PDF of Triangular Distribution for a =4.5, b = 7.2, c = 5.5 & x = 12
Triangular distribution example problem workout with steps & calculation summary for the probabilty of random variable x = 12 lies between two points a = 4.5 & b = 7.2 and mid-point c = 5.5, along with mean, median, mode and variance estimation in statistical experiments.
| Calculation Summary | |
|---|---|
| a | 4.5 |
| b | 7.2 |
| c | 5.5 |
| x | 12 |
| 0 | |
| Mean µ | 5.7333 |
| Median | 5.6851 |
| Mode | 5.5 |
| Variance σ2 | 0.3106 |
Work with steps for a = 4.5, b = 7.2, c = 5.5 & x = 12
Question:
Find the probability density function P(x) for random variable x = 12 which follows Triangular distribution having the lower limit a = 4.5, upper limit b = 7.2 and height c = 5.5
workout :Find the probability density function P(x) for random variable x = 12 which follows Triangular distribution having the lower limit a = 4.5, upper limit b = 7.2 and height c = 5.5
step 1 Address the formula input parameters & values
a = 4.5
b = 7.2
c = 5.5
x = 12
step 2Find PDF value
x > b value
PDF = 0
step 3 Find Mean using a, b & c values
Mean = a + b + c/2
= 4.5 + 7.2 + 5.5/2
= 17.2/3
Mean µ = 5.7333
step 4 Find Median value using a, b & c values
Median = b -√(b - a)(b - c)/2
= 7.2 - √(7.2 - 4.5)(7.2 - 5.5)/2
= 7.2 - √(2.7)(1.7)/2
= 7.2 - √4.59/2
= 7.2 - √2.295
= 7.2 - 1.5149
Median = 5.6851
step 5 Find Mode value
Mode = c
Mode = 5.5
step 6 Find variance using a, b and c values
Variance σ2 = a² + b² + c² - ab - ac - bc/18
= (4.5)² + (7.2)² + (5.5)² - (4.5 x 7.2) - (4.5 x 5.5) - (7.2 x 5.5)/18
= 20.25 + 51.84 + 30.25 - 32.4 - 24.75 - 39.6/18
= 102.34 - 96.75/18
= 5.59/18
Variance σ² = 0.3106
