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you hace a median of 8 and you have a even number of boxes what are the other numbers.

the question asks you to add -12 to the numbers in the squares but yet there aren't any numbers other than 8 to work with.

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Probability and Sampling/Distributions - Andrews University

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Free Spreadsheet on the App Store - iTunes

May 11, 2016 · ... Free Spreadsheet has the number crunching power you need ... with many other ... PROD, SUMIF, FACT, MEAN, MEDIAN, MODE ...
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Combo with "Volume 7 - Paramedic Program" and 1 other ...

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Suggested Questions And Answer :

you hace a median of 8 and you have a even number of boxes what are the other numbers.

the "median" in krat speek is the middel number av a list av numbers but, wen hav even number av numbers, median=averaej av 2 numbers in the middel so yuno 2 middel numbers add up tu 16 Yu DONT no nuthun bout wot the numbers mite be
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How do I find the median when the only information I am given is the mean and the standard deviation

"meen" is krat werd for averaej=sum/(number av numbers) the "median" is the middel number...yu gotta SORT the numbers & find the 1 in the middel.  But if hav even number av numbers, gotta get 2 in middel & get averaej av em 2. yu hav averaej=94 & std deviashun=7.25 Me thank yu DONT hav nuf info... dont hav number av numbers prosess...step 1...get varians=sum av the square av (number -averaej) step 2: varians=abuv sum/(number av numbers) step3: std deviashun=root(varians) even if yu had (number av numbers) me thank yu stil hav a problem the "median" gotta bi 1 av the input numbers or averaej av 2 av the input numbers how yu gonna get that wen yu dont hav NE input numbers?
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Add a number to this data set so that the mode is 4 and the median is 6. 3 4 4 8 10

The mode is already 4 because it is the only repeated number. The median is the central value of the dataset has an odd number of elements or the average of the central pair if it has an even number of elements. The median is the average of 4 and 8, i.e., 6. So any number that doesn't equal any of the others and doesn't disturb the median of 6 is valid. Let's pick 9.
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what is the median for 180 171 180 162 163 177 203 155 165 191 196 178 165 168 175 195

what is the median for 180 171 180 162 163 177 203 155 165 191 196 178 165 168 175 195 rewrite the number lowest to highest 155 162 163 165 165 168 171 175 177 178 180 180 191 195 196 203 the median is the middle number there are 16 values/2 = 8 count over 8 175 and 177 are used because the number of values is even (175 + 177)/2 = 176 the median is 176 if it was odd number of values divide by 2 and take the center number
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If the Mean is 40, the mode is 10, the range is 10, and the median is 5, what is the number sequence

If the data is in order and is represented by a1, a2, ..., an for dataset size n, where n is an odd number then a[(n+1)/2]=5. Also an=a1+10. It's also clear that a1<5 and an<15 since 5 is the median and the range is 10. If n is even then the median is the average of a[n/2] and a[(n+2)/2]. The mode is 10 and that implies at least two tens in the dataset. Mean, median and mode are different versions of the average, but mean=40. This is not consistent with the requirements, particularly because the range is 10 and the lowest datum has to be less than 5, the median. All the data values "left" of the median must be less than the median and those to the "right" of the median must be greater, by definition of the median. If we assume that the range at best is approximately 5 to 15, then the mean=40 lies outside the range which suggests an error in the question. The mean, or average, has to be within the range of the data. TENTATIVE SOLUTIONS Let us suppose that 40 is the sum of the data rather than the mean, which is the sum of the data divided by the size of the dataset. If the least of the data is a1 then the greatest is a1+10. We know that a1<5 so a1+10<15 and ≥10 so a1≥0. There have to be at least 2 tens in the data because the mode is 10. The minimum size of the data is 7 consisting of a1, a2, a3, 5, 10, 10, a1+10. We assumed the sum was 40 so 2a1+a2+a3+35=40 making 2a1+a2+a3=5. If a1=0, then a2+a3=5 and we know a3>a2>a1, so a2 and a3 could be 1 and 4, 2 and 3, 1.5 and 3.5, etc. This gives us the data: 0, 1, 4, 5, 10, 10, 10 where the median is 5, the mode is 10, the range is 10 and the mean is 40/7. We could also have: 0, 2, 3, 5, 10, 10, 10, etc. If we put a1=0.5 then a2 and a3 could be 1 and 3, 1.5 and 2.5, etc., giving us, for example: 0.5, 1.5, 2.5, 5, 10, 10, 10.5. This also meets all requirements with a mean of 40/7. If we keep the mean at 40 then we need to adjust the range. For the sake of illustration let the range be 100, so the greatest data value is 100+a1. More to follow... ​
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Create a set of data in which the mean is 83 the median is 81 the mode is 80 and the range is 26

The minimum size for the required dataset appears to be 6, made up of data which, when ordered would be represented by the letters a to f. The sum of the data = 6*83=498, so that the mean will be 83. Because there is an even number of data in the set, the median has to be the average of c and d, so (c+d)/2=81; also we need a mode, which requires a minimum of two data values of 80, so b and c can be 80 and d needs to be 82 so (80+82)/2=81, the required median. f=a+26 so that the range is 26. Now we can add all the elements together: a+80+80+82+e+(a+26)=498. 2a+e=498-(80+80+82+26)=498-268=230. e>82 and needs to be an even number, so the smallest is 84, leaving 2a=146, so a=73. The ordered dataset is { 73 80 80 82 84 99 }. The data doesn't need to be ordered, so we could have { 99 80 84 73 80 82 }, for example.
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I need to find 8 numbers that equal 126

8 numbers with sum=126, so averaej=126/8=15.75 . . . for raenj=22, biggest number -smallest number=22 . . . median=11 meen middel number=11 . . . but yu hav even number av numbers, so averaeg av 2 middel numbs=11 . . . sins averaej=15.75 bigger than median, gotta hav sum big numbers . . . chuez a smallest number, sae 8, then big=8+22=30 . . . start with sumthun simpel such as...8, 9, 10, 11, 11 ,?, ?, 30 . . . sum=79 . . . 2 dontno numbers gotta add up tu 126-79=47 . . . so yukan yuze 23+24
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shortcut to finding the median without a computer

Question: shortcut to finding the median without a computer. The median is the middle value. If you have an array of values, then the median is that value that separates the upper half of the array from the lower half. If the number of elements in your array A[ ] is odd, i.e. N = 2n+1, then the median, M = A[n+1] If the number of elements in your array A[ ] is even, i.e. N = 2n, then the median, M = (A[n-1]+A[n+1])/2 i.e. M is the mean of the two middle values. If you don't want to use a computer, then you will have to visually inspect the arrray. Once you know the number of elements in the array, N, then you will have to manually count along until you reach the middle element, which is your median (or two middle ements, the mean of which is your median)
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Median, proportion

"median"=fansee guvt krat werd for the middel number first yu gotta sort the numbers, then pik out the won in the middel zampel...1,2,3...median=2 if hav even number av numbers, take averaej av 2 middel numbers this impli half (50%) av em hav more inkum
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how do you find the median for an even number

if yu oenlee hav 1 number, its the anser tu all quesshuns sum=number averaej=number median=number most other krat-invented werds=number
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