You entered a number set X of {300,45,20}
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
20, 45, 300
Rank Ascending
20 is the 1st lowest/smallest number
45 is the 2nd lowest/smallest number
300 is the 3rd lowest/smallest number
300, 45, 20
Rank Descending
300 is the 1st highest/largest number
45 is the 2nd highest/largest number
20 is the 3rd highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 20 | 45 | 300 |
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 20,45,300
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 = 202 + 452 + 3002
A = 400 + 2025 + 90000
A = 92425
RMS = | √92425 |
√3 |
RMS = | 304.01480227121 |
1.7320508075689 |
RMS = 175.52302792891
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 = | 20 + 45 + 300 |
3 |
μ = | 365 |
3 |
μ = 121.66666666667
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:
(20,45,300)
Median = 45
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 |
1/20 + 1/45 + 1/300 |
Harmonic Mean = | 3 |
0.05 + 0.022222222222222 + 0.0033333333333333 |
Harmonic Mean = | 3 |
0.075555555555556 |
Harmonic Mean = 39.705882352941
Geometric Mean = (x1 * x2 * x3)1/N
Geometric Mean = (20 * 45 * 300)1/3
Geometric Mean = 2700000.33333333333333
Geometric Mean = 64.633040700956
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 300 + 20 |
2 |
Mid-Range = | 320 |
2 |
Mid-Range = 160
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
{300,45,20}
Stem | Leaf |
---|---|
3 | 00 |
4 | 5 |
2 | 0 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 20 + 45 + 300 |
3 |
μ = | 365 |
3 |
μ = 121.66666666667
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (20 - 121.66666666667)2 = -101.666666666672 = 10336.111111111
(X2 - μ)2 = (45 - 121.66666666667)2 = -76.6666666666672 = 5877.7777777778
(X3 - μ)2 = (300 - 121.66666666667)2 = 178.333333333332 = 31802.777777778
ΣE(Xi - μ)2 = 10336.111111111 + 5877.7777777778 + 31802.777777778
ΣE(Xi - μ)2 = 48016.666666667
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 16005.555555556 | Variance: σs2 = 24008.333333333 | ||||||||
Standard Deviation: σp = √σp2 = √16005.555555556 | Standard Deviation: σs = √σs2 = √24008.333333333 | ||||||||
Standard Deviation: σp = 126.5131 | Standard Deviation: σs = 154.9462 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 73.0424 | SEM = 89.4582 |
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 = (20 - 121.66666666667)3 = -101.666666666673 = -1050837.962963
(X2 - μ)3 = (45 - 121.66666666667)3 = -76.6666666666673 = -450629.62962963
(X3 - μ)3 = (300 - 121.66666666667)3 = 178.333333333333 = 5671495.3703704
ΣE(Xi - μ)3 = -1050837.962963 + -450629.62962963 + 5671495.3703704
ΣE(Xi - μ)3 = 4170027.7777778
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 4170027.7777778 |
(3 - 1)126.51313 |
Skewness = | 4170027.7777778 |
(2)2024913.5785532 |
Skewness = | 4170027.7777778 |
4049827.1571065 |
Skewness = 1.0296804322773
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |20 - 121.66666666667| = |-101.66666666667| = 101.66666666667
|X2 - μ| = |45 - 121.66666666667| = |-76.666666666667| = 76.666666666667
|X3 - μ| = |300 - 121.66666666667| = |178.33333333333| = 178.33333333333
Σ|Xi - μ| = 101.66666666667 + 76.666666666667 + 178.33333333333
Σ|Xi - μ| = 356.66666666667
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 356.66666666667 |
3 |
Average Deviation = 118.88889
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:
(20,45,300)
Median = 45
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 = 300 - 20
Range = 280
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(121.66666666667 - N/A) |
126.5131 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(121.66666666667 - 45) |
126.5131 |
PSC2 = | 3 x 76.666666666667 |
126.5131 |
PSC2 = | 230 |
126.5131 |
PSC2 = 1.818
Entropy = Ln(n)
Entropy = Ln(3)
Entropy = 1.0986122886681
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 300 + 20 |
2 |
Mid-Range = | 320 |
2 |
Mid-Range = 160
We need to sort our number set from lowest to highest shown below:
{20,45,300}
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 300
20,45,300V = | 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 20
20,45,300
IQR = UQ - LQ
IQR = 300 - 20
IQR = 280
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 20 - 1.5 x 280
Lower Inner Fence (LIF) = 20 - 420
Lower Inner Fence (LIF) = -400
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 300 + 1.5 x 280
Upper Inner Fence (UIF) = 300 + 420
Upper Inner Fence (UIF) = 720
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 20 - 3 x 280
Lower Outer Fence (LOF) = 20 - 840
Lower Outer Fence (LOF) = -820
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 300 + 3 x 280
Upper Outer Fence (UOF) = 300 + 840
Upper Outer Fence (UOF) = 1140
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 -820 < v < -400 and 720 < v < 1140
20,45,300
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 < -820 or v > 1140
20,45,300
20, 45, 300
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 = | 20 x 0.2 + 45 x 0.4 + 300 x 0.6 |
3 |
Weighted Average = | 4 + 18 + 180 |
3 |
Weighted Average = | 202 |
3 |
Weighted Average = 67.333333333333
Show the freqency distribution table for this number set
20, 45, 300
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 300 - 20 = 280
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 280 |
2 |
Add 1 to this giving us 140 + 1 = 141
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
20 - 161 | 19.5 - 161.5 | 2 | 2 | 2/3 = 66.67% | 2/3 = 66.67% |
161 - 302 | 160.5 - 302.5 | 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
20:45 → 0.4444
45:300 → 0.15
Successive Ratio = 20:45,45:300 or 0.4444,0.15