(12 points)
To compare the yields of an old and a new breeding method, a survey was conducted on 100 aquaculture farms, recording the yield (in kg) of a certain aquatic product. The frequency distribution histogram is given.
(1) Let $A$ denote the event ``the yield using the old breeding method is less than 50 kg''. Estimate the probability of $A$.
(2) Complete the contingency table below, and use the chi-squared test to determine whether we can be 99\% confident that yield is related to breeding method.

	Yield $< 50 \text{ kg}$	Yield $\geq 50 \text{ kg}$
Old breeding method
New breeding method

$P(K^2 \geq k)$	0.050	0.010	0.001
$k$	3.841	6.635	10.828

$$K^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$$

A student interest group randomly surveyed the air quality level and the number of people exercising in a certain park each day over 100 days in a city. The data is organized in the following table (unit: days):

Air Quality Level	$[ 0,200 ]$	$( 200,400 ]$	$( 400,600 ]$
1 (Excellent)	2	16	25
2 (Good)	5	10	12
3 (Slight Pollution)	6	7	8
4 (Moderate Pollution)	7	2	0

(1) Estimate the probability that the air quality level on a given day in the city is 1, 2, 3, or 4 respectively;
(2) Find the estimated average number of people exercising in the park on a given day (use the midpoint of each interval as the representative value for data in that interval);
(3) If the air quality level on a given day is 1 or 2, the day is called ``good air quality''; if the air quality level is 3 or 4, the day is called ``poor air quality''. Based on the given data, complete the following $2 \times 2$ contingency table and determine whether there is 95\% confidence to conclude that the number of people exercising in the park on a given day is related to the air quality of the city on that day.

	Number of people $\leqslant 400$	Number of people $> 400$
Good air quality
Poor air quality

Attachment: $K ^ { 2 } = \frac { n ( a d - b c ) ^ { 2 } } { ( a + b ) ( c + d ) ( a + c ) ( b + d ) }$,

$P \left( K ^ { 2 } \geqslant k \right)$	0.050	0.010	0.001
$k$	3.841	6.635	10.828

.

A student interest group randomly surveyed the air quality level and the number of people exercising in a certain park on each of 100 days in a certain city. The organized data is shown in the table below (unit: days):

\backslashbox{Air Quality Level}{Number of Exercisers}	[0, 200]	(200, 400]	(400, 600]
1 (Excellent)	2	16	25
2 (Good)	5	10	12
3 (Slight Pollution)	6	7	8
4 (Moderate Pollution)	7	2	0

(1) Estimate the probability that the air quality level on a given day in the city is 1, 2, 3, or 4 respectively;
(2) Find the estimated value of the average number of people exercising in the park on a given day (use the midpoint of each interval as the representative value for data in that interval);
(3) If the air quality level on a given day is 1 or 2, the day is called ``good air quality''; if the air quality level is 3 or 4, the day is called ``poor air quality''. Based on the given data, complete the following $2 \times 2$ contingency table, and based on the contingency table, determine whether there is 95\% confidence to conclude that the number of people exercising in the park on a given day is related to the air quality level of the city on that day?

17. Two machine tools, Machine A and Machine B, produce the same type of product. Products are classified by quality into first-grade and second-grade products. To compare the quality of products from the two machines, 200 products were produced by each machine. The quality statistics are shown in the table below:

	First-grade	Second-grade	Total
Machine A	150	50	200
Machine B	120	80	200
Total	270	130	400

(1) What are the frequencies of first-grade products produced by Machine A and Machine B, respectively?
(2) Can we conclude with 99\% confidence that there is a difference in product quality between Machine A and Machine B? Attachment: $\mathrm { K } ^ { 2 } = \frac { n ( a d - b c ) ^ { 2 } } { ( a + b ) ( c + d ) ( a + c ) ( b + d ) }$,

$\mathrm { P } \left( \mathrm { K } ^ { 2 } \geqslant k \right)$	0.050	0.010	0.001
$k$	3.841	6.635	10.828

17. (12 points) Two machine tools, A and B, produce the same type of product. Products are classified by quality into first-grade and second-grade products. To compare the quality of products from the two machine tools, 200 products were produced by each machine tool. The quality statistics are shown in the table below:

	First-grade	Second-grade	Total
Machine A	150	50	200
Machine B	120	80	200
Total	270	130	400

(1) What are the proportions of first-grade products produced by machine A and machine B respectively?
(2) Can we conclude with 99\% confidence that there is a difference in product quality between machine A and machine B? Attachment: $k^2 = \frac{n(ac - bd)^2}{(a+b)(c+d)(a+c)(b+d)}$,

$P(K^2 \geq k)$	0.050	0.010	0.001
$k$	3.841	6.635	10.828

20. (12 points)
A medical team conducted a study on the relationship between a certain endemic disease in a region and the hygiene habits of local residents (hygiene habits are classified as either good or not sufficiently good). Among patients with the disease, 100 cases were randomly surveyed (called the case group), and among people without the disease, 100 people were randomly surveyed (called the control group). The following data were obtained:

	Not Sufficiently Good	Good
Case Group	40	60
Control Group	10	90

(1) Can we conclude with 99\% confidence that there is a difference in hygiene habits between the group with the disease and the group without the disease?
(2) From the population of the region, one person is randomly selected. Let $A$ denote the event ``the selected person has not sufficiently good hygiene habits'' and $B$ denote the event ``the selected person has the disease''. The ratio $\frac { P ( B \mid A ) } { P ( \bar { B } \mid A ) }$ to $\frac { P ( B \mid \bar { A } ) } { P ( \bar { B } \mid \bar { A } ) }$ is a measure of the risk level of the disease associated with not sufficiently good hygiene habits. Let this measure be denoted as $R$.
(i) Prove that $R = \frac { P ( A \mid B ) } { P ( \bar { A } \mid B ) } \

To study the relationship between a certain disease and ultrasound examination results, 1000 people who had undergone ultrasound examination were randomly surveyed, yielding the following contingency table:

\backslashbox{Category}{Ultrasound Result}	Normal	Abnormal	Total
Has Disease	20	180	200
No Disease	780	20	800
Total	800	200	1000

(1) Let $P$ denote the probability that a person with abnormal ultrasound results has the disease. Find the estimated value of $P$.
(2) Based on the significance level $\alpha = 0.001$ for independence testing, analyze whether the ultrasound examination result is related to having the disease.
Attachment: $\chi^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$,

$P(\chi^2 \geq k)$	0.005	0.010	0.001
$k$	3.841	6.635	10.828

(13 points) To study the relationship between a certain disease and ultrasound examination results, 1000 people who had undergone ultrasound examination were randomly surveyed, yielding the following contingency table:

	Normal	Abnormal	Total
Has disease	20	180	200
Does not have disease	780	20	800
Total	800	200	1000

(1) Let $p$ denote the probability that a person with abnormal ultrasound examination results has the disease. Find the estimated value of $p$.
(2) Based on the significance level $\alpha = 0.001$ for the independence test, analyze whether the ultrasound examination result is related to having the disease. Attachment: $\chi^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$,

$P(\chi^2 \geq k)$	0.050	0.010	0.001
$k$	3.841	6.635	10.828

.

The value of $\tan \left( 2 \tan ^ { - 1 } \left( \frac { 3 } { 5 } \right) + \sin ^ { - 1 } \left( \frac { 5 } { 13 } \right) \right)$ is equal to:
(1) $\frac { - 181 } { 69 }$
(2) $\frac { 220 } { 21 }$
(3) $\frac { - 291 } { 76 }$
(4) $\frac { 151 } { 63 }$