Calculators & Converters

    PDF of Triangular Distribution for a =3, b = 8.7, c = 2.9 & x = 5

    Triangular Distribution Calculator

    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

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