Let $n$ be a natural number, $p$ a real number between 0 and 1, and $X_n$ a random variable following a binomial distribution with parameters $n$ and $p$. We denote $F_n = \frac{X_n}{n}$ and $f$ a value taken by $F_n$. We recall that, for $n$ sufficiently large, the interval $\left[p - \frac{1}{\sqrt{n}} ; p + \frac{1}{\sqrt{n}}\right]$ contains the frequency $f$ with probability at least equal to 0.95.
Deduce that the interval $\left[f - \frac{1}{\sqrt{n}} ; f + \frac{1}{\sqrt{n}}\right]$ contains $p$ with probability at least equal to 0.95.

We seek to study the number of students knowing the meaning of the acronym URSSAF. For this, we survey them by proposing a multiple choice questionnaire. Each student must choose from three possible answers, denoted $A$, $B$ and $C$, the correct answer being $A$. We denote by $r$ the probability that a student knows the correct answer. Any student knowing the correct answer responds $A$, otherwise they respond at random (equiprobably).

1. We survey a student at random. We denote: $A$ the event ``the student responds $A$'', $B$ the event ``the student responds $B$'', $C$ the event ``the student responds $C$'', $R$ the event ``the student knows the answer'', $\bar{R}$ the complementary event of $R$. a. Translate this situation using a probability tree. b. Show that the probability of event $A$ is $P(A) = \frac{1}{3}(1 + 2r)$. c. Express as a function of $r$ the probability that a person who chose $A$ knows the correct answer.
2. To estimate $r$, we survey 400 people and denote by $X$ the random variable counting the number of correct answers. We will assume that surveying 400 students at random is equivalent to performing sampling with replacement of 400 students from the set of all students. a. Give the distribution of $X$ and its parameters $n$ and $p$ as a function of $r$. b. In an initial survey, we observe that 240 students respond $A$, out of 400 surveyed. Give a 95\% confidence interval for the estimate of $p$. Deduce a 95\% confidence interval for $r$. c. In what follows, we assume that $r = 0.4$. Given the large number of students, we will consider that $X$ follows a normal distribution. i. Give the parameters of this normal distribution. ii. Give an approximate value of $P(X \leqslant 250)$ to $10^{-2}$ precision.

The store manager wishes to estimate the proportion of defective padlocks in his stock of budget padlocks. For this, he takes a random sample of 500 budget padlocks, among which 39 prove to be defective.
Give a confidence interval for this proportion at the $95\%$ confidence level.

An institute conducts a survey to determine, in a given population, the proportion of people who are in favour of a territorial development project. A random sample of people from this population is interviewed, and one question is asked to each person.
Part A: Number of people who agree to answer the survey
We admit in this part that the probability that a person interviewed agrees to answer the question is equal to 0.6.

1. The survey institute interviews 700 people. We denote by $X$ the random variable corresponding to the number of people interviewed who agree to answer the question asked. a. What is the distribution of the random variable $X$? Justify the answer. b. What is the best approximation of $P(X \geqslant 400)$ among the following numbers? 0.92 0.93 0.94 0.95.
   
2. How many people must the institute interview at minimum to guarantee, with a probability greater than 0.9, that the number of people answering the survey is greater than or equal to 400?

Part B: Proportion of people in favour of the project in the population
In this part, we assume that $n$ people have answered the question, and we admit that these people constitute a random sample of size $n$ (where $n$ is a natural number greater than 50). Among these people, 29\% are in favour of the development project.

1. Give a confidence interval, at the 95\% confidence level, for the proportion of people who are in favour of the project in the total population.
   
2. Determine the minimum value of the integer $n$ so that the confidence interval, at the 95\% confidence level, has an amplitude less than or equal to 0.04.

Part C: Correction due to insincere responses
In this part, we assume that, among the surveyed people who agreed to answer the question asked, 29\% claim that they are in favour of the project. The survey institute also knows that some people are not sincere and answer the opposite of their true opinion. Based on experience, the institute estimates at 15\% the rate of insincere responses among the people who responded, and admits that this rate is the same regardless of the opinion of the person interviewed.
A file of a person who responded is randomly selected, and we define:

• $F$ the event ``the person is actually in favour of the project'';
• $\bar{F}$ the event ``the person is actually opposed to the project'';
• $A$ the event ``the person claims that they are in favour of the project'';
• $\bar{A}$ the event ``the person claims that they are opposed to the project''.

Thus, according to the data, we have $p(A) = 0.29$.

1. By interpreting the data in the statement, indicate the values of $P_F(A)$ and $P_{\bar{F}}(A)$.
   
2. We set $x = P(F)$. a. Reproduce on your paper and complete the probability tree. b. Deduce an equality satisfied by $x$.
   
