From cfc4c2a8ddd4d3e6f84c1d601c6aba78f448aa0d Mon Sep 17 00:00:00 2001 From: "weiye.wang" Date: Sat, 3 Jun 2023 19:59:18 +0800 Subject: [PATCH] =?UTF-8?q?=E5=BD=95=E5=85=A5=E6=9D=A8=E9=AB=982023?= =?UTF-8?q?=E5=B1=8A=E6=A6=82=E7=8E=87=E7=BB=9F=E8=AE=A1=E5=A4=8D=E4=B9=A0?= =?UTF-8?q?=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 工具/批量收录题目.py | 2 +- 题库0.3/Problems.json | 1300 +++++++++++++++++++++++++++++++++++++++++ 2 files changed, 1301 insertions(+), 1 deletion(-) diff --git a/工具/批量收录题目.py b/工具/批量收录题目.py index 9728a4a2..9d26a9a4 100644 --- a/工具/批量收录题目.py +++ b/工具/批量收录题目.py @@ -1,5 +1,5 @@ #修改起始id,出处,文件名 -starting_id = 17465 +starting_id = 17486 raworigin = "" filename = r"C:\Users\weiye\Documents\wwy sync\临时工作区\自拟题目12.tex" editor = "20230602\t王伟叶" diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json index 278c4a74..0e22c85d 100644 --- a/题库0.3/Problems.json +++ b/题库0.3/Problems.json @@ -451718,6 +451718,1306 @@ "space": "4em", "unrelated": [] }, + "017486": { + "id": "017486", + "content": "某班有男生$26$人, 女生$9$人, 已知男生中 A 档的有$8$人, 女生中 A 档的有$3$人. 现从班级中随机挑选一人, 再选出同性别的另一人, 求在第一次选出的是 A 档生的情况下, 第二次选出的是非 A 档生的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题1", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017487": { + "id": "017487", + "content": "由频率分布直方图绘制概率密度曲线, 小明认为形状接近钟形曲线, 可以近似为正态分布$N(\\mu, \\sigma^2)$, 则该正态分布是$N(70.9,14^2)$, 若该抽象基本合理, 并用此正态分布估计本校学生的本次测试得分情况, 则随机抽取本校一名学生, 求该生得分高于$84.9$分的概率.\n(已知$\\Phi(1) \\approx 0.8413$, $\\Phi(2) \\approx 0.9772$, $\\Phi(3) \\approx 0.9987$.$\\Phi(x)$表示标准正态分布的密度函数从$-\\infty$到$x$的累计面积)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题2", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017488": { + "id": "017488", + "content": "某盒子甲中有$5$个红球, $10$个黑球\n(1) 求第一次摸出红球的情况下, 第二次摸出黑球的概率\n(2) 求第二次摸球, 摸出黑球的概率\n(3) 摸出$2$个球, 用随机变量$X$表示该两个球中红球的个数, 求$X$的分布.\n(4) 另一个盒子乙中有$7$个红球, $8$个墨球, 从两个盒子中随机挑选出一个, 并从该盒子中先后随机抽取两个球, 求在第一次取出的是红球的情况下, 第二次取出的是黑球的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题3", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017489": { + "id": "017489", + "content": "在抛掷一枚均匀硬币两次的试验中, 恰一次朝上的概率为\\blank{50}; 至少有一枚正面朝上的概率为\\blank{50}; 在至少有一枚反面朝上的条件下, 第一次反面朝上的概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题4", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017490": { + "id": "017490", + "content": "从$1,2,3,4,5$中随机取出两个不同的数, 则其和为奇数的概率是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题5", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017491": { + "id": "017491", + "content": "甲乙两人射击同一个标靶, 每人一发, 其中甲命中概率为$\\dfrac{2}{3}$, 乙命中概率为$\\dfrac{1}{2}$, 若甲乙的射击相互独立, 则标靬被击中的概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题6", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017492": { + "id": "017492", + "content": "某种疾病的患病率为$0.50$, 