You entered a number set X of {170,280,1300}
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
170, 280, 1300
Rank Ascending
170 is the 1st lowest/smallest number
280 is the 2nd lowest/smallest number
1300 is the 3rd lowest/smallest number
1300, 280, 170
Rank Descending
1300 is the 1st highest/largest number
280 is the 2nd highest/largest number
170 is the 3rd highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 170 | 280 | 1300 |
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 170,280,1300
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 = 1702 + 2802 + 13002
A = 28900 + 78400 + 1690000
A = 1797300
RMS = | √1797300 |
√3 |
RMS = | 1340.6341782903 |
1.7320508075689 |
RMS = 774.01550372069
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 = | 170 + 280 + 1300 |
3 |
μ = | 1750 |
3 |
μ = 583.33333333333
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:
(170,280,1300)
Median = 280
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 |
1/170 + 1/280 + 1/1300 |
Harmonic Mean = | 3 |
0.0058823529411765 + 0.0035714285714286 + 0.00076923076923077 |
Harmonic Mean = | 3 |
0.010223012281836 |
Harmonic Mean = 293.45558014543
Geometric Mean = (x1 * x2 * x3)1/N
Geometric Mean = (170 * 280 * 1300)1/3
Geometric Mean = 618800000.33333333333333
Geometric Mean = 395.53364820434
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 1300 + 170 |
2 |
Mid-Range = | 1470 |
2 |
Mid-Range = 735
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
{1300,280,170}
Stem | Leaf |
---|---|
1 | 70,300 |
2 | 80 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 170 + 280 + 1300 |
3 |
μ = | 1750 |
3 |
μ = 583.33333333333
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (170 - 583.33333333333)2 = -413.333333333332 = 170844.44444444
(X2 - μ)2 = (280 - 583.33333333333)2 = -303.333333333332 = 92011.111111111
(X3 - μ)2 = (1300 - 583.33333333333)2 = 716.666666666672 = 513611.11111111
ΣE(Xi - μ)2 = 170844.44444444 + 92011.111111111 + 513611.11111111
ΣE(Xi - μ)2 = 776466.66666667
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 258822.22222222 | Variance: σs2 = 388233.33333333 | ||||||||
Standard Deviation: σp = √σp2 = √258822.22222222 | Standard Deviation: σs = √σs2 = √388233.33333333 | ||||||||
Standard Deviation: σp = 508.7457 | Standard Deviation: σs = 623.0837 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 293.7245 | SEM = 359.7375 |
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 = (170 - 583.33333333333)3 = -413.333333333333 = -70615703.703704
(X2 - μ)3 = (280 - 583.33333333333)3 = -303.333333333333 = -27910037.037037
(X3 - μ)3 = (1300 - 583.33333333333)3 = 716.666666666673 = 368087962.96296
ΣE(Xi - μ)3 = -70615703.703704 + -27910037.037037 + 368087962.96296
ΣE(Xi - μ)3 = 269562222.22222
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 269562222.22222 |
(3 - 1)508.74573 |
Skewness = | 269562222.22222 |
(2)131674674.83744 |
Skewness = | 269562222.22222 |
263349349.67488 |
Skewness = 1.0235917520017
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |170 - 583.33333333333| = |-413.33333333333| = 413.33333333333
|X2 - μ| = |280 - 583.33333333333| = |-303.33333333333| = 303.33333333333
|X3 - μ| = |1300 - 583.33333333333| = |716.66666666667| = 716.66666666667
Σ|Xi - μ| = 413.33333333333 + 303.33333333333 + 716.66666666667
Σ|Xi - μ| = 1433.3333333333
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 1433.3333333333 |
3 |
Average Deviation = 477.77778
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:
(170,280,1300)
Median = 280
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 = 1300 - 170
Range = 1130
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(583.33333333333 - N/A) |
508.7457 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(583.33333333333 - 280) |
508.7457 |
PSC2 = | 3 x 303.33333333333 |
508.7457 |
PSC2 = | 910 |
508.7457 |
PSC2 = 1.7887
Entropy = Ln(n)
Entropy = Ln(3)
Entropy = 1.0986122886681
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 1300 + 170 |
2 |
Mid-Range = | 1470 |
2 |
Mid-Range = 735
We need to sort our number set from lowest to highest shown below:
{170,280,1300}
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 1300
170,280,1300V = | 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 170
170,280,1300
IQR = UQ - LQ
IQR = 1300 - 170
IQR = 1130
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 170 - 1.5 x 1130
Lower Inner Fence (LIF) = 170 - 1695
Lower Inner Fence (LIF) = -1525
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 1300 + 1.5 x 1130
Upper Inner Fence (UIF) = 1300 + 1695
Upper Inner Fence (UIF) = 2995
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 170 - 3 x 1130
Lower Outer Fence (LOF) = 170 - 3390
Lower Outer Fence (LOF) = -3220
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 1300 + 3 x 1130
Upper Outer Fence (UOF) = 1300 + 3390
Upper Outer Fence (UOF) = 4690
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 -3220 < v < -1525 and 2995 < v < 4690
170,280,1300
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 < -3220 or v > 4690
170,280,1300
170, 280, 1300
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 = | 170 x 0.2 + 280 x 0.4 + 1300 x 0.6 |
3 |
Weighted Average = | 34 + 112 + 780 |
3 |
Weighted Average = | 926 |
3 |
Weighted Average = 308.66666666667
Show the freqency distribution table for this number set
170, 280, 1300
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 1300 - 170 = 1130
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 1130 |
2 |
Add 1 to this giving us 565 + 1 = 566
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
170 - 736 | 169.5 - 736.5 | 2 | 2 | 2/3 = 66.67% | 2/3 = 66.67% |
736 - 1302 | 735.5 - 1302.5 | 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
170:280 → 0.6071
280:1300 → 0.2154
Successive Ratio = 170:280,280:1300 or 0.6071,0.2154