The article "Assessment of Dermopharmacokinetic Approach in the Bioequivalence Determination of Topical Tretinoin Gel Products" (L. Pershing, J. Nelson, et al„ J Am Acad Dermatol 2003:740‒751) reports that the amount of a certain antifungal ointment that is absorbed into the skin can be modeled with a lognormal distribution. Assume that the amount (in ng/cm2) of active ingredient in the skin two hours after application is log normally distributed with μ = 2.2 and σ = 2.1.

a. Find the mean amount absorbed.

b. Find the median amount absorbed.

c. Find the probability that the amount absorbed is more than 100 ng/cm2.

d. Find the probability that the amount absorbed is less than 50 ng/cm2.

e. Find the 80th percentile of the amount absorbed.

f. Find the standard deviation of the amount absorbed.

Solution 2E

Step1 of 7:

Let us consider a random variable X it presents the amount of active ingredient absorbed.

And X follows lognormal distribution with mean and standard deviation .

That is XN(2.2, 2.12)

Here our goal is:

a).We need to find the mean amount absorbed.

b).We need to find the median amount absorbed.

c).We need to find the probability that the amount absorbed is more than 100 ng/cm2.

d).We need to find the probability that the amount absorbed is less than 50 ng/cm2.

e).We need to find the 80th percentile of the amount absorbed.

f).We need to find the standard deviation of the amount absorbed.

Step2 of 7:

a).

The probability density function of lognormal distribution is given by

Where,

mean

standard deviation

Z = standard normal variable.

Now, the mean amount absorbed is given by

=

=

=

= 81.8591

Hence, E(X) = 81.8591

Step3 of 7:

b).

Here we need to find median of an amount of active ingredient absorbed. Let us consider “m” be the median lifetime of an electric component. And we know that median divides the data into two parts that is 0.5.

Now, median of the amount of active ingredient absorbed is given by

We know that X N(2.2, 2.12)

Take

Hence, m = 9.0250.

Step4 of 7:

c).

The probability that the amount absorbed is more than 100 ng/cm2 is given by

=

=

=

=

=

=

Here is obtained from statistical table(area under normal curve)

(In a standard normal table we have to see in row 1.1 under column 0.04)

= 1 - 0.8729

= 0.1271

Hence, =0.1271.

Step5 of 7:

d).

The probability that the amount absorbed is less than 50 ng/cm2 is given by

Here is obtained from statistical table(area under normal curve)

(In a standard normal table we have to see in row 0.8 under column 0.01)

= 0.7910

Hence, =0.7910.

Step6 of 7:

e).

Let us consider is a 80th percentile of the amount of active ingredient absorbed.

That is P(X) = 0.80

Now,

Here

= 0.84

is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.80 value falls, it falls in row 0.8 under column 0.04.

Hence,

80th percentile of the amount of active ingredient absorbed is given by

=

= 3.964

= 52.66

Therefore, 80th percentile of the amount of active ingredient absorbed is 52.66.

Step6 of 7:

f).

The standard deviation of the amount absorbed is given by

= 737.9516

Hence, =737.9516.