患该种疾病且血检呈阳性的概率为$0.49$, 则已知在患该种疾病的条件下血检呈阳性的概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题7", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017493": { + "id": "017493", + "content": "已知不透明的袋子中有$3$个相同的白球和$3$个相同的黑球, 分别求下列概率: \\\\\n(1) 有放回地依次摸$3$个球, 每次摸一个, 则恰摸到$2$个白球的概率为\\blank{50}, 若$3$次摸到的白球数记为$X$, 则$E[X]=$\\blank{50}, $D[X]=$\\blank{50};\\\\\n(2)不放回地依次摸$3$个球, 每次摸一个, 则第三次摸到白球的概率为恰摸到$2$个白球的概率为若$3$次摸到的白球总数记为$Y$, 则$E[Y]=$\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题8", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017494": { + "id": "017494", + "content": "已知$X\\sim N(1,2^2)$, 若$P(X>3)=\\alpha$, 则$P(-1 \\leq X \\leq 3)=$(用$a$表示)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题9", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017495": { + "id": "017495", + "content": "高三(1)班甲、乙两同学报名参加$A, B, C$三所高校的自主招生考试, 因为三所高校考试时间相同, 所以甲、乙只能随机报考其中一所高校.\\\\\n(1) 写出合适的样本空间;\\\\\n(2) 求甲、乙两人报考不同高校的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题10", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017496": { + "id": "017496", + "content": "有$5$条线段, 长度分别为$1,3,5,7,9$. 从这五条线段中任取三条, 求所取线段长能构成三角形三边长的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题11", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017497": { + "id": "017497", + "content": "任取$m \\in\\{2,5,8,9\\}$, $n \\in\\{1,3,5,7\\}$, 求方程$\\dfrac{x^2}{m}+\\dfrac{y^2}{n}=1$表示的焦点在$x$轴上的椭圆的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题12", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017498": { + "id": "017498", + "content": "两个篮球运动员甲和乙罚球时命中的概率分别是$0.7$和$0.6$, 两人各设一次, 假设事件``甲命中''与``乙命中'', 是独立的. 求至少一人命中的概率.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题13", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017499": { + "id": "017499", + "content": "甲乙二人争夺一场围棋比赛的冠军, 若比赛为``三局两胜''制 (无平局), 甲在每局比赛中获胜的概率均为$\\dfrac{2}{3}$, 且各局比赛结果相互独立, 则在甲获得冠军的条件下, 比赛进行了三局的概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题14", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017500": { + "id": "017500", + "content": "设盒中装有$5$只球, 其中$3$只是黑球, $2$只是白球, 现从盒中随机地摸出两只, 并换进$2$只黑球之后, 再从盒中摸出$2$只, 第二次摸出的$2$只全是黑球概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题15", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017501": { + "id": "017501", + "content": "某品牌冰柜由甲、乙、丙三个工厂生产, 其中甲厂占$25 \\%$, 乙厂占$35 \\%$, 丙厂占$40 \\%$, 且各厂的次品率分别为$5 \\%, 4 \\%, 2 \\%$. 如果某人已经买到一台次品冰柜, 则该冰柜由甲厂生产的概率为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题16", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017502": { + "id": "017502", + "content": "某中学数学竞赛培训共开设有初等代数、初等几何、初等数论和微积分初步共四门课程, 要求初等代数、初等几何都要合格, 且初等数论和微积分初步至少有一门合格, 则能取得参加数学竞赛赛的资格, 现有甲、乙、丙三位同学报名参加数学竞赛培训, 每一位同学对这四门课程考试是否合格相互独立, 其合格的概率均相同, (见下表), 且每一门课程是否合格相互独立.