You entered a number set X of {110,220,5}
From the 3 numbers you entered, we want to calculate the mean, variance, standard deviation, standard error of the mean, skewness, average deviation (mean absolute deviation), median, mode, range, Pearsons Skewness Coefficient of that number set, entropy, mid-range
5, 110, 220
Rank Ascending
5 is the 1st lowest/smallest number
110 is the 2nd lowest/smallest number
220 is the 3rd lowest/smallest number
220, 110, 5
Rank Descending
220 is the 1st highest/largest number
110 is the 2nd highest/largest number
5 is the 3rd highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 5 | 110 | 220 |
Rank | 1 | 2 | 3 |
Since we have 3 numbers in our original number set,
we assign ranks from lowest to highest (1 to 3)
Our original number set in unsorted order was 5,110,220
Our respective ranked data set is 1,2,3
Root Mean Square = | √A |
√N |
where A = x12 + x22 + x32 and N = 3 number set items
A = 52 + 1102 + 2202
A = 25 + 12100 + 48400
A = 60525
RMS = | √60525 |
√3 |
RMS = | 246.01829200285 |
1.7320508075689 |
RMS = 142.03872711342
Central tendency contains:
Mean, median, mode, harmonic mean,
geometric mean, mid-range, weighted-average:
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 5 + 110 + 220 |
3 |
μ = | 335 |
3 |
μ = 111.66666666667
Since our number set contains 3 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1,n2,n3)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(4)
Median Number = n2
Our median is entry 2 of our number set highlighted in red:
(5,110,220)
Median = 110
The highest frequency of occurence in our number set is 1 times
by the following numbers in green:
()
Since the maximum frequency of any number is 1, no mode exists.
Mode = N/A
Harmonic Mean = | N |
1/x1 + 1/x2 + 1/x3 |
With N = 3 and each xi a member of the number set you entered, we have:
Harmonic Mean = | 3 |
1/5 + 1/110 + 1/220 |
Harmonic Mean = | 3 |
0.2 + 0.0090909090909091 + 0.0045454545454545 |
Harmonic Mean = | 3 |
0.21363636363636 |
Harmonic Mean = 14.042553191489
Geometric Mean = (x1 * x2 * x3)1/N
Geometric Mean = (5 * 110 * 220)1/3
Geometric Mean = 1210000.33333333333333
Geometric Mean = 49.460874432487
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 220 + 5 |
2 |
Mid-Range = | 225 |
2 |
Mid-Range = 112.5
Take the first digit of each value in our number set
Use this as our stem value
Use the remaining digits for our leaf portion
{220,110,5}
Stem | Leaf |
---|---|
2 | 20 |
1 | 10 |
5 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 5 + 110 + 220 |
3 |
μ = | 335 |
3 |
μ = 111.66666666667
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (5 - 111.66666666667)2 = -106.666666666672 = 11377.777777778
(X2 - μ)2 = (110 - 111.66666666667)2 = -1.66666666666672 = 2.7777777777778
(X3 - μ)2 = (220 - 111.66666666667)2 = 108.333333333332 = 11736.111111111
ΣE(Xi - μ)2 = 11377.777777778 + 2.7777777777778 + 11736.111111111
ΣE(Xi - μ)2 = 23116.666666667
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 7705.5555555556 | Variance: σs2 = 11558.333333333 | ||||||||
Standard Deviation: σp = √σp2 = √7705.5555555556 | Standard Deviation: σs = √σs2 = √11558.333333333 | ||||||||
Standard Deviation: σp = 87.7813 | Standard Deviation: σs = 107.5097 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 50.6806 | SEM = 62.0708 |
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Let's evaluate the square difference from the mean of each term (Xi - μ)3:
(X1 - μ)3 = (5 - 111.66666666667)3 = -106.666666666673 = -1213629.6296296
(X2 - μ)3 = (110 - 111.66666666667)3 = -1.66666666666673 = -4.6296296296297
(X3 - μ)3 = (220 - 111.66666666667)3 = 108.333333333333 = 1271412.037037
ΣE(Xi - μ)3 = -1213629.6296296 + -4.6296296296297 + 1271412.037037
ΣE(Xi - μ)3 = 57777.777777777
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 57777.777777777 |
(3 - 1)87.78133 |
Skewness = | 57777.777777777 |
(2)676403.77817781 |
Skewness = | 57777.777777777 |
1352807.5563556 |
Skewness = 0.042709532118099
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |5 - 111.66666666667| = |-106.66666666667| = 106.66666666667
|X2 - μ| = |110 - 111.66666666667| = |-1.6666666666667| = 1.6666666666667
|X3 - μ| = |220 - 111.66666666667| = |108.33333333333| = 108.33333333333
Σ|Xi - μ| = 106.66666666667 + 1.6666666666667 + 108.33333333333
Σ|Xi - μ| = 216.66666666667
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 216.66666666667 |
3 |
Average Deviation = 72.22222
Since our number set contains 3 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1,n2,n3)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(4)
Median Number = n2
Our median is entry 2 of our number set highlighted in red:
(5,110,220)
Median = 110
The highest frequency of occurence in our number set is 1 times
by the following numbers in green:
()
Since the maximum frequency of any number is 1, no mode exists.