3. Determine, among the people who responded to the survey, the proportion of those who are actually in favour of the project.

The company sought to improve the quality of its production and claims that there are no more than $6\%$ defective light bulbs in its production. A consumer association conducts a test on a sample and obtains 71 defective light bulbs out of 1000.

1. In the case where there would be exactly $6\%$ defective light bulbs, determine an asymptotic confidence interval at the $95\%$ level for the frequency of defective light bulbs in a random sample of size 1000.
2. Do we have reasons to question the company's claim?

Improvements made to line A had the effect of increasing the proportion $p$ of defect-free components. To estimate this proportion, a random sample of 400 components manufactured by line A is taken. In this sample, the observed frequency of defect-free components is 0.92.

1. Determine a confidence interval for the proportion $p$ at the $95\%$ confidence level.
2. What should be the minimum sample size so that such a confidence interval has a maximum amplitude of 0.02?

We denote by $p$ the unknown proportion of young people aged 16 to 24 years who practice illegal downloading on the internet at least once a week.
A young person participating in protocol $( \mathscr { P } )$ is randomly selected. The protocol $( \mathscr { P } )$ is as follows: each young person rolls a fair 6-sided die; if the result is even, the young person answers sincerely; if the result is ``1'', the young person must answer ``Yes''; if the result is ``3 or 5'', the young person must answer ``No''.
We denote: $R$ the event ``the result of the roll is even'', $O$ the event ``the young person answered Yes''.

1. Probability calculations

Reproduce and complete the weighted tree diagram.
Deduce that the probability $q$ of the event ``the young person answered Yes'' is: $$q = \frac { 1 } { 2 } p + \frac { 1 } { 6 }$$

2. Confidence interval

a. At the request of Hadopi, a polling institute conducts a survey according to protocol $( \mathscr { P } )$. On a sample of size 1500, it counts 625 ``Yes'' responses. Give a confidence interval, at the 95\% confidence level, for the proportion $q$ of young people who answer ``Yes'' to such a survey, among the population of young French people aged 16 to 24 years. b. What can be concluded about the proportion $p$ of young people who practice illegal downloading on the internet at least once a week?

The company wishes to estimate the proportion of people over 20 years old among its customers, at a confidence level of $95\%$, with an interval amplitude less than 0.05. It questions for this purpose a random sample of customers.
What is the minimum number of customers to question?
Answer a: 40 Answer b: 400 Answer c: 1600 Answer d: 20

The tourist office wishes to conduct a survey to estimate the proportion of clients satisfied with the services offered at the ski resort. For this, it uses a confidence interval of length 0.04 with a confidence level of 0.95. The number of clients to interview is: a. 50 b. 2500 c. 25 d. 625

An electoral research institute receives an order in which the margin of error should be at most 2 percentage points (0.02).
The institute has 5 recent surveys, P1 to P5, on the subject of the order and will use the one with an error smaller than requested.
The data on the surveys are as follows:

Survey	$\boldsymbol{\sigma}$	$\boldsymbol{N}$	$\sqrt{\boldsymbol{N}}$
P1	0.5	1764	42
P2	0.4	784	28
P3	0.3	576	24
P4	0.2	441	21
P5	0.1	64	8

The error $e$ can be expressed by
$$|e| < 1.96 \frac{\sigma}{\sqrt{N}}$$
where $\sigma$ is a parameter and $N$ is the number of people interviewed by the survey. Which survey should be used?
(A) P1
(B) P2
(C) P3
(D) P4
(E) P5

The following explains the relationship between confidence interval, confidence level, and sample size.
There is a population following a normal distribution $N \left( m , \sigma ^ { 2 } \right)$. When a sample of size $n$ is randomly extracted from this population, the sample mean follows a normal distribution $\square$ (a). Using the distribution of this sample mean, let the confidence interval for the population mean $m$ with confidence level $\alpha$ be $a \leqq m \leqq b$. When the sample size is fixed at $n$ and the confidence level is set higher than $\alpha$, let the confidence interval be $c \leqq m \leqq d$. Then $d - c$ is $\square$ (b) than $b - a$. On the other hand, when the confidence level is fixed at $\alpha$ and the sample size is $2 n$, let the confidence interval be $e \leqq m \leqq f$. Then $f - e$ is $\square$ (c) times $b - a$.
What are the correct values for (a), (b), and (c) in the above process? [3 points]

