The Quest for a Five-star Trustpilot Score

Apparently, Trustpilot starts everyone off with seven hidden 3.5-star reviews. That way, someone getting just one review of five stars doesn’t give them a higher score than someone with 10 five-star reviews and one four-star review (the second case would be 4.9 stars). Trustpilot rounds the average score to one decimal place. Trustpilot explains this on their website (see the image of the screenshot)

Information from Trustpilot about how a TrustScore is calculated.

In order to get an average of 4.95 stars (which is the smallest number that rounds to 5.0 to one decimal place) one would require 203 five-star reviews (and no other reviews)!

The mathematics

The total number of reviews used in the calculation of the mean rating would be 7 + N, so the mean rating would be given by:

4.95=7×3.5+N×5.07+N4.95 = \frac{7 \cross 3.5 + N \cross 5.0}{7 + N}

Where N is the number of five-star reviews. Rearranged to make N the subject in stages. First, multiple by 7 + N:

4.95×(7+N)=7×3.5+N×5.04.95 \cross \left(7 + N \right) = 7 \cross 3.5 + N \cross 5.0

Then, multiply out the parentheses on the left:

4.95×7+4.95×N=7×3.5+N×5.04.95 \cross 7 + 4.95 \cross N = 7 \cross 3.5 + N \cross 5.0

Then, gather all terms with N on one side, and all those without N on the other:

4.95×77×3.5=N×5.04.95×N4.95 \cross 7 – 7 \cross 3.5 = N \cross 5.0 – 4.95 \cross N

Then, factor out the N:

4.95×77×3.5=N×(5.04.95)4.95 \cross 7 – 7 \cross 3.5 = N \cross \left(5.0 – 4.95 \right)

Then, divide both sides by 5.0 − 4.95:

N=4.95×77×3.55.04.95=203N = \frac{4.95 \times 7 – 7 \times 3.5}{5.0 – 4.95} = 203

I currently have ten five-star reviews (and no other reviews except the seven hidden 3.5-star reviews), giving me an average of 4.38 stars (rounded to 4.4).

100% five-star reviews,
giving an average score of 4.4 stars.

Physics With Keith's trustpilot score, 100% 5-star!

I’d be so grateful if 193 other folk could be kind enough to review what I offer! I would really help push Physics With Keith up the rankings, which would make it easier for more people to find me, which would mean I could help even more people!

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