# Probability

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• #53002

Woferia
Member

I have annual customer satisfaction scores from the past 11 years. The average of those scores is 75.7, the standard dev is 5.1, and I would like to know how to identify the probability of achieving a score of 80% on our next survey.
Appreciate any guidance that’s out there – not too technical, please.  Just need to be pointed in the right direction.

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#187371

Mundorff
Member

You do not say how the customer satisfaction measure is expressed.  Is it on a scale of 1 to 100?  Are there any trends over the past 11 years, or only random variation around the grand mean of 75.7?  Has there been or will there be any intervention or improvement project prior to the next survey (the most crucial element)?
Customer satisfaction scores, despite their possibly being expressed as a quasi-continuous scale, are by their nature ordinal data.  They are entirely subjective: there is no specification which mandates what deserves a rating of 80, nor any definition of what constitutes a one-point difference (e.g., 74 and 75).  So (1) the answer is essentially unknowable; and (2) the calculations are not very relevant to determining real-world customer service satisfaction.
Better to use something like quartiles/deciles to determine the distribution of scores.  You want to know who is really happy, and who is really unhappy, not the great gray middle.  Conventional wisdom (supported by much empirical experience) says that every dissatisfied customer will tell four or five times as many people about their bad experience than satisfied customers will tell people about their good experience.

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#187517

Woferia
Member

Thank you for your response.  It validated what I was beginning to think anyway.

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#187520

Mundorff
Member

Our satisfaction surveys use a 5-point or 7-point scale.  What we generally do — in addition to a frequency distribution by response — is transform these data into a binary variable where (for example) if 5 is “Excellent”, use a p-chart to graph the percentage of 5/Excellent over time.  Same for 1 = “Poor”.  This also works with dichotomous Likert-type (agree/disagree) survey results.

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