You entered a number set X of {2,15,15,18,30}
From the 5 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, midrange
2, 15, 15, 18, 30
30, 18, 15, 15, 2
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value  2  15  15  18  30 
Rank  1  2  3  4  5 
Since we have 5 numbers in our original number set,
we assign ranks from lowest to highest (1 to 5)
Our original number set in unsorted order was 2,15,15,18,30
Our respective ranked data set is 1,3,3,4,5
Root Mean Square =  √A 
√N 
where A = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2} and N = 5 number set items
A = 2^{2} + 15^{2} + 15^{2} + 18^{2} + 30^{2}
A = 4 + 225 + 225 + 324 + 900
A = 1678
RMS =  √1678 
√5 
RMS =  40.963398296528 
2.2360679774998 
RMS = 18.319388636087
Central tendency contains:
Mean, median, mode, harmonic mean,
geometric mean, midrange, weightedaverage:
μ =  Sum of your number Set 
Total Numbers Entered 
μ =  ΣX_{i} 
n 
μ =  2 + 15 + 15 + 18 + 30 
5 
μ =  80 
5 
μ = 16
Since our number set contains 5 elements which is an odd number,
our median number is determined as follows:
Number Set = (n_{1},n_{2},n_{3},n_{4},n_{5})
Median Number = Entry ½(n + 1)
Median Number = Entry ½(6)
Median Number = n_{3}
Our median is entry 3 of our number set highlighted in red:
(2,15,15,18,30)
Median = 15
The highest frequency of occurence in our number set is 2 times
by the following numbers in green:
()
Our mode is denoted as: 15
Mode = 15
Harmonic Mean =  N 
1/x_{1} + 1/x_{2} + 1/x_{3} + 1/x_{4} + 1/x_{5} 
With N = 5 and each x_{i} a member of the number set you entered, we have:
Harmonic Mean =  5 
1/2 + 1/15 + 1/15 + 1/18 + 1/30 
Harmonic Mean =  5 
0.5 + 0.066666666666667 + 0.066666666666667 + 0.055555555555556 + 0.033333333333333 
Harmonic Mean =  5 
0.72222222222222 
Harmonic Mean = 6.9230769230769
Geometric Mean = (x_{1} * x_{2} * x_{3} * x_{4} * x_{5})^{1/N}
Geometric Mean = (2 * 15 * 15 * 18 * 30)^{1/5}
Geometric Mean = 243000^{0.2}
Geometric Mean = 11.943215116605
MidRange =  Maximum Value in Number Set + Minimum Value in Number Set 
2 
MidRange =  30 + 2 
2 
MidRange =  32 
2 
MidRange = 16
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
{30,18,15,15,2}
Stem  Leaf 

3  0 
1  5,5,8 
2 
Mean, Variance, Standard Deviation, Median, Mode
μ =  Sum of your number Set 
Total Numbers Entered 
μ =  ΣX_{i} 
n 
μ =  2 + 15 + 15 + 18 + 30 
5 
μ =  80 
5 
μ = 16
Let's evaluate the square difference from the mean of each term (X_{i}  μ)^{2}:
(X_{1}  μ)^{2} = (2  16)^{2} = 14^{2} = 196
(X_{2}  μ)^{2} = (15  16)^{2} = 1^{2} = 1
(X_{3}  μ)^{2} = (15  16)^{2} = 1^{2} = 1
(X_{4}  μ)^{2} = (18  16)^{2} = 2^{2} = 4
(X_{5}  μ)^{2} = (30  16)^{2} = 14^{2} = 196
ΣE(X_{i}  μ)^{2} = 196 + 1 + 1 + 4 + 196
ΣE(X_{i}  μ)^{2} = 398
Population  Sample  


 

 
Variance: σ_{p}^{2} = 79.6  Variance: σ_{s}^{2} = 99.5  
Standard Deviation: σ_{p} = √σ_{p}^{2} = √79.6  Standard Deviation: σ_{s} = √σ_{s}^{2} = √99.5  
Standard Deviation: σ_{p} = 8.9219  Standard Deviation: σ_{s} = 9.975 
Population  Sample  


 

 