	(a)	(b)
(1)	$N \left( m , \sigma ^ { 2 } \right)$	larger
(2)	$N \left( m , \sigma ^ { 2 } \right)$	smaller
(3)	$N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$	smaller
(4)	$N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$	larger
(5)	$N \left( m , \frac { \sigma ^ { 2 } } { n } \right)$	smaller

	(c)
(1)	$\frac{1}{2}$
(2)	$\frac{1}{2}$
(3)	$\frac{1}{\sqrt{2}}$
(4)	$\sqrt{2}$
(5)	$\frac{1}{\sqrt{2}}$

A population follows a normal distribution $\mathrm { N } \left( m , 2 ^ { 2 } \right)$. Let $\overline { X _ { A } }$ and $\overline { X _ { B } }$ be the sample means of samples of size 7 and size 10, respectively, drawn from this population. Using the distributions of $\overline { X _ { A } }$ and $\overline { X _ { B } }$, the 95\% confidence intervals for the population mean $m$ are $[a, b]$ and $[c, d]$, respectively. Choose all correct statements from the given options. [4 points]
Options ㄱ. The variance of $\overline { X _ { A } }$ is greater than the variance of $\overline { X _ { B } }$. ㄴ. $\mathrm { P } \left( \overline { X _ { A } } \leqq m + 2 \right) < \mathrm { P } \left( \overline { X _ { B } } \leqq m + 2 \right)$ ㄷ. $d - c < b - a$
(1) ㄱ
(2) ㄷ
(3) ㄱ, ㄴ
(4) ㄴ, ㄷ
(5) ㄱ, ㄴ, ㄷ

(Probability and Statistics) To determine the proportion $p$ of students arriving before 8 AM at a certain high school, 300 students were randomly sampled from the school on a certain day, and the sample proportion $\hat{p}$ of students arriving before 8 AM was obtained. Using the sample proportion $\hat{p}$, the 95\% confidence interval for the proportion $p$ is $[ 0.701, 0.799 ]$. Find the number of students among the 300 randomly sampled students who arrived before 8 AM. (Note: When $Z$ follows the standard normal distribution, $\mathrm { P } ( | Z | \leqq 1.96 ) = 0.95$.) [4 points]

[Probability and Statistics] A survey of 100 randomly selected people from city A regarding the safest mode of transportation found that 20 people chose express buses. Using this result, a 95\% confidence interval for the proportion of people who chose express buses was found to be $[ a , b ]$. For city B, a 95\% confidence interval is to be constructed for the proportion of people who think express buses are the safest mode of transportation based on a random sample of $n$ people. Find the minimum value of $n$ such that the maximum allowable sampling error of this confidence interval is at most $\frac { b - a } { 2 }$. [4 points]

We want to investigate the proportion of antibody possession for a specific disease among Korean adults. Let $p$ be the proportion of antibody possession in the population, and let $\hat { p }$ be the proportion of antibody possession in a sample of $n$ people randomly selected from the population. Find the minimum value of $n$ such that the probability that $| \hat { p } - p | \leq 0.16 \sqrt { \hat { p } ( 1 - \hat { p } ) }$ is at least 0.9544. (where $Z$ is a random variable following the standard normal distribution and $\mathrm { P } ( 0 \leq Z \leq 2 ) = 0.4772$.) [4 points]

The calcium content in one bottle of beverage produced by a certain company follows a normal distribution with population mean $m$ and population standard deviation $\sigma$. When 16 bottles of beverage produced by this company were randomly sampled and the calcium content was measured, the sample mean was 12.34. When the 95\% confidence interval for the population mean $m$ of the calcium content in one bottle of beverage produced by this company is $11.36 \leq m \leq a$, what is the value of $a + \sigma$? (Note: When $Z$ follows the standard normal distribution, $\mathrm { P } ( 0 \leq Z \leq 1.96 ) = 0.4750$, and the unit of calcium content is mg.) [3 points]
(1) 14.32
(2) 14.82
(3) 15.32
(4) 15.82
(5) 16.32

The lifespan of monitors produced by a certain company follows a normal distribution. From a random sample of 100 monitors produced by this company, the sample mean is $\bar{x}$ and the sample standard deviation is 500. Using this result, the confidence interval for the mean lifespan of monitors produced by this company at a confidence level of $95\%$ is $[\bar{x} - c, \bar{x} + c]$. Find the value of $c$. (Here, $Z$ is a random variable following the standard normal distribution, and $\mathrm{P}(0 \leq Z \leq 1.96) = 0.4750$.) [3 points]