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline 课程 & 初等代数 & 初等几何 & 初等数论 & 微积分初步 \\\\\n\\hline 合格的概率 &$\\dfrac{3}{4}$&$\\dfrac{2}{3}$&$\\dfrac{2}{3}$&$\\dfrac{1}{2}$\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n(1) 求甲同学取得参加数学竞赛复赛的资格的概率;\\\\\n(2) 记$\\xi$表示三位同学中取得参加数学竞赛复赛的资格的人数, 求$\\xi$的分布及期望$E[\\xi]$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题17", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017503": { + "id": "017503", + "content": "生产方提供$50$箱的一批产品, 其中有$2$箱不合格产品. 采购方接收该批产品的准则是: 从该批产品中任取$5$箱产品进行检测, 若至多有$1$箱不合格产品, 便接收该批产品. 问: 该批产品被接收的概率是多少?", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题18", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017504": { + "id": "017504", + "content": "当前, 以``立德树人''为目标的课程改革正在有序推进. 高中联招对初三毕业学生进行体育测试, 是激发学生、家长和学校积极开展体育活动, 保证学生健康成长的有效措施. ``某地区 2019 年初中毕业生升学体育考试规定, 考生必须参加立定跳远、投实心球、 1 分钟跳绳三项测试, 三项考试满分为$50$分, 其中立定跳远$15$分, 投实心球$15$分, 1 分钟跳绳$20$分. 某学校在初三上学期开始时要掌握全年级学生每分钟跳绳的情况, 随机抽取了$100$名学生进行测试, 得到如下频率分布直方图, 且规定计分规则如表:\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.1, yscale = 100]\n\\draw [->] (150,0) -- (152,0) -- (153,0.002) -- (155,-0.002) -- (156,0) -- (225,0) node [below right] {每分钟跳绳个数};\n\\draw [->] (150,0) -- (150,0.065) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\foreach \\i/\\j in {165/0.005,175/0.009,185/0.05,195/0.03,205/0.006}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (10,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {165/0.005/below left,175/0.009/left,185/0.05/left,195/0.03/left,205/0.006/left}\n{\\draw [dashed] (\\i,\\j) -- (150,\\j) node [\\k] {$\\j$};};\n\\draw (215,0) node [below] {$215$};\n\\end{tikzpicture}\n\\end{center}\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline 每分钟咇绳令数 & {$[165,175)$} & {$[175,185)$} & {$[185,195)$} & {$[195,205)$} & {$[205,215)$} \\\\\n\\hline 得分 & 16 & 17 & 18 & 19 & 20 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n(1) 现从样本的$100$名学生中, 任意选取$2$人, 求两人得分之和不大于$33$分的概率;\\\\\n(2) 若该校初三年级所有学生的跳绳个数$X$服从正态分布$N(\\mu , \\sigma^2)$, 用样本数据的平均值和方差估计总体的期望和方差 (结果四舍五入到整数), 已知样本方差$S^2 \\approx 77.8$(各组数据用中点值代替). 根据往年经验, 该校初三年级学生经过一年的训练, 正式测试时每人每分钟跳绳个数都有明显进步, 假设明年正式测试时每人每分钟跳绳个数比初三上学期开始时个数增加$10$个, 利用现所得正态分布模型:\\\\ (I) 预估全年级恰好有$1000$名学生, 正式测试时每分钟跳$193$个以上的人数. (结果四舍五入到整数);\\\\\n(II) 若在该地区$2020$年所有初三毕业生中任意选取$3$人, 记正式测试时每分钟跳$202$个以上的人数为$\\xi$, 求随机变量$\\xi$的分布和期望.\\\\\n附: 若随机变量$X$服从正态分布$N(\\mu, \\sigma^2)$, $\\sigma=\\sqrt{77.8} \\approx 9$, 则$P(\\mu-\\sigma=latex]\n\\draw (0,0) node [draw] (A) {元件A};\n\\draw (0,1) node [draw] (A1) {元件A};\n\\draw (0,-1) node [draw] (A2) {元件A};\n\\draw (2,0) node [draw] (B) {元件B};\n\\draw (A)--(B);\n\\draw (A) --++ (-1.5,0) (B)--++ (1,0);\n\\draw (A1) --++ (-1,0) --++ (0,-2) -- (A2); \n\\draw (A1) --++ (1,0) --++ (0,-2) -- (A2); \n\\end{tikzpicture}\n\\end{center}\n(1) 求该装置正常工作超过$10000$小时的概率;\\\\\n(2) 某城市$5G$基站建设需购进$1200$台该装置, 估计该批装置能正常工作超过$10000$小时的件数.