Exercise 3 -- Common to all candidates

Part A: Study of the lifespan of a household appliance

Statistical studies have made it possible to model the lifespan, in months, of a type of dishwasher by a random variable $X$ following a normal distribution $\mathscr{N}(\mu, \sigma^2)$ with mean $\mu = 84$ and standard deviation $\sigma$. Furthermore, we have $P(X \leqslant 64) = 0.16$.

1. a. By exploiting the graph, determine $P(64 \leqslant X \leqslant 104)$. b. What approximate integer value of $\sigma$ can we propose?
2. We denote by $Z$ the random variable defined by $Z = \dfrac{X - 84}{\sigma}$. a. What is the probability distribution followed by $Z$? b. Justify that $P(X \leqslant 64) = P\!\left(Z \leqslant \dfrac{-20}{\sigma}\right)$. c. Deduce the value of $\sigma$, rounded to $10^{-3}$.
3. In this question, we consider that $\sigma = 20.1$.
   The probabilities requested will be rounded to $10^{-3}$. a. Calculate the probability that the lifespan of the dishwasher is between 2 and 5 years. b. Calculate the probability that the dishwasher has a lifespan greater than 10 years.

Part B: Study of the warranty extension offered by El'Ectro

The dishwasher is guaranteed free of charge for the first two years. The company El'Ectro offers its customers a warranty extension of 3 additional years. Statistical studies conducted on customers who take the warranty extension show that $11.5\%$ of them use the warranty extension.

1. We randomly choose 12 customers among those who have taken the warranty extension (this choice can be treated as random sampling with replacement given the large number of customers). a. What is the probability that exactly 3 of these customers use this warranty extension? Detail the approach by specifying the probability distribution used. Round to $10^{-3}$. b. What is the probability that at least 6 of these customers use this warranty extension? Round to $10^{-3}$.
2. The warranty extension offer is as follows: for 65 euros additional, El'Ectro will reimburse the customer the initial value of the dishwasher, namely 399 euros, if an irreparable breakdown occurs between the beginning of the third year and the end of the fifth year. The customer cannot use this warranty extension if the breakdown is repairable.
   We randomly choose a customer among those who have subscribed to the warranty extension, and we denote by $Y$ the random variable representing the algebraic gain in euros realized on this customer by the company El'Ectro, thanks to the warranty extension. a. Justify that $Y$ takes the values 65 and $-334$ then give the probability distribution of $Y$. b. Is this warranty extension offer financially advantageous for the company? Justify.

[Probability and Statistics] One method to determine whether a certain bone fossil belongs to animal A or animal B is to use the length of a specific part. The length of this part in animal A follows a normal distribution $\mathrm { N } \left( 10,0.4 ^ { 2 } \right)$, and the length of this part in animal B follows a normal distribution $\mathrm { N } \left( 12,0.6 ^ { 2 } \right)$. If the length of this part is less than $d$, it is judged to be a fossil of animal A, and if it is at least $d$, it is judged to be a fossil of animal B. Find the value of $d$ such that the probability of judging an animal A fossil as an animal A fossil equals the probability of judging an animal B fossil as an animal B fossil. (The unit of length is cm.) [4 points]
(1) 10.4
(2) 10.5
(3) 10.6
(4) 10.7
(5) 10.8

The daily production of employees at a certain company varies depending on their length of service. The daily production of an employee with $n$ months of service ( $1 \leqq n \leqq 100$ ) follows a normal distribution with mean $a n + 100$ ( $a$ is a constant) and standard deviation 12. When the probability that the daily production of an employee with 16 months of service is 84 or less is 0.0228, find the probability that the daily production of an employee with 36 months of service is at least 100 and at most 142 using the standard normal distribution table.

$z$	$\mathrm { P } ( 0 \leqq Z \leqq z )$
1.0	0.3413
1.5	0.4332
2.0	0.4772
2.5	0.4938

[3 points]
(1) 0.7745
(2) 0.8185
(3) 0.9104
(4) 0.9270
(5) 0.9710

A random variable $X$ follows a normal distribution $\mathrm { N } \left( m , \sigma ^ { 2 } \right)$ and satisfies the following conditions.
(a) $\mathrm { P } ( X \geq 64 ) = \mathrm { P } ( X \leq 56 )$
(b) $\mathrm { E } \left( X ^ { 2 } \right) = 3616$ What is the value of $\mathrm { P } ( X \leq 68 )$ obtained using the table on the right? [3 points]
(1) 0.9104
(2) 0.9332
(3) 0.9544
(4) 0.9772
(5) 0.9938

$x$	$\mathrm { P } ( m \leq X \leq x )$
$m + 1.5 \sigma$	0.4332
$m + 2 \sigma$	0.4772
$m + 2.5 \sigma$	0.4938

