You entered a number set X of {180,230,150,170,230}
From the 5 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
150, 170, 180, 230, 230
Rank Ascending
150 is the 1st lowest/smallest number
170 is the 2nd lowest/smallest number
180 is the 3rd lowest/smallest number
230 is the 4th lowest/smallest number
230 is the 5th lowest/smallest number
230, 230, 180, 170, 150
Rank Descending
230 is the 1st highest/largest number
230 is the 2nd highest/largest number
180 is the 3rd highest/largest number
170 is the 4th highest/largest number
150 is the 5th highest/largest number
Sort our number set in ascending order
and assign a ranking to each number:
Number Set Value | 150 | 170 | 180 | 230 | 230 |
Rank | 1 | 2 | 3 | 4 | 5 |
Since we have 5 numbers in our original number set,
we assign ranks from lowest to highest (1 to 5)
Our original number set in unsorted order was 150,170,180,230,230
Our respective ranked data set is 1,2,3,5,5
Root Mean Square = | √A |
√N |
where A = x12 + x22 + x32 + x42 + x52 and N = 5 number set items
A = 1502 + 1702 + 1802 + 2302 + 2302
A = 22500 + 28900 + 32400 + 52900 + 52900
A = 189600
RMS = | √189600 |
√5 |
RMS = | 435.43082114154 |
2.2360679774998 |
RMS = 194.73058311421
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 = | 150 + 170 + 180 + 230 + 230 |
5 |
μ = | 960 |
5 |
μ = 192
Since our number set contains 5 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1,n2,n3,n4,n5)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(6)
Median Number = n3
Our median is entry 3 of our number set highlighted in red:
(150,170,180,230,230)
Median = 180
The highest frequency of occurence in our number set is 2 times
by the following numbers in green:
()
Our mode is denoted as: 230
Mode = 230
Harmonic Mean = | N |
1/x1 + 1/x2 + 1/x3 + 1/x4 + 1/x5 |
With N = 5 and each xi a member of the number set you entered, we have:
Harmonic Mean = | 5 |
1/150 + 1/170 + 1/180 + 1/230 + 1/230 |
Harmonic Mean = | 5 |
0.0066666666666667 + 0.0058823529411765 + 0.0055555555555556 + 0.0043478260869565 + 0.0043478260869565 |
Harmonic Mean = | 5 |
0.026800227337312 |
Harmonic Mean = 186.56558159262
Geometric Mean = (x1 * x2 * x3 * x4 * x5)1/N
Geometric Mean = (150 * 170 * 180 * 230 * 230)1/5
Geometric Mean = 2428110000000.2
Geometric Mean = 189.25774950314
Mid-Range = | Maximum Value in Number Set + Minimum Value in Number Set |
2 |
Mid-Range = | 230 + 150 |
2 |
Mid-Range = | 380 |
2 |
Mid-Range = 190
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
{230,230,180,170,150}
Stem | Leaf |
---|---|
2 | 30,30 |
1 | 50,70,80 |
Mean, Variance, Standard Deviation, Median, Mode
μ = | Sum of your number Set |
Total Numbers Entered |
μ = | ΣXi |
n |
μ = | 150 + 170 + 180 + 230 + 230 |
5 |
μ = | 960 |
5 |
μ = 192
Let's evaluate the square difference from the mean of each term (Xi - μ)2:
(X1 - μ)2 = (150 - 192)2 = -422 = 1764
(X2 - μ)2 = (170 - 192)2 = -222 = 484
(X3 - μ)2 = (180 - 192)2 = -122 = 144
(X4 - μ)2 = (230 - 192)2 = 382 = 1444
(X5 - μ)2 = (230 - 192)2 = 382 = 1444
ΣE(Xi - μ)2 = 1764 + 484 + 144 + 1444 + 1444
ΣE(Xi - μ)2 = 5280
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
| ||||||
Variance: σp2 = 1056 | Variance: σs2 = 1320 | ||||||||
Standard Deviation: σp = √σp2 = √1056 | Standard Deviation: σs = √σs2 = √1320 | ||||||||
Standard Deviation: σp = 32.4962 | Standard Deviation: σs = 36.3318 |
Population | Sample | ||||||||
---|---|---|---|---|---|---|---|---|---|
|
|
|
|
|
| ||||
SEM = 14.5327 | SEM = 16.2481 |
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 = (150 - 192)3 = -423 = -74088
(X2 - μ)3 = (170 - 192)3 = -223 = -10648
(X3 - μ)3 = (180 - 192)3 = -123 = -1728
(X4 - μ)3 = (230 - 192)3 = 383 = 54872
(X5 - μ)3 = (230 - 192)3 = 383 = 54872
ΣE(Xi - μ)3 = -74088 + -10648 + -1728 + 54872 + 54872
ΣE(Xi - μ)3 = 23280
Skewness = | E(Xi - μ)3 |
(n - 1)σ3 |
Skewness = | 23280 |
(5 - 1)32.49623 |
Skewness = | 23280 |
(4)34316.085157845 |
Skewness = | 23280 |
137264.34063138 |
Skewness = 0.1695997656268
AD = | Σ|Xi - μ| |
n |
Evaluate the absolute value of the difference from the mean