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题33", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017519": { + "id": "017519", + "content": "2023 年春节期间, 我国高速公路继续执行``节假日高速免费政策'', 某路桥公司为掌握春节期间车辆出行的高皘情况, 在某高速收费点处记录了大年初三上午$9: 20 \\sim 10: 40$这一时间段内通过的车辆数, 统计发现这一时间段内共有$600$辆车通过该收费点, 它们通过该收费点的时刻的频率分布直方图如图所示, 其中时间段$9: 20 \\sim 9: 40$记作区间$[20,40)$, $9: 40 \\sim 10: 00$记作$[40,60)$, $10: 00 \\sim 10: 20$记作$[60, 80)$, $10: 20 \\sim 10: 40$记作$[80, 100]$, 比方: $10$点$04$分, 记作时刻$64$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.06, yscale = 100]\n\\draw [->] (0,0) -- (120,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,0.03) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (0,0) node [below left] {$O$};\n\\foreach \\i/\\j in {20/0.005,40/0.015,60/0.020,80/0.010}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (20,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {20/0.005,40/0.015,60/0.020,80/0.010}\n{\\draw [dashed] (\\i,\\j) -- (0,\\j) node [left] {$\\k$};};\n\\draw (100,0) node [below] {$100$};\n\\end{tikzpicture}\n\\end{center}\n(1) 估计这$600$辆车在$9: 20 \\sim 10: 40$时间段内通过该收费点的时刻的平均值(同一组中的数据用该组区间的中点值代表);\\\\\n(2) 为了对数据进行分析, 现采用分层抽样的方法从这$600$辆车中抽取$10$辆, 再从这$10$辆车中随机抽取$4$辆, 记$X$为$9: 20\\sim 10: 00$之间通过的车辆数, 求$X$的分布列与数学期望;\\\\\n(3) 由大数据分析可知, 车辆在春节期间每天通过该收费点的时刻$T$服从正态分布$N(\\mu, \\sigma^2)$, 其中$\\mu$可用这$600$辆车在$9: 20\\sim 10: 40$之间通过该收点的时刻的平均值近似代替, $\\sigma^2$可用样本的方差近似代替 (同一组中的数据用该组区间的中点值代表), 已知大年初五全天共有$1000$辆车通过该收费点, 估计在 $9: 46\\sim 10: 40$之间通过的车辆数 (结果保留到整数).\\\\\n参考数据: 若$T \\sim N(\\mu, a^2)$; 则$P(\\mu-\\sigmap_2>p_1>0$, 记该棋手连胜两盘的概率为$p$, 则\\bracket{20}.\n\\twoch{$p$与该棋手和甲、乙、丙的比赛次序无关}{该棋手在第二盘与甲比赛, $p$最大}{该棋手在第二盘与乙比赛, $p$最大}{该棋手在第二盘与丙比赛, $p$最大}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题38", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017524": { + "id": "017524", + "content": "一种微生物群体可以经过自身繁殖不断生存下来, 设一个这种微生物为第$0$代, 经过一次繁蝒后为第$1$代, 再经过一次繁殖后为第$2$代, $\\cdots$, 该微生物每代繁殖的个数是相互独立的且有相同的分布列, 设$X$㕈示$1$个微生物个体繁殖下一代的个数, $P(X=i)=p_t$($i=0,1,2,3$).\\\\\n(1) 已知$p_0=0.4$, $p_1=0.3$, $p_2=0.2$, $p_3=0.1$, 求$E[X]$;\\\\\n(2) 设$p$表示该种微生物经过多代繁殖后临近灭绝的概率, $p$是关于$x$的方程: $p_0+p_1 x+p_2 x^2+p_3 x^3=x$的最小正实根, 求证: 当$E[X] \\leq 1$时, $p=1$, 当$E[X]>1$时, $p<1$;\\\\\n(3) 根据你的理解说明(2)问结论的实际含义.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题39", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017525": { + "id": "017525", + "content": "从某企业生产的某种产品中随机抽取$100$件, 测量这些产品的一项质量指标值, 由测量表得如下频数分布表:\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline 质量指标值 & {$[70,80)$} & {$[80,90)$} & {$[90,100)$} & {$[100,110)$} &$[110,120]$\\\\\n\\hline 频数 & 14 & 20 & 36 & 18 & 12 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n估计这种产品质量指标值的平均数为(同一组中的数据用该组区间的中点值作代表) \\bracket{20}.