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# what do you do if there are multiple modes in a set of data?

Im in 7th grade math, and i forgot the rules of finding the mode, if there are mulitple modes.

## Research, Knowledge and Information :

### Math Forum - Ask Dr. Math

More Than One Mode? ... You can have multiple modes in a data set. ... When no number occurs more than once in a data set, there is no mode.

### How Do You Find the Mode of a Data Set When ... - Virtual Nerd

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### www.virtualnerd.com

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### How to Calculate the Mode or Modal Value - Math is Fun- Mode

So there are two modes: ... Having more than two modes is called "multimodal". ... as the mode. You could use different groupings and get a different answer! ...

### How to Calculate Mode Using Excel: 10 Steps (with Pictures)

How to Calculate Mode ... will output the same as MODE.SNGL. If you set your spreadsheet ... MODE.MULT displays multiple modes. For a data set of 2,1,3,4,3,2 ...

### The Mode of a Set of Data

The mode of a set of data is presented in this ... the modes are 18 and 24 miles ... Since each value occurs only once in the data set, there is no mode for this ...

### Mode (statistics) - Wikipedia

The mode of a set of data values is the ... distribution has multiple local maxima it is common to ... , while a set with more than two modes may be ...

### The Excel MODE.MULT Function - ThoughtCo

The Excel MODE.MULT Function. ... used to determine if the data contains multiple modes ... will tell you if there are multiple values - or multiple modes ...

## Suggested Questions And Answer :

### what do you do if there are multiple modes in a set of data?

thro out all yu dont understand so yer left with null

### 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.

### 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.

### What is the median for a five number set when the second and third number are the same 6.5,7.,7.,10.,11.

The median is always the middle value in an ordered set of data where the number of data is odd. So the median is the third datum in a set of 5: 7. This value is also the mode in this case.

### Gina spent \$4,\$5,\$7,\$7,and \$6 over the past 5 days buying lunch. Is the mean,media,mode,or range the most useful way to describe this data set?Explain.

I think so, because we can find the mean, median, mode and range for the dataset. The mean is \$29/5=\$14.5; the median is \$6 because it's the central value of all the values when ordered: \$4, \$5, \$6, \$7, \$7; the mode is 2 because \$7 is repeated; and the range is \$3 ranging from 4\$ to \$7. From these other stats can be derived.

### what is mode

each time, write 2 diffrent data sets with 6 numbers The mode is 100. The median and the mean are equal. The mode is 100. The mean is less than the median.

### 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... ​