Calculate the standard deviation from confidence interval, sample size, and mean, when confidence interval is not symmetric around mean

Question

I have the following data:

• 300 customers were asked to participate in a survey, and each customer was able to submit his/her response within the first two weeks.
• The mean time to respond to the survey was 107 hours.

Using this data, a statistician was able to conclude that the mean time to respond to the survey for future customers would be between 96.12 and 120.65 hours.

Assuming that this is 95% confidence interval, how can we figure out the type of distribution and the standard deviation from this data?

Thanks in advance for your help!

PS: I tried to use t-statistic, but realized that the 95% confidence interval (96.12,120.65) is not symmetric by the mean (107). This means that the data is not normally distributed.

Answer 1

If the distribution of response times is $\mathsf{Exp}(\mathrm{rate}=1/107)$ then the probability a customer responds within two weeks (336 hrs) is $P(X_i < 336) = 0.9567,$ So it is seems possible most or all customers may have responded within two weeks. (Three weeks might be better.)

pexp(336, 1/107)
 0.9567253

Then $\frac{\bar X_{300}}{\mu} \sim \mathsf{Gamma}(300,300).$ So a 95% CI for $\mu$ is about $(96, 120).$

107/qgamma(c(.975,.025), 300,300)
[1]  95.85392 120.22054

In summary, it seems that the distribution of response times is $\mathsf{Exp}(1/107),$ with mean $\mu = \sigma = 107.$

I am aware that not everything matches perfectly, but I think I'm on the right track. [One source of slight discrepancies would come from using printed chi-squared tables (instead of gamma distributions in R) for probabilities involving the sample mean.]