You entered a number set X of {1200,280,175}
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
175, 280, 1200
Rank Ascending
175 is the 1st lowest/smallest number
280 is the 2nd lowest/smallest number
1200 is the 3rd lowest/smallest number
1200, 280, 175
Rank Descending
1200 is the 1st highest/largest number
280 is the 2nd highest/largest number
175 is the 3rd highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 175 | 280 | 1200 |
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 175,280,1200
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 = 1752 + 2802 + 12002
A = 30625 + 78400 + 1440000
A = 1549025
RMS = | √1549025 |
√3 |
RMS = | 1244.5983287792 |
1.7320508075689 |
RMS = 718.56918015363
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 = | 175 + 280 + 1200 |
3 |
μ = | 1655 |
3 |
μ = 551.66666666667
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:
(175,280,1200)
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 |
1/175 + 1/280 + 1/1200 |
Harmonic Mean = | 3 |
0.0057142857142857 + 0.0035714285714286 + 0.00083333333333333 |
Harmonic Mean = | 3 |
0.010119047619048 |
Harmonic Mean = 296.47058823529
Geometric Mean = (x1 * x2 * x3)1/N
Geometric Mean = (175 * 280 * 1200)1/3
Geometric Mean = 588000000.33333333333333
Geometric Mean = 388.85925700148
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 1200 + 175 |
2 |
Mid-Range = | 1375 |
2 |
Mid-Range = 687.5
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
{1200,280,175}
Stem | Leaf |
---|---|
1 | 75,200 |
2 | 80 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 175 + 280 + 1200 |
3 |
μ = | 1655 |
3 |
μ = 551.66666666667
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (175 - 551.66666666667)2 = -376.666666666672 = 141877.77777778
(X2 - μ)2 = (280 - 551.66666666667)2 = -271.666666666672 = 73802.777777778
(X3 - μ)2 = (1200 - 551.66666666667)2 = 648.333333333332 = 420336.11111111
ΣE(Xi - μ)2 = 141877.77777778 + 73802.777777778 + 420336.11111111
ΣE(Xi - μ)2 = 636016.66666667
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 212005.55555556 | Variance: σs2 = 318008.33333333 | ||||||||
Standard Deviation: σp = √σp2 = √212005.55555556 | Standard Deviation: σs = √σs2 = √318008.33333333 | ||||||||
Standard Deviation: σp = 460.4406 | Standard Deviation: σs = 563.9223 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 265.8355 | SEM = 325.5807 |
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 = (175 - 551.66666666667)3 = -376.666666666673 = -53440629.62963
(X2 - μ)3 = (280 - 551.66666666667)3 = -271.666666666673 = -20049754.62963
(X3 - μ)3 = (1200 - 551.66666666667)3 = 648.333333333333 = 272517912.03704
ΣE(Xi - μ)3 = -53440629.62963 + -20049754.62963 + 272517912.03704
ΣE(Xi - μ)3 = 199027527.77778
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 199027527.77778 |
(3 - 1)460.44063 |
Skewness = | 199027527.77778 |
(2)97615960.86267 |
Skewness = | 199027527.77778 |
195231921.72534 |
Skewness = 1.0194415237984
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |175 - 551.66666666667| = |-376.66666666667| = 376.66666666667
|X2 - μ| = |280 - 551.66666666667| = |-271.66666666667| = 271.66666666667
|X3 - μ| = |1200 - 551.66666666667| = |648.33333333333| = 648.33333333333
Σ|Xi - μ| = 376.66666666667 + 271.66666666667 + 648.33333333333
Σ|Xi - μ| = 1296.6666666667
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 1296.6666666667 |
3 |
Average Deviation = 432.22222
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:
(175,280,1200)
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 = 1200 - 175
Range = 1025
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(551.66666666667 - N/A) |
460.4406 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(551.66666666667 - 280) |
460.4406 |
PSC2 = | 3 x 271.66666666667 |
460.4406 |
PSC2 = | 815 |
460.4406 |
PSC2 = 1.77
Entropy = Ln(n)
Entropy = Ln(3)
Entropy = 1.0986122886681
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 1200 + 175 |
2 |
Mid-Range = | 1375 |
2 |
Mid-Range = 687.5
We need to sort our number set from lowest to highest shown below:
{175,280,1200}
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 1200
175,280,1200V = | 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 175
175,280,1200
IQR = UQ - LQ
IQR = 1200 - 175
IQR = 1025
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 175 - 1.5 x 1025
Lower Inner Fence (LIF) = 175 - 1537.5
Lower Inner Fence (LIF) = -1362.5
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 1200 + 1.5 x 1025
Upper Inner Fence (UIF) = 1200 + 1537.5
Upper Inner Fence (UIF) = 2737.5
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 175 - 3 x 1025
Lower Outer Fence (LOF) = 175 - 3075
Lower Outer Fence (LOF) = -2900
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 1200 + 3 x 1025
Upper Outer Fence (UOF) = 1200 + 3075
Upper Outer Fence (UOF) = 4275
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 -2900 < v < -1362.5 and 2737.5 < v < 4275
175,280,1200
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 < -2900 or v > 4275
175,280,1200
175, 280, 1200
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 = | 175 x 0.2 + 280 x 0.4 + 1200 x 0.6 |
3 |
Weighted Average = | 35 + 112 + 720 |
3 |
Weighted Average = | 867 |
3 |
Weighted Average = 289
Show the freqency distribution table for this number set
175, 280, 1200
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 1200 - 175 = 1025
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 1025 |
2 |
Add 1 to this giving us 512 + 1 = 513
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
175 - 688 | 174.5 - 688.5 | 2 | 2 | 2/3 = 66.67% | 2/3 = 66.67% |
688 - 1201 | 687.5 - 1201.5 | 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
175:280 → 0.625
280:1200 → 0.2333
Successive Ratio = 175:280,280:1200 or 0.625,0.2333