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

(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]

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]

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

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

A population with mean $m$ and standard deviation 5 follows a normal distribution. A sample of size 36 is randomly extracted, and the 99\% confidence interval for the population mean $m$ obtained using the sample mean is $1.2 \leq m \leq a$. What is the value of $a$? (When $Z$ is a random variable following the standard normal distribution, calculate using $\mathrm { P } ( | Z | \leq 2.58 ) = 0.99$.) [3 points]
(1) 5.1
(2) 5.2
(3) 5.3
(4) 5.4
(5) 5.5