You entered a number set X of {6,24}
6, 24
Rank Ascending
6 is the 1st lowest/smallest number
24 is the 2nd lowest/smallest number
24, 6
Rank Descending
24 is the 1st highest/largest number
6 is the 2nd highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 6 | 24 |
Rank | 1 | 2 |
Since we have 2 numbers in our original number set,
we assign ranks from lowest to highest (1 to 2)
Our original number set in unsorted order was 6,24
Our respective ranked data set is 1,2
Root Mean Square = | √A |
√N |
where A = x12 + x22 and N = 2 number set items
A = 62 + 242
A = 36 + 576
A = 612
RMS = | √612 |
√2 |
RMS = | 24.738633753706 |
1.4142135623731 |
RMS = 17.492855684536
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 = | 6 + 24 |
2 |
μ = | 30 |
2 |
μ = 15
Since our number set contains 2 elements which is an even number,
our median number is determined as follows
Number Set = (n1,n2)
Median Number 1 = ½(n)
Median Number 1 = ½(2)
Median Number 1 = Number Set Entry 1
Median Number 2 = Median Number 1 + 1
Median Number 2 = Number Set Entry 1 + 1
Median Number 2 = Number Set Entry 2
Median = ½(n1 + n2)
Our median is the average of entry 1 and entry 2 of our number set highlighted in red:
(6,24)
Median = ½(6 + 24)
Median = ½(30)
Median = 15
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 |
With N = 2 and each xi a member of the number set you entered, we have:
Harmonic Mean = | 2 |
1/6 + 1/24 |
Harmonic Mean = | 2 |
0.16666666666667 + 0.041666666666667 |
Harmonic Mean = | 2 |
0.20833333333333 |
Harmonic Mean = 9.6
Geometric Mean = (x1 * x2)1/N
Geometric Mean = (6 * 24)1/2
Geometric Mean = 1440.5
Geometric Mean = 12
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 24 + 6 |
2 |
Mid-Range = | 30 |
2 |
Mid-Range = 15
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
{24,6}
Stem | Leaf |
---|---|
2 | 4 |
6 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 6 + 24 |
2 |
μ = | 30 |
2 |
μ = 15
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (6 - 15)2 = -92 = 81
(X2 - μ)2 = (24 - 15)2 = 92 = 81
ΣE(Xi - μ)2 = 81 + 81
ΣE(Xi - μ)2 = 162
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 81 | Variance: σs2 = 162 | ||||||||
Standard Deviation: σp = √σp2 = √81 | Standard Deviation: σs = √σs2 = √162 | ||||||||
Standard Deviation: σp = 9 | Standard Deviation: σs = 12.7279 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 6.364 | SEM = 9 |
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 = (6 - 15)3 = -93 = -729
(X2 - μ)3 = (24 - 15)3 = 93 = 729
ΣE(Xi - μ)3 = -729 + 729
ΣE(Xi - μ)3 = 0
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 0 |
(2 - 1)93 |
Skewness = | 0 |
(1)729 |
Skewness = | 0 |
729 |
Skewness = 0
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |6 - 15| = |-9| = 9
|X2 - μ| = |24 - 15| = |9| = 9
Σ|Xi - μ| = 9 + 9
Σ|Xi - μ| = 18
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 18 |
2 |
Average Deviation = 9
Since our number set contains 2 elements which is an even number,
our median number is determined as follows
Number Set = (n1,n2)
Median Number 1 = ½(n)
Median Number 1 = ½(2)
Median Number 1 = Number Set Entry 1
Median Number 2 = Median Number 1 + 1
Median Number 2 = Number Set Entry 1 + 1
Median Number 2 = Number Set Entry 2
Median = ½(n1 + n2)
Our median is the average of entry 1 and entry 2 of our number set highlighted in red:
(6,24)
Median = ½(6 + 24)
Median = ½(30)
Median = 15
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 = 24 - 6
Range = 18
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(15 - N/A) |
9 |
Since no mode exists, we do not have a Pearsons Skewness Coefficient 1
PSC2 = | μ - Median |
σ |
PSC1 = | 3(15 - 15) |
9 |
PSC2 = | 3 x 0 |
9 |
PSC2 = | 0 |
9 |
PSC2 = 0
Entropy = Ln(n)
Entropy = Ln(2)
Entropy = 0.69314718055995
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 24 + 6 |
2 |
Mid-Range = | 30 |
2 |
Mid-Range = 15
We need to sort our number set from lowest to highest shown below:
{6,24}
V = | y(n + 1) |
100 |
V = | 75(2 + 1) |
100 |
V = | 75(3) |
100 |
V = | 225 |
100 |
V = 2 ← Rounded down to the nearest integer
Upper quartile (UQ) point = Point # 2 in the dataset which is 24
6,24V = | y(n + 1) |
100 |
V = | 25(2 + 1) |
100 |
V = | 25(3) |
100 |
V = | 75 |
100 |
V = 1 ← Rounded up to the nearest integer
Lower quartile (LQ) point = Point # 1 in the dataset which is 6
6,24
IQR = UQ - LQ
IQR = 24 - 6
IQR = 18
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 6 - 1.5 x 18
Lower Inner Fence (LIF) = 6 - 27
Lower Inner Fence (LIF) = -21
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 24 + 1.5 x 18
Upper Inner Fence (UIF) = 24 + 27
Upper Inner Fence (UIF) = 51
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 6 - 3 x 18
Lower Outer Fence (LOF) = 6 - 54
Lower Outer Fence (LOF) = -48
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 24 + 3 x 18
Upper Outer Fence (UOF) = 24 + 54
Upper Outer Fence (UOF) = 78
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 -48 < v < -21 and 51 < v < 78
6,24
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 < -48 or v > 78
6,24
6, 24
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 |
n |
Weighted Average = | 6 x 0.2 + 24 x 0.4 |
2 |
Weighted Average = | 1.2 + 9.6 |
2 |
Weighted Average = | 10.8 |
2 |
Weighted Average = 5.4
Show the freqency distribution table for this number set
6, 24
We need to choose the smallest integer k such that 2k ≥ n where n = 2
For k = 1, we have 21 = 2 ← Use this since it is greater than our n value of 2
Therefore, we use 1 intervals
Our maximum value in our number set of 24 - 6 = 18
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 18 |
1 |
Add 1 to this giving us 18 + 1 = 19
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
6 - 25 | 5.5 - 25.5 | 2 | 2 | 2/2 = 100% | 2/2 = 100% |
2 | 100% |
Go through our 2 numbers
Determine the ratio of each number to the next one
6:24 → 0.25
Successive Ratio = 6:24 or 0.25