Solution to Exercise 6.06.

  Unfortunately, the data for this problem is not on the CD in the back of the book. If I had realized that, I wouldn't have assigned it. I will have to check for that in the future.

(a) Anyway, after inputting the data into minitab, I took logarithms and then got the mean and standard deviation of those. The parameter estimates are

$$\hat{\mu} \; = \; 5.1017 , \quad \hat{\sigma} \; = \; 0.4961 .$$

(b) According to the formula on p. 181,

$$E[X] \; = \; \exp[ \mu + \sigma^2/2 ] .$$

Thus, our estimate of E[X] is

$$\exp[ 5.1017 + (0.4961^2)/2 ] \; = \; 185.8161$$

Interestingly, the sample mean (of the raw stream flow - not taking logarithms) is 185.4, which is almost exactly the same. In this case, one can show mathematically that the estimate based on the $\hat{\mu}$and the $\hat{\sigma}$ above is in fact more accurate than the sample mean as an estimate of the ``population'' mean E[X], provided the random sample from the log-normal model is correct. A normal probability plot of the logarithm of flow is shown below.