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

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

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

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

6. The measurement result of a certain physical quantity follows a normal distribution $N \left( 10 , \sigma ^ { 2 } \right)$. Which of the following conclusions is incorrect? ( )
A. The smaller $\sigma$ is, the greater the probability that the physical quantity falls in $( 9.9,10.1 )$ in a single measurement.
B. The smaller $\sigma$ is, the probability that the physical quantity is greater than 10 in a single measurement is 0.5.
C. The smaller $\sigma$ is, the probability that the physical quantity is less than 9.99 equals the probability that it is greater than 10.01 in a single measurement.
D. The smaller $\sigma$ is, the probability that the physical quantity falls in $( 9.9,10.2 )$ equals the probability that it falls in $( 10,10.3 )$ in a single measurement.
【Answer】D

【Solution】

【Analysis】Use the properties of the normal distribution density curve to judge each option. 【Detailed Solution】For option A, $\sigma ^ { 2 }$ is the variance of the data, so the smaller $\sigma$ is, the more concentrated the data is around $\mu = 10$. Therefore, the probability that the measurement result falls in $( 9.9,10.1 )$ increases, so A is correct.
For option B, by the symmetry of the normal distribution density curve, the probability that the physical quantity is greater than 10 in a single measurement is 0.5, so B is correct.
For option C, by the symmetry of the normal distribution density curve, the probability that the physical quantity is greater than 10.01 equals the probability that it is less than 9.99 in a single measurement, so C is correct.
For option D, since the probability that the physical quantity falls in $( 9.9,10.0 )$ does not equal the probability that it falls in $( 10.2,10.3 )$, the probability that the measurement result falls in $( 9.9,10.2 )$ does not equal the probability that it falls in $( 10,10.3 )$, so D is incorrect. Therefore, the answer is: D.

The measurement result of a certain physical quantity follows a normal distribution $N \left( 10 , \sigma ^ { 2 } \right)$. Which of the following conclusions is incorrect?
A. The smaller $\sigma$ is, the greater the probability that the physical quantity falls in $( 9.9,10.1 )$ in a single measurement.
B. The smaller $\sigma$ is, the probability that the physical quantity is greater than 10 in a single measurement is 0.5.
C. The smaller $\sigma$ is, the probability that the physical quantity is less than 9.99 equals the probability that it is greater than 10.01 in a single measurement.
D. The smaller $\sigma$ is, the probability that the physical quantity falls in $( 9.9,10.2 )$ equals the probability that it falls in $( 10,10.3 )$ in a single measurement.

To understand the per-acre income (in units of 10,000 yuan) after promoting exports, a sample was taken from the planting area. The sample mean of per-acre income after promoting exports is $\bar { x } = 2.1$ , and the sample variance is $s ^ { 2 } = 0.01$ . The historical per-acre income $X$ in the planting area follows a normal distribution $N \left( 1.8 , ~ 0.1 ^ { 2 } \right)$ . Assume that the per-acre income $Y$ after promoting exports follows a normal distribution $N \left( \bar { x } , s ^ { 2 } \right)$ . Then (if a random variable $Z$ follows a normal distribution $N \left( \mu , \sigma ^ { 2 } \right)$ , then $P ( Z < \mu + \sigma ) \approx 0.8413$ )
A. $P ( X > 2 ) > 0.2$
B. $P ( X > 2 ) < 0.5$
C. $P ( Y > 2 ) > 0.5$
D. $P ( Y > 2 ) < 0.8$

Deduce a simple equivalent of $1 - \Phi ( x )$ as $x$ tends to $+ \infty$, where $\Phi ( x ) = \int _ { - \infty } ^ { x } \varphi ( t ) \mathrm { d } t$ and $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n } : \mathbb { R } \rightarrow \mathbb { R }$ is defined by $$\forall x \in ] - \infty , - \sqrt { n } - \frac { 1 } { \sqrt { n } } [ , \quad B _ { n } ( x ) = 0$$ $$\forall k \in \llbracket 0 , n \rrbracket , \forall x \in \left[ x _ { n , k } - \frac { 1 } { \sqrt { n } } , x _ { n , k } + \frac { 1 } { \sqrt { n } } [ , \right. \quad B _ { n } ( x ) = \frac { \sqrt { n } } { 2 } \binom { n } { k } \frac { 1 } { 2 ^ { n } }$$ $$\forall x \in \left[ \sqrt { n } + \frac { 1 } { \sqrt { n } } , + \infty [ , \right. \quad B _ { n } ( x ) = 0$$ and $\Delta _ { n } = \sup _ { x \in \mathbb { R } } \left| B _ { n } ( x ) - \varphi ( x ) \right|$ where $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.
Show that, for all $n \in \mathbb { N } ^ { * }$, we have the equality $$\Delta _ { n } = \sup _ { x \geqslant 0 } \left| B _ { n } ( x ) - \varphi ( x ) \right| .$$

For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, we set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B _ { n }$ is defined as in Q19, and $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$. We fix $\varepsilon > 0$ and $\ell \in \mathbb{R}^+$ such that $\varphi(\ell) \leqslant \frac{\varepsilon}{2}$.
Show that there exists a natural number $n _ { 1 }$ such that, for all integers $n \geqslant n _ { 1 }$, $$\sup _ { x \in [ 0 , \ell ] } \left| B _ { n } ( x ) - \varphi ( x ) \right| \leqslant \frac { \varepsilon } { 2 }$$

The function $B _ { n }$ is defined as in Q19, and $\varphi ( x ) = \frac { 1 } { \sqrt { 2 \pi } } \mathrm { e } ^ { - x ^ { 2 } / 2 }$.
For all $\ell > 0$, show that there exists a natural number $n _ { 2 }$, such that, for all $n \geqslant n _ { 2 }$, $$B _ { n } ( \ell ) \leqslant 2 \varphi ( \ell )$$