Step 1

The data for the urinary AGT level $\left( {\mu g} \right)$ denoted by $X$ for a sample of adults with chronic kidney disease is given below:

X = 2.6, 6.2, 7.4, 9.6, 11.5, 13.5, 14.5, 17.0, 20.0, 28.8, 29.5, 29.5, 41.7, 45.7, 56.2, 56.2, 66.1, 66.1, 67.6, 74.1, 97.7, 141.3, 147.9, 177.8, 186.2, 186.2, 190.6, 208.9, 229.1, 229.1, 288.4, 288.4, 346.7, 407.4, 426.6, 575.4, 616.6, 724.4, 812.8, 1122.0

The AGT level $X$ follows lognormal distribution with parameter $\mu \;and\;{\sigma ^2}$.

The total number of observations is $n = 40$.

 
Step 2

It is known that if $X$ has lognormal distribution with parameter $\mu \;and\;{\sigma ^2}$ then the logarithmic transformation of $X$ that is $\log \left( X \right)$ has a normal distribution with mean $\mu $ and variance ${\sigma ^2}$.

Let $Y = \log \left( X \right)$ and must have normal distribution with mean $\mu $ and variance ${\sigma ^2}$.

The sample mean and sample standard deviation is unbiased estimator of population mean and population standard deviation respectively.

The parameters are calculated using the table below:

X	Y=log (X)	${\left( {{Y_i} - \bar Y} \right)^2}$
2.6	0.956	12.072
6.2	1.825	6.788
7.4	2.001	5.898
9.6	2.262	4.701
11.5	2.442	3.951
13.5	2.603	3.339
14.5	2.674	3.083
17	2.833	2.550
20	2.996	2.057
28.8	3.360	1.144
29.5	3.384	1.093
29.5	3.384	1.093
41.7	3.731	0.489
45.7	3.822	0.370
56.2	4.029	0.161
56.2	4.029	0.161
66.1	4.191	0.057
66.1	4.191	0.057
67.6	4.214	0.047
74.1	4.305	0.016
97.7	4.582	0.023
141.3	4.951	0.271
147.9	4.997	0.321
177.8	5.181	0.563
186.2	5.227	0.635
186.2	5.227	0.635
190.6	5.250	0.673
208.9	5.342	0.831
229.1	5.434	1.008
229.1	5.434	1.008
288.4	5.664	1.524
288.4	5.664	1.524
346.7	5.848	2.012
407.4	6.010	2.496
426.6	6.056	2.643
575.4	6.355	3.706
616.6	6.424	3.977
724.4	6.585	4.646
812.8	6.700	5.155
1122	7.023	6.723
Total	177.187	89.501

The sample mean is calculated as follows:

\[\begin{array}{c} \bar y = \frac{1}{n}\sum\limits_{i = 1}^{40} {{y_i}} \\ = \frac{{177.187}}{{40}}\\ = 4.430 \end{array}\]

The sample variance is calculated as follows:

\[\begin{array}{c} {s^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^{40} {{{\left( {{y_i} - \bar y} \right)}^2}} \\ = \frac{1}{{40 - 1}}\left( {89.501} \right)\\ = 2.295 \end{array}\]

The sample standard deviation is calculated as follows:

\[\begin{array}{c} s = \sqrt {2.295} \\ = 1.515 \end{array}\]

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The estimated parameters of the distribution of ACT level are, $\hat \mu = 4.430\;and\;\hat \sigma = 1.515$.

 
Step 3

(b)

The estimated expected value of AGT level is calculated as follows:

\[\begin{array}{c} \hat E\left( X \right) = {e^{\hat \mu + \frac{{{{\hat \sigma }^2}}}{2}}}\\ = {e^{4.430 + \frac{{{{\left( {1.515} \right)}^2}}}{2}}}\\ = {e^{4.430 + 1.148}}\\ = 264.542\mu g \end{array}\]