Applies to: @RISK 5.x and Up

Question:

Data set is as follows:

The mean was calculated in Excel by summing the product of each data value multiplied by its probability

• The result is 107.5.

When a fit is done on the data in @RISK:

• the Input column in the Fit tab shows the mean as 183.74.

What's going on?

Answer:

The 107.5 came from treating the 5 points as discrete outcomes with given probabilities, which is correct for discrete data.

When you fit in @RISK, it assumed a continuous distribution connecting your points linearly, effectively treating it as a RiskGeneral distribution.

Continuous distributions compute the mean from the area under the density curve, not just the listed points, which is why @RISK calculated a mean of 183.74.

Solution:

• If your 5 points are meant as discrete outcomes, choose “Discrete Sample Data (Counted Format)” in the fit dialog.

• Convert probabilities to counts (multiply by a common factor, e.g., 100) so @RISK sees integer counts.

• The new mean will match the manual calculation (107.5).

Elaboration:

The manual mean is correct for discrete probability, but the fit was treating the data as continuous, so it calculated a different mean.

Suppose your 5 discrete points are:

Value	Probability
0	0.10
50	0.20
100	0.40
200	0.20
500	0.10

---

Discrete mean (manual method)

[
\text{Mean} = \sum (\text{Value} \times \text{Probability})
= 0\cdot0.1 + 50\cdot0.2 + 100\cdot0.4 + 200\cdot0.2 + 500\cdot0.1
= 107.5
]

---

Continuous mean (linear interpolation, RiskGeneral)

• @RISK treats the 5 points as connected by straight lines, so density exists between points.

• The mean is the integral of (x \cdot f(x)) over the range 0–500.

• Result: 183.74

  • It’s higher than the discrete mean because the linear interpolation weights higher values more.

Takeaway:

• Discrete data → manual mean or Discrete Sample fit → 107.5

• Continuous fit → @RISK interpolates → 183.74