 
SEM = 3.99  SEM = 4.461 
Skewness =  E(X_{i}  μ)^{3} 
(n  1)σ^{3} 
Let's evaluate the square difference from the mean of each term (X_{i}  μ)^{3}:
(X_{1}  μ)^{3} = (2  16)^{3} = 14^{3} = 2744
(X_{2}  μ)^{3} = (15  16)^{3} = 1^{3} = 1
(X_{3}  μ)^{3} = (15  16)^{3} = 1^{3} = 1
(X_{4}  μ)^{3} = (18  16)^{3} = 2^{3} = 8
(X_{5}  μ)^{3} = (30  16)^{3} = 14^{3} = 2744
ΣE(X_{i}  μ)^{3} = 2744 + 1 + 1 + 8 + 2744
ΣE(X_{i}  μ)^{3} = 6
Skewness =  E(X_{i}  μ)^{3} 
(n  1)σ^{3} 
Skewness =  6 
(5  1)8.9219^{3} 
Skewness =  6 
(4)710.18591309046 
Skewness =  6 
2840.7436523618 
Skewness = 0.0021121229981493
AD =  ΣX_{i}  μ 
n 
Evaluate the absolute value of the difference from the mean
X_{i}  μ:
X_{1}  μ = 2  16 = 14 = 14
X_{2}  μ = 15  16 = 1 = 1
X_{3}  μ = 15  16 = 1 = 1
X_{4}  μ = 18  16 = 2 = 2
X_{5}  μ = 30  16 = 14 = 14
ΣX_{i}  μ = 14 + 1 + 1 + 2 + 14
ΣX_{i}  μ = 32
Calculate average deviation (mean absolute deviation)
AD =  ΣX_{i}  μ 
n 
AD =  32 
5 
Average Deviation = 6.4
Since our number set contains 5 elements which is an odd number,
our median number is determined as follows:
Number Set = (n_{1},n_{2},n_{3},n_{4},n_{5})
Median Number = Entry ½(n + 1)
Median Number = Entry ½(6)
Median Number = n_{3}
Our median is entry 3 of our number set highlighted in red:
(2,15,15,18,30)
Median = 15
The highest frequency of occurence in our number set is 2 times
by the following numbers in green:
()
Our mode is denoted as: 15
Mode = 15
Range = Largest Number in the Number Set  Smallest Number in the Number Set
Range = 30  2
Range = 28
PSC1 =  μ  Mode 
σ 
PSC1 =  3(16  15) 
8.9219 
PSC1 =  3 x 1 
8.9219 
PSC1 =  3 
8.9219 
PSC1 = 0.3363
PSC2 =  μ  Median 
σ 
PSC1 =  3(16  15) 
8.9219 
PSC2 =  3 x 1 
8.9219 
PSC2 =  3 
8.9219 
PSC2 = 0.3363
Entropy = Ln(n)
Entropy = Ln(5)
Entropy = 1.6094379124341
MidRange =  Smallest Number in the Set + Largest Number in the Set 
2 
MidRange =  30 + 2 
2 
MidRange =  32 
2 
MidRange = 16
We need to sort our number set from lowest to highest shown below:
{2,15,15,18,30}
V =  y(n + 1) 
100 
V =  75(5 + 1) 
100 
V =  75(6) 
100 
V =  450 
100 
V = 4 ← Rounded down to the nearest integer
Upper quartile (UQ) point = Point # 4 in the dataset which is 18
2,15,15,18,30V =  y(n + 1) 
100 
V =  25(5 + 1) 
100 
V =  25(6) 
100 
V =  150 
100 
V = 2 ← Rounded up to the nearest integer
Lower quartile (LQ) point = Point # 2 in the dataset which is 15
2,15,15,18,30
IQR = UQ  LQ
IQR = 18  15
IQR = 3
Lower Inner Fence (LIF) = LQ  1.5 x IQR
Lower Inner Fence (LIF) = 15  1.5 x 3
Lower Inner Fence (LIF) = 15  4.5
Lower Inner Fence (LIF) = 10.5
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 18 + 1.5 x 3
Upper Inner Fence (UIF) = 18 + 4.5
Upper Inner Fence (UIF) = 22.5
Lower Outer Fence (LOF) = LQ  3 x IQR
Lower Outer Fence (LOF) = 15  3 x 3
Lower Outer Fence (LOF) = 15  9
Lower Outer Fence (LOF) = 6
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 18 + 3 x 3
Upper Outer Fence (UOF) = 18 + 9
Upper Outer Fence (UOF) = 27
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 6 < v < 10.5 and 22.5 < v < 27
2,15,15,18,30
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 < 6 or v > 27
2,15,15,18,30
Array
Multiply each value by each probability amount
We do this by multiplying each X_{i} x p_{i} to get a weighted score Y
Weighted Average =  
n 
Weighted Average =  
0 
Weighted Average =  
0 
Weighted Average =  0 
0 
Weighted Average = NAN
Show the freqency distribution table for this number set
2, 15, 15, 18, 30
We need to choose the smallest integer k such that 2^{k} ≥ n where n = 0
Therefore, we use 0 intervals
Our maximum value in our number set of 30  2 = 28
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size =  28 
0 
Add 1 to this giving us INF + 1 = INF
Class Limits  Class Boundaries  FD  CFD  RFD  CRFD 

0  100% 
Go through our 5 numbers
Determine the ratio of each number to the next one
2:15 → 0.1333
15:15 → 1
15:18 → 0.8333
18:30 → 0.6
Successive Ratio = 2:15,15:15,15:18,18:30 or 0.1333,1,0.8333,0.6