A random variable $X$ follows a normal distribution with mean $m$ and standard deviation $\sigma$, and $$\mathrm { P } ( X \leq 3 ) = \mathrm { P } ( 3 \leq X \leq 80 ) = 0.3$$ Find the value of $m + \sigma$. (Here, when $Z$ is a random variable following the standard normal distribution, $\mathrm { P } ( 0 \leq Z \leq 0.25 ) = 0.1 , \mathrm { P } ( 0 \leq Z \leq 0.52 ) = 0.2$ for calculation purposes.) [4 points]

The random variable $X$ follows the normal distribution $\mathrm { N } \left( 10,2 ^ { 2 } \right)$, and the random variable $Y$ follows the normal distribution $\mathrm { N } \left( m , 2 ^ { 2 } \right)$. The probability density functions of $X$ and $Y$ are $f ( x )$ and $g ( x )$ respectively.
$$f ( 12 ) \leq g ( 20 )$$
For $m$ satisfying this condition, what is the maximum value of $\mathrm { P } ( 21 \leq Y \leq 24 )$ using the standard normal distribution table on the right? [4 points]

$z$	$\mathrm { P } ( 0 \leq Z \leq z )$
0.5	0.1915
1.0	0.3413
1.5	0.4332
2.0	0.4772

(1) 0.5328
(2) 0.6247
(3) 0.7745
(4) 0.8185
(5) 0.9104

A random variable $X$ follows a normal distribution with mean 8 and standard deviation 3, and a random variable $Y$ follows a normal distribution with mean $m$ and standard deviation $\sigma$. If the two random variables $X$ and $Y$ satisfy $$\mathrm { P } ( 4 \leq X \leq 8 ) + \mathrm { P } ( Y \geq 8 ) = \frac { 1 } { 2 }$$ find the value of $\mathrm { P } \left( Y \leq 8 + \frac { 2 \sigma } { 3 } \right)$ using the standard normal distribution table below.

$z$	$\mathrm { P } ( 0 \leq Z \leq z )$
1.0	0.3413
1.5	0.4332
2.0	0.4772
2.5	0.4938

[4 points]
(1) 0.8351
(2) 0.8413
(3) 0.9332
(4) 0.9772
(5) 0.9938

The random variable $X$ follows a normal distribution with mean 8 and standard deviation 3, and the random variable $Y$ follows a normal distribution with mean $m$ and standard deviation $\sigma$. The two random variables $X$ and $Y$ satisfy $$\mathrm { P } ( 4 \leq X \leq 8 ) + \mathrm { P } ( Y \geq 8 ) = \frac { 1 } { 2 }$$ Find the value of $\mathrm { P } \left( Y \leq 8 + \frac { 2 \sigma } { 3 } \right)$ using the standard normal distribution table on the right.

$z$	$\mathrm { P } ( 0 \leq Z \leq z )$
1.0	0.3413
1.5	0.4332
2.0	0.4772
2.5	0.4938

[3 points]
(1) 0.8351
(2) 0.8413
(3) 0.9332
(4) 0.9772
(5) 0.9938

For a positive number $t$, the random variable $X$ follows a normal distribution $\mathrm{N}(1, t^2)$. $$\mathrm{P}(X \leq 5t) \geq \frac{1}{2}$$ For all positive numbers $t$ satisfying this condition, find the maximum value of $\mathrm{P}(t^2 - t + 1 \leq X \leq t^2 + t + 1)$ using the standard normal distribution table below, and let this value be $k$. Find the value of $1000 \times k$. [4 points]

$z$	$\mathrm{P}(0 \leq Z \leq z)$
0.6	0.226
0.8	0.288
1.0	0.341
1.2	0.385
1.4	0.419

A random variable $X$ follows a normal distribution $\mathrm{N}\left(m_{1}, \sigma_{1}^{2}\right)$ and a random variable $Y$ follows a normal distribution $\mathrm{N}\left(m_{2}, \sigma_{2}^{2}\right)$, satisfying the following conditions. For all real numbers $x$, $\mathrm{P}(X \leq x) = \mathrm{P}(X \geq 40 - x)$ and $\mathrm{P}(Y \leq x) = \mathrm{P}(X \leq x + 10)$. When $\mathrm{P}(15 \leq X \leq 20) + \mathrm{P}(15 \leq Y \leq 20) = 0.4772$ using the standard normal distribution table below, what is the value of $m_{1} + \sigma_{2}$? (Given: $\sigma_{1}$ and $\sigma_{2}$ are positive.) [4 points]

$z$	$\mathrm{P}(0 \leq Z \leq z)$
0.5	0.1915
1.0	0.3413
1.5	0.4332
2.0	0.4772