For a normal distribution with known standard deviation $\sigma$, a sample of size $n$ is randomly extracted from the population. The 95\% confidence interval for the population mean obtained from this sample is [100.4, 139.6]. Using the same sample, find the number of natural numbers contained in the 99\% confidence interval for the population mean. (Given that when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( 0 \leq Z \leq 1.96 ) = 0.475$ and $\mathrm { P } ( 0 \leq Z \leq 2.58 ) = 0.495$.) [3 points]

To determine the proportion of residents in a certain city who have experience using the central park, $n$ residents of the city were randomly sampled and surveyed. The result showed that 80\% had experience using the central park. Using this result, the 95\% confidence interval for the proportion of residents in the entire city who have experience using the central park is $[ a , b ]$. When $b - a = 0.098$, find the value of $n$. (Here, when $Z$ is a random variable following the standard normal distribution, calculate using $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95$.) [4 points]

The weight of pomegranates produced at a certain farm follows a normal distribution with mean $m$ and standard deviation 40. A sample of size 64 was taken from the pomegranates produced at this farm, and the sample mean of the pomegranate weights was $\bar { x }$. Using this result, the 99\% confidence interval for the mean $m$ of the pomegranate weights produced at this farm is $\bar { x } - c \leq m \leq \bar { x } + c$. What is the value of $c$? (Here, the unit of weight is g, and when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( 0 \leq Z \leq 2.58 ) = 0.495$.) [4 points]
(1) 25.8
(2) 21.5
(3) 17.2
(4) 12.9
(5) 8.6

The weight of watermelons harvested in a certain village follows a normal distribution with mean $m$ kg and standard deviation 1.4 kg. When 49 watermelons are randomly sampled from this village and a 95\% confidence interval for the mean weight $m$ is constructed using the sample mean, the interval is $a \leq m \leq 7.992$. What is the value of $a$? (Here, when $Z$ is a random variable following the standard normal distribution, calculate using $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95$.) [3 points]
(1) 7.198
(2) 7.208
(3) 7.218
(4) 7.228
(5) 7.238

The daily leisure activity time of residents in a certain region follows a normal distribution with mean $m$ minutes and standard deviation $\sigma$ minutes. When 16 residents are randomly sampled and the sample mean of daily leisure activity time is 75 minutes, the 95\% confidence interval for the population mean $m$ is $a \leq m \leq b$. When 16 residents are randomly sampled again and the sample mean of daily leisure activity time is 77 minutes, the 99\% confidence interval for the population mean $m$ is $c \leq m \leq d$. Find the value of $\sigma$ that satisfies $d - b = 3.86$. (Here, when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95 , \mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [4 points]

A certain automobile company produces electric vehicles whose driving range on a single charge follows a normal distribution with mean $m$ and standard deviation $\sigma$.
When a sample of 100 electric vehicles produced by this company is randomly selected and the sample mean of the driving range is $\overline { x _ { 1 } }$, the 95\% confidence interval for the population mean $m$ is $a \leq m \leq b$.
When a sample of 400 electric vehicles produced by this company is randomly selected and the sample mean of the driving range is $\overline { x _ { 2 } }$, the 99\% confidence interval for the population mean $m$ is $c \leq m \leq d$.
If $\overline { x _ { 1 } } - \overline { x _ { 2 } } = 1.34$ and $a = c$, what is the value of $b - a$? (Here, the unit of driving range is km, and when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95 , \mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [3 points]
(1) 5.88
(2) 7.84
(3) 9.80
(4) 11.76
(5) 13.72

A company produces shampoo where the volume of 1 bottle follows a normal distribution $\mathrm { N } \left( m , \sigma ^ { 2 } \right)$. A sample of 16 bottles was randomly selected from the company's shampoo, and the 95\% confidence interval for $m$ using the sample mean is $746.1 \leq m \leq 755.9$. When a sample of $n$ bottles is randomly selected and a 99\% confidence interval for $m$ is constructed as $a \leq m \leq b$, what is the minimum natural number $n$ such that $b - a \leq 6$? (Here, the unit of volume is mL, and when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( | Z | \leq 1.96 ) = 0.95$ and $\mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [3 points]
(1) 70
(2) 74
(3) 78
(4) 82
(5) 86

From a population following a normal distribution $\mathrm{N}(m, 5^2)$, a sample of size 49 is randomly extracted, and the sample mean is $\bar{x}$. The 95\% confidence interval for the population mean $m$ is $a \leq m \leq \frac{6}{5}a$. Find the value of $\bar{x}$. (Here, if $Z$ is a random variable following the standard normal distribution, $\mathrm{P}(|Z| \leq 1.96) = 0.95$.) [3 points]
(1) 15.2
(2) 15.4
(3) 15.6
(4) 15.8
(5) 16.0