Select a number, x, at random.
Find z = 11 - x = 2*5.5 - x.
Then the mean of x and z is 5.5
Add another pair, and another until you get fed up.
Or, you could select the numbers x and y at random and calculate z = 16.5 - (x + y) = 3*5.5 - (x + y). Then the triplet x, y, z will have mean 5.5
You can combine pair and triplets (and quartets, quintets and so on).
To create a set of data with a mean of 5.5, you can start by determining the number of data points you want in your set. Then, distribute the values in a way that their sum equals the product of the mean and the number of data points. For example, if you want 10 data points, you can distribute values such as 4, 5, 6, 5, 6, 5, 4, 6, 5, and 4, which sum up to 55, the product of 5.5 and 10.
Say this is your set of data: 22, 33, 33, 12, 33, 45, 65, and 77. ^ ^ ^ 33 occurs 3 times, which is more than any other number in this set, making it the mode. Say this is your set of data: 33, 55, 55, 33, 12, 115, 11, 155 ^ ^ ^ ^ Both 33 and 55 occur twice, which is more than any of the other numbers, meaning that the two modes are 33 and 55. Say this is your set of data: 1, 2, 3, 12, 124, 45, 576 No number(s) occurs more than the others, meaning there is NO mode in this set of data. Hope this helps, and recommend.
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If you mean: 55/100 then it is 0.55 as a decimal
If you mean 30/55 then it is 6/11 in its lowest terms
For this set of numbers, 45 87 61 73 55 78 69 80: σ=14.0204
The range of a data set is the difference between the largest and smallest number in your set of data. Median is the number that comes in the middle. 54, 55, 56 has a range of 54-56 and a median of 55. The set 53, 55, 57 has a median of 55 also!
28, 35, 43, 44, 55, 68, 77 Mean: 60 Median: 44 There is no mode.
It is 0.5
Compute the variance (or its square root , standard deviation) of each of the data set. Set 1: standard deviation = 10.121 Set 2: standard deviation = 12.09 Set 2 shows more variation around the mean. Check the link below
Say this is your set of data: 22, 33, 33, 12, 33, 45, 65, and 77. ^ ^ ^ 33 occurs 3 times, which is more than any other number in this set, making it the mode. Say this is your set of data: 33, 55, 55, 33, 12, 115, 11, 155 ^ ^ ^ ^ Both 33 and 55 occur twice, which is more than any of the other numbers, meaning that the two modes are 33 and 55. Say this is your set of data: 1, 2, 3, 12, 124, 45, 576 No number(s) occurs more than the others, meaning there is NO mode in this set of data. Hope this helps, and recommend.
No, because the average is all of the numbers added up and divided by the number of numbers. For example: 25,40, and 55 add up to 120. When divided in this case, the mean is 40. But: Say you have 70,90, and 101 when added it comes to 261. When divided, the answer is 87. Since 87 is not one of the values, the mean is not always equal to a value in the data set.
for the data set shown below find the interwar range IQR. 300,280,245,290,268,288,270,292,279,282
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The mean is the average: (2+15+21+27+31+42+55) divided by the number of terms (7). The mean is 193/7 = 27.6 The median is the number from the set which is in the middle, when listed lowest to highest, which you have already done. With your odd numbered set of 7 values, three numbers will be below the median, and three numbers above. The median is 27. If you had an even-numbered set, the median would be half-way between the two middle values of the set. In your example, there is no mode. The mode in a set of data is the value that occurs most often. No element in your set occurs more than once, and so there is no mode.
Mean: 55 and 1/11 Median: 56 Mode: 57
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