Mode = N/A
Range = Largest Number in the Number Set - Smallest Number in the Number Set
Range = 220 - 5
Range = 215
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(111.66666666667 - N/A) |
87.7813 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(111.66666666667 - 110) |
87.7813 |
PSC2 = | 3 x 1.6666666666667 |
87.7813 |
PSC2 = | 5 |
87.7813 |
PSC2 = 0.057
Entropy = Ln(n)
Entropy = Ln(3)
Entropy = 1.0986122886681
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 220 + 5 |
2 |
Mid-Range = | 225 |
2 |
Mid-Range = 112.5
We need to sort our number set from lowest to highest shown below:
{5,110,220}
V = | y(n + 1) |
100 |
V = | 75(3 + 1) |
100 |
V = | 75(4) |
100 |
V = | 300 |
100 |
V = 3 ← Rounded down to the nearest integer
Upper quartile (UQ) point = Point # 3 in the dataset which is 220
5,110,220V = | y(n + 1) |
100 |
V = | 25(3 + 1) |
100 |
V = | 25(4) |
100 |
V = | 100 |
100 |
V = 1 ← Rounded up to the nearest integer
Lower quartile (LQ) point = Point # 1 in the dataset which is 5
5,110,220
IQR = UQ - LQ
IQR = 220 - 5
IQR = 215
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 5 - 1.5 x 215
Lower Inner Fence (LIF) = 5 - 322.5
Lower Inner Fence (LIF) = -317.5
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 220 + 1.5 x 215
Upper Inner Fence (UIF) = 220 + 322.5
Upper Inner Fence (UIF) = 542.5
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 5 - 3 x 215
Lower Outer Fence (LOF) = 5 - 645
Lower Outer Fence (LOF) = -640
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 220 + 3 x 215
Upper Outer Fence (UOF) = 220 + 645
Upper Outer Fence (UOF) = 865
Suspect Outliers are values between the inner and outer fences
We wish to mark all values in our dataset (v) in red below such that -640 < v < -317.5 and 542.5 < v < 865
5,110,220
Highly Suspect Outliers are values outside the outer fences
We wish to mark all values in our dataset (v) in red below such that v < -640 or v > 865
5,110,220
5, 110, 220
Multiply each value by each probability amount
We do this by multiplying each Xi x pi to get a weighted score Y
Weighted Average = | X1p1 + X2p2 + X3p3 |
n |
Weighted Average = | 5 x 0.2 + 110 x 0.4 + 220 x 0.6 |
3 |
Weighted Average = | 1 + 44 + 132 |
3 |
Weighted Average = | 177 |
3 |
Weighted Average = 59
Show the freqency distribution table for this number set
5, 110, 220
We need to choose the smallest integer k such that 2k ≥ n where n = 3
For k = 1, we have 21 = 2
For k = 2, we have 22 = 4 ← Use this since it is greater than our n value of 3
Therefore, we use 2 intervals
Our maximum value in our number set of 220 - 5 = 215
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 215 |
2 |
Add 1 to this giving us 107 + 1 = 108
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
5 - 113 | 4.5 - 113.5 | 2 | 2 | 2/3 = 66.67% | 2/3 = 66.67% |
113 - 221 | 112.5 - 221.5 | 1 | 2 + 1 = 3 | 1/3 = 33.33% | 3/3 = 100% |
3 | 100% |
Go through our 3 numbers
Determine the ratio of each number to the next one
5:110 → 0.0455
110:220 → 0.5
Successive Ratio = 5:110,110:220 or 0.0455,0.5