|Xi - μ|:
|X1 - μ| = |150 - 192| = |-42| = 42
|X2 - μ| = |170 - 192| = |-22| = 22
|X3 - μ| = |180 - 192| = |-12| = 12
|X4 - μ| = |230 - 192| = |38| = 38
|X5 - μ| = |230 - 192| = |38| = 38
Σ|Xi - μ| = 42 + 22 + 12 + 38 + 38
Σ|Xi - μ| = 152
Calculate average deviation (mean absolute deviation)
AD = | Σ|Xi - μ| |
n |
AD = | 152 |
5 |
Average Deviation = 30.4
Since our number set contains 5 elements which is an odd number,
our median number is determined as follows:
Number Set = (n1,n2,n3,n4,n5)
Median Number = Entry ½(n + 1)
Median Number = Entry ½(6)
Median Number = n3
Our median is entry 3 of our number set highlighted in red:
(150,170,180,230,230)
Median = 180
The highest frequency of occurence in our number set is 2 times
by the following numbers in green:
()
Our mode is denoted as: 230
Mode = 230
Range = Largest Number in the Number Set - Smallest Number in the Number Set
Range = 230 - 150
Range = 80
PSC1 = | μ - Mode |
σ |
PSC1 = | 3(192 - 230) |
32.4962 |
PSC1 = | 3 x -38 |
32.4962 |
PSC1 = | -114 |
32.4962 |
PSC1 = -3.5081
PSC2 = | μ - Median |
σ |
PSC1 = | 3(192 - 180) |
32.4962 |
PSC2 = | 3 x 12 |
32.4962 |
PSC2 = | 36 |
32.4962 |
PSC2 = 1.1078
Entropy = Ln(n)
Entropy = Ln(5)
Entropy = 1.6094379124341
Mid-Range = | Smallest Number in the Set + Largest Number in the Set |
2 |
Mid-Range = | 230 + 150 |
2 |
Mid-Range = | 380 |
2 |
Mid-Range = 190
We need to sort our number set from lowest to highest shown below:
{150,170,180,230,230}
V = | y(n + 1) |
100 |
V = | 75(5 + 1) |
100 |
V = | 75(6) |
100 |
V = | 450 |
100 |
V = 4 ← Rounded down to the nearest integer
Upper quartile (UQ) point = Point # 4 in the dataset which is 230
150,170,180,230,230V = | y(n + 1) |
100 |
V = | 25(5 + 1) |
100 |
V = | 25(6) |
100 |
V = | 150 |
100 |
V = 2 ← Rounded up to the nearest integer
Lower quartile (LQ) point = Point # 2 in the dataset which is 170
150,170,180,230,230
IQR = UQ - LQ
IQR = 230 - 170
IQR = 60
Lower Inner Fence (LIF) = LQ - 1.5 x IQR
Lower Inner Fence (LIF) = 170 - 1.5 x 60
Lower Inner Fence (LIF) = 170 - 90
Lower Inner Fence (LIF) = 80
Upper Inner Fence (UIF) = UQ + 1.5 x IQR
Upper Inner Fence (UIF) = 230 + 1.5 x 60
Upper Inner Fence (UIF) = 230 + 90
Upper Inner Fence (UIF) = 320
Lower Outer Fence (LOF) = LQ - 3 x IQR
Lower Outer Fence (LOF) = 170 - 3 x 60
Lower Outer Fence (LOF) = 170 - 180
Lower Outer Fence (LOF) = -10
Upper Outer Fence (UOF) = UQ + 3 x IQR
Upper Outer Fence (UOF) = 230 + 3 x 60
Upper Outer Fence (UOF) = 230 + 180
Upper Outer Fence (UOF) = 410
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 -10 < v < 80 and 320 < v < 410
150,170,180,230,230
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 < -10 or v > 410
150,170,180,230,230
150, 170, 180, 230, 230
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 + X4p4 + X5p5 |
n |
Weighted Average = | 150 x 0.2 + 170 x 0.4 + 180 x 0.6 + 230 x 0.8 + 230 x 0.9 |
5 |
Weighted Average = | 30 + 68 + 108 + 184 + 207 |
5 |
Weighted Average = | 597 |
5 |
Weighted Average = 119.4
Show the freqency distribution table for this number set
150, 170, 180, 230, 230
We need to choose the smallest integer k such that 2k ≥ n where n = 5
For k = 1, we have 21 = 2
For k = 2, we have 22 = 4
For k = 3, we have 23 = 8 ← Use this since it is greater than our n value of 5
Therefore, we use 3 intervals
Our maximum value in our number set of 230 - 150 = 80
Each interval size is the difference of the maximum and minimum value divided by the number of intervals
Interval Size = | 80 |
3 |
Add 1 to this giving us 26 + 1 = 27
Class Limits | Class Boundaries | FD | CFD | RFD | CRFD |
---|---|---|---|---|---|
150 - 177 | 149.5 - 177.5 | 2 | 2 | 2/5 = 40% | 2/5 = 40% |
177 - 204 | 176.5 - 204.5 | 1 | 2 + 1 = 3 | 1/5 = 20% | 3/5 = 60% |
204 - 231 | 203.5 - 231.5 | 2 | 2 + 1 + 2 = 5 | 2/5 = 40% | 5/5 = 100% |
5 | 100% |
Go through our 5 numbers
Determine the ratio of each number to the next one
150:170 → 0.8824
170:180 → 0.9444
180:230 → 0.7826
230:230 → 1
Successive Ratio = 150:170,170:180,180:230,230:230 or 0.8824,0.9444,0.7826,1