EXERCISE A - Natural logarithm functionPart A:In a country, a disease affects the population with a probability of 0.05. There is a screening test for this disease. We consider a sample of $n$ people ($n \geqslant 20$) taken at random from the population, assimilated to a draw with replacement. The sample is tested using this method: the blood of these $n$ individuals is mixed, the mixture is tested. If the test is positive, an individual analysis of each person is performed. Let $X_n$ be the random variable that gives the number of analyses performed.
- Show that $X_n$ takes the values 1 and $(n+1)$.
- Prove that $P(X_n = 1) = 0.95^n$.
Establish the distribution of $X_n$ by copying on the answer sheet and completing the following table:
| $x_i$ | 1 | $n+1$ |
| $P(X_n = x_i)$ | | |
\setcounter{enumi}{2} - What does the expectation of $X_n$ represent in the context of the experiment?
Show that $E(X_n) = n + 1 - n \times 0.95^n$.
Part B: - Consider the function $f$ defined on $[20;+\infty[$ by $f(x) = \ln(x) + x\ln(0.95)$.
Show that $f$ is decreasing on $[20;+\infty[$.
\setcounter{enumi}{1} - We recall that $\lim_{x\rightarrow+\infty} \frac{\ln x}{x} = 0$. Show that $\lim_{x\rightarrow+\infty} f(x) = -\infty$.
- Show that $f(x) = 0$ has a unique solution $a$ on $[20;+\infty[$. Give an approximation to 0.1 of this solution.
- Deduce the sign of $f$ on $[20;+\infty[$.
Part C:We seek to compare two types of screening. The first method is described in Part A, the second, more classical, consists of testing all individuals. The first method makes it possible to reduce the number of analyses as soon as $E(X_n) < n$. Using Part B, show that the first method reduces the number of analyses for samples containing at most 87 people.