \(= [an+n(n1)/2]/n\) [{1+2...{n1} this is a simple AP series with 1st term 1, last term n1, common difference 1 and number of terms n1. so you can easily calculate the sum in terms of n] so average of the set \(= a+\frac{(n1)}{2}\) to calculate standard deviation we need to reduce each element in the set by the average, then square it, then take the average of the resultant no and finally take the square root Step 1: \(aa\frac{(n1)}{2}\), \(a+1a\frac{(n1)}{2}\),.................., \(a+(n1)a\frac{(n1)}{2}\) \(= \frac{(n1)}{2}\), \(1\frac{(n1)}{2}\)........., \((n1)\frac{(n1)}{2}\) Step 2: now square each of the elements and add. On squaring & adding each of the elements, you will get something in terms of \(n^2\). let's call it \(kn^2\), where \(k\) is any constant resulting from the summation of the AP series Step 3: take the average of Step 2 \(= \frac{kn^2}{n} = kn\) Step 4: standard deviation \(= \sqrt{kn}\) so we have \(\sqrt{kn}=\sqrt{2}\) hence \(n=\frac{2}{k}\). So we get the value of \(n\) but we don't know any of the elements in the set. InsufficientCombining 1 & 2: we get all the required parameters. Statement 1 provides the 1st element in the set and Statement 2 provides the number of elements in the set. Hence CkunalsinghNS, instead of going for the mathematical derivation, you can test by using simple nos, for e.g 1,2,3 etc. to know that if you are given st. deviation/variance, then you can calculate the number of terms in the set because each element is consecutive. I have skipped some calculation because working with variables was becoming cumbersome . but the point is statement 2 will give you the number of elements in the set as the set is consecutive. Hope this helps



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Re: Set A consists of consecutive integers. What is the median of all the
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09 Nov 2017, 08:10
This seems to be a pretty long calculation !! approximation is the best option i guess ! but thank you !!



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Re: Set A consists of consecutive integers. What is the median of all the
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09 Nov 2017, 08:18
kunalsinghNS wrote: This seems to be a pretty long calculation !! approximation is the best option i guess ! but thank you !! Yup. But you only need to visualise the process and need to note that if st. Deviation of consecutive nos is given we can calculate the number of terms in the set. Mathematical derivation is just for academics purposes Posted from my mobile device



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Re: Set A consists of consecutive integers. What is the median of all the
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23 Feb 2019, 02:37
we also need to divide the summation of the squares by the total no of elements. Can someone point out where are we dividing the summation before taking the square root?
Thanks in advance



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Re: Set A consists of consecutive integers. What is the median of all the
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23 Feb 2019, 03:15
Debashis Roy wrote: we also need to divide the summation of the squares by the total no of elements. Can someone point out where are we dividing the summation before taking the square root?
Thanks in advance Hello , In this step niks18 divided by n ( after approximation ) Step 3: take the average of Step 2 =kn2n=kn



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Set A consists of consecutive integers. What is the median of all the
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27 Feb 2019, 23:33
chetan2u VeritasKarishmaHi, In your explanation : √=1^2+0+1^2−−−−−−−−−√2=1^2+0+1^2, so 3 elements... shouldnt we divide the sum of the squares of the terms by the no of terms also... In gthat case for 3 terms 4,5,6... SD= √[(1^2+0+1^2)/3]...gives √(2/3)... Please explain..



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Re: Set A consists of consecutive integers. What is the median of all the
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28 Feb 2019, 20:48
Debashis Roy wrote: chetan2u VeritasKarishmaHi, In your explanation : √=1^2+0+1^2−−−−−−−−−√2=1^2+0+1^2, so 3 elements... shouldnt we divide the sum of the squares of the terms by the no of terms also... In gthat case for 3 terms 4,5,6... SD= √[(1^2+0+1^2)/3]...gives √(2/3)... Please explain.. Yes, the SD will be \(\sqrt{2} = \sqrt{\frac{(2)^2 + (1)^2 + 0 + 1^2 + 2^2}{5}}\) In any case, the answer doesn't change. Each unique distribution will have a unique SD. For median, you will need the exact position where the distribution is placed on the number line so you need both statements to answer the question. Answer (C)
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Re: Set A consists of consecutive integers. What is the median of all the
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28 Feb 2019, 22:57
VeritasKarishmaTo quote an above explanation: (2) The standard deviation of all the numbers in set A is (2)½(2)½ since theterms are consecutive, we can know the number of elements in set.. 2√=12+0+12−−−−−−−−−√2=12+0+12, so 3 elements but where are these 3 elements... insuff combined 3 elements starting with 4.. 4,5,6 median 5 suff How come the SD is √2 for the set {4,5,6}... I am not getting this part as written in the explanation by chetan2u



