You entered a number set X of {62}
From the 1 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
62
Rank Ascending
62 is the 1st lowest/smallest number
62
Rank Descending
62 is the 1st highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 62 |
Rank | 1 |
Since we have 1 numbers in our original number set,
we assign ranks from lowest to highest (1 to 1)
Our original number set in unsorted order was 62
Our respective ranked data set is 1
Root Mean Square = | √A |
√N |
where A = x12 and N = 1 number set items
A = 622
A = 3844
A = 3844
RMS = | √3844 |
√1 |
RMS = | 62 |
1 |
RMS = 62
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 = | 62 |
1 |
μ = | 62 |
1 |
μ = 62
Since our number set contains 1 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(2)
Median Number = n1
Our median is entry 1 of our number set highlighted in red:
(62)
Median = 62
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 |
With N = 1 and each xi a member of the number set you entered, we have:
Harmonic Mean = | 1 |
1/62 |
Harmonic Mean = | 1 |
0.016129032258065 |
Harmonic Mean = | 1 |
0.016129032258065 |
Harmonic Mean = 62
Geometric Mean = (x1)1/N
Geometric Mean = (62)1/1
Geometric Mean = 621
Geometric Mean = 62
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 62 + 62 |
2 |
Mid-Range = | 124 |
2 |
Mid-Range = 62
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
{62}
Stem | Leaf |
---|---|
6 | 2 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 62 |
1 |
μ = | 62 |
1 |
μ = 62
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (62 - 62)2 = 02 = 0
ΣE(Xi - μ)2 = 0
ΣE(Xi - μ)2 = 0
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 0 | Variance: σs2 = NAN | ||||||||
Standard Deviation: σp = √σp2 = √0 | Standard Deviation: σs = √σs2 = √NAN | ||||||||
Standard Deviation: σp = 0 | Standard Deviation: σs = NAN |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 0 | SEM = NAN |
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 = (62 - 62)3 = 03 = 0
ΣE(Xi - μ)3 = 0
ΣE(Xi - μ)3 = 0
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 0 |
(1 - 1)03 |
Skewness = | 0 |
(0)0 |
Skewness = | 0 |
0 |
Skewness = NAN
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |62 - 62| = |0| = 0
Σ|Xi - μ| = 0
Σ|Xi - μ| = 0
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 0 |
1 |
Average Deviation = 0
Since our number set contains 1 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(2)
Median Number = n1
Our median is entry 1 of our number set highlighted in red:
(62)
Median = 62
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 = 62 - 62
Range = 0
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(62 - N/A) |
0 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(62 - 62) |
0 |
PSC2 = | 3 x 0 |
0 |
PSC2 = | 0 |
0 |
PSC2 = NAN
Entropy = Ln(n)
Entropy = Ln(1)
Entropy = 0
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 62 + 62 |
2 |
Mid-Range = | 124 |
2 |
Mid-Range = 62
We need to sort our number set from lowest to highest shown below:
{62}
V = | y(n + 1) |
100 |
V = | 75(1 + 1) |
100 |
V = | 75(2) |
100 |
V = | 150 |
100 |
V = 1 ← Rounded down to the nearest integer
Upper quartile (UQ) point = Point # 1 in the dataset which is 62
62V = | y(n + 1) |
100 |
V = | 25(1 + 1) |
100 |
V = | 25(2) |
100 |
V = | 50 |
100 |
V = 1 ← Rounded up to the nearest integer
Lower quartile (LQ) point = Point # 1 in the dataset which is 62
62
IQR = UQ - LQ
IQR = 62 - 62
IQR = 0
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 62 - 1.5 x 0
Lower Inner Fence (LIF) = 62 - 0
Lower Inner Fence (LIF) = 62
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 62 + 1.5 x 0
Upper Inner Fence (UIF) = 62 + 0
Upper Inner Fence (UIF) = 62
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 62 - 3 x 0
Lower Outer Fence (LOF) = 62 - 0
Lower Outer Fence (LOF) = 62
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 62 + 3 x 0
Upper Outer Fence (UOF) = 62 + 0
Upper Outer Fence (UOF) = 62
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 62 < v < 62 and 62 < v < 62
62
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 < 62 or v > 62
62
62
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 |
n |
Weighted Average = | 62 x |
1 |
Weighted Average = | 0 |
1 |
Weighted Average = | 0 |
1 |
Weighted Average = 0
Show the freqency distribution table for this number set
62
We need to choose the smallest integer k such that 2k ≥ n where n = 1
Therefore, we use 0 intervals
Our maximum value in our number set of 62 - 62 = 0
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 0 |
0 |
Add 1 to this giving us NAN + 1 = NAN
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
1 | 100% |
Go through our 1 numbers
Determine the ratio of each number to the next one
Successive Ratio = or