Step 1 :The problem is asking for the score that separates the bottom 5% of customers from the rest. This is a problem of finding a percentile in a normal distribution.
Step 2 :The z-score for the 5th percentile in a standard normal distribution is approximately -1.645. We can use this z-score, the given mean and standard deviation to find the corresponding score in the questionnaire distribution.
Step 3 :Given that the mean is 49.5 and the standard deviation is 16, we can use the formula for converting a z-score to a raw score in a distribution: \(X = \mu + Z \sigma\), where \(X\) is the raw score, \(\mu\) is the mean, \(Z\) is the z-score, and \(\sigma\) is the standard deviation.
Step 4 :Substituting the given values into the formula, we get \(X = 49.5 + (-1.645)(16)\).
Step 5 :Calculating the above expression, we find that \(X \approx 23.18\).
Step 6 :Rounding to one decimal place, we get \(X \approx 23.2\).
Step 7 :Final Answer: The questionnaire score that separates customers who are considered risk averse from those who are not is approximately \(\boxed{23.2}\).