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Re: Set A consists of consecutive integers. What is the median of all the
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14 Jun 2019, 00:06
chetan2u wrote: souvonik2k wrote: Set A consists of consecutive integers. What is the median of all the numbers in set A? (1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \((2)^½\)
Please give kudos, if u liked my post! hi.. (1) The smallest number in set A is 4. we just know the smallest number . we require to know the number of items or the largest number to know the median insuff (2) The standard deviation of all the numbers in set A is \((2)^½\) since theterms are consecutive, we can know the number of elements in set.. \(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements but where are these 3 elements... insuff combined 3 elements starting with 4.. 4,5,6 median 5 suff c souvonik2k, in response to your query mentioned in post below the consecutive numbers have a difference of 1 ... so two cases .. 1) ODD numbers in the list.. MEDIAN is also the MEAN.. SD depends on how each element is away from median.. If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\) If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\) and so on, it will keep increasing with more elements added 2) EVEN numbers in the list median=mean = average of centre two numbers if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)hope it helps Hi chetan2u, I agree with your explanation but if we go by the SD formula then it will be \(\sqrt{2}\)/3. 3 being the number of terms. Hence, answer should be E instead of C?



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Re: Set A consists of consecutive integers. What is the median of all the
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30 Jun 2019, 08:56
Hi chetan2uAs per your solution posted above , SD for 4,5,6 will never be root(2). It will be root(2/3). SD of 4,5,6,7,8 will be root(2). SD of any 5 consecutive positive integers is root(2). Thus median will be 6 not 5



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Re: Set A consists of consecutive integers. What is the median of all the
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07 Aug 2019, 09:52
chetan2u wrote: souvonik2k wrote: Set A consists of consecutive integers. What is the median of all the numbers in set A? (1) The smallest number in set A is 4. (2) The standard deviation of all the numbers in set A is \((2)^½\)
Please give kudos, if u liked my post! hi.. (1) The smallest number in set A is 4. we just know the smallest number . we require to know the number of items or the largest number to know the median insuff (2) The standard deviation of all the numbers in set A is \((2)^½\) since theterms are consecutive, we can know the number of elements in set.. \(\sqrt{2}=\sqrt{1^2+0+1^2}\), so 3 elements but where are these 3 elements... insuff combined 3 elements starting with 4.. 4,5,6 median 5 suff c souvonik2k, in response to your query mentioned in post below the consecutive numbers have a difference of 1 ... so two cases .. 1) ODD numbers in the list.. MEDIAN is also the MEAN.. SD depends on how each element is away from median.. If 3 elements, say 5,6,7,.... middle is 0 away from mean, the smaller and bigger are 1 away from mean so SD = \(\sqrt{1^2+0+1^2}=\sqrt{2}\) If 5 elements say 2,3,4,5,6... middle is 0 away, the biggest and smallest are 2 away from mean and other two 1 away, so SD =\(\sqrt{2^2+1^2+0+1^2+2^2}=\sqrt{10}\) and so on, it will keep increasing with more elements added 2) EVEN numbers in the list median=mean = average of centre two numbers if 2 elements.. 3,4...SD = \(\sqrt{(1/2)^2+(1/2)^2}=\sqrt{1/2}\)hope it helps Please see that formula used for SD is incorrect. Statment 1: clearly insufficient Statement 2: for SD to be root2, we need any 5 consecutive integers. But we still dont know the starting point. Not sufficient Combined: we know the starting point and we know that there are 5 integers Sufficient. Answer C




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