PDF of Triangular Distribution for a =3, b = 8.7, c = 2.9 & x = 5 Triangular distribution example problem workout with steps & calculation summary for the probabilty of random variable x = 5 lies between two points a = 3 & b = 8.7 and mid-point c = 2.9, along with mean, median, mode and variance estimation in statistical experiments.

Calculation Summary
a 3
b8.7
c2.9
x5
PDF 0.2238
Mean µ4.8667
Median 4.6343
Mode 2.9
Variance σ21.8372

Work with steps for a = 3, b = 8.7, c = 2.9 & x = 5

Question:
A random variable x = 5 which follows the triangular distribution with lower limit a = 3, upper limit b = 8.7 and height c = 2.9. Find the probability density function of x to fall between the interval or limits.
workout :
step 1 Address the formula input parameters & values
a = 3
b = 8.7
c = 2.9
x = 5

step 2 Find PDF value using a, b, c & x values
PDF = 2 (b - x)/(b - a)(b - c)
= 2 x (8.7 - 5)/(8.7 - 3)(8.7 - 2.9)
= 2 x 3.7/(5.7 x 5.8)
= 7.4/33.06
PDF = 0.2238

step 3 Find Mean using a, b & c values
Mean = a + b + c/2
= 3 + 8.7 + 2.9/2
= 14.6/3
Mean µ = 4.8667

step 4 Find Median value using a, b & c values
Median = b -(b - a)(b - c)/2
= 8.7 - (8.7 - 3)(8.7 - 2.9)/2
= 8.7 - (5.7)(5.8)/2
= 8.7 - 33.06/2
= 8.7 - √16.53
= 8.7 - 4.0657
Median = 4.6343

step 5 Find Mode value
Mode = c
Mode = 2.9

step 6 Find variance using a, b and c values
Variance σ2 = a² + b² + c² - ab - ac - bc/18
= (3)² + (8.7)² + (2.9)² - (3 x 8.7) - (3 x 2.9) - (8.7 x 2.9)/18
= 9 + 75.69 + 8.41 - 26.1 - 8.7 - 25.23/18
= 93.1 - 60.03/18
= 33.07/18
Variance σ² = 1.8372 