\n\\fourch{$100$}{$98.8$}{$96.6$}{$94.4$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题40", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017526": { + "id": "017526", + "content": "某公司生产三种型号的轿车, 产量分别为$120$辆、 $600$辆和$200$辆.为检验该公司的产品质量, 先用分层抽样的方法抽取$23$辆进行检验, 这三种型号的轿车应分别抽取\\blank{50}, \\blank{50}和\\blank{50}辆.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题41", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017527": { + "id": "017527", + "content": "为了解某校高三学生的视力情况, 随机抽查了该校$100$名高三学生的视力情况, 得到频率分布直方图如图所示, 由于不慎将部分数据丢失, 仅知道后$5$组的频数和为$62$. 设视力在$4.6$到$4.8$之间的学生数为$m$, 最大频率为$0.32$, 则$m$的值为\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 7, yscale = 1]\n\\draw [->] (4.2,0) -- (4.25,0) -- (4.26,0.2) -- (4.28,-0.2) -- (4.29,0)-- (5.3,0) node [below] {视力};\n\\draw [->] (4.2,0) -- (4.2,3.6) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (4.2,0) node [below left] {$O$};\n\\foreach \\i/\\j in {4.4/0.5,4.5/1.1,4.6/2.2,4.7/3.2,4.8/1.6,4.9/1.1,5.0/0.5,5.1/0.2}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (0.1,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {4.4/0.5,4.5/1.1}\n{\\draw [dashed] (\\i,\\j) -- (4.2,\\j) node [left] {$\\k$};};\n\\draw (5.2,0) node [below] {$5.2$};\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$27$}{$48$}{$54$}{$64$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题42", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017528": { + "id": "017528", + "content": "如图为甲、乙两位同学在$5$次数学测试中得分的茎叶图, 则平均成绩较小的那位同学的成绩的方差为\\bracket{20}.\n\\begin{center}\n\\begin{tabular}{ccc|c|ccc}\n\\multicolumn{3}{r|}{甲} & & \\multicolumn{3}{l}{乙} \\\\\n& 9 & 8 & 8 & 7 & 9\\\\\n2 & 1 & 0 & 9 & 0 & 1 & 8\n\\end{tabular}\n\\end{center}\n\\fourch{$1$}{$2$}{$3$}{$4$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题43", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017529": { + "id": "017529", + "content": "下列数据的第$10$百分位数是\\blank{50}; 第$35$百分位数是\\blank{50}.\n\\begin{center}\n\\begin{tabular}{ccccc}\n10 & 10 & 10 & 12 & 14 \\\\\n15 & 15 & 23 & 35 & 60\n\\end{tabular}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题44", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017530": { + "id": "017530", + "content": "变量$x, y$之间的一组相关数据如表所示: 若$x, y$之间的线性回归方程为$y=\\hat{a} x+12.28$, 则$\\hat{a}$的值为\\bracket{20}.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline$x$& 4 & 5 & 6 & 7 \\\\\n\\hline$y$& 8.2 & 7.8 & 6.6 & 5.4 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\fourch{$-0.92$}{$-0.94$}{$-0.96$}{$-0.98$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题45", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017531": { + "id": "017531", + "content": "某校抽取了$50$名高三年级学生, 测量他们的身高数据. 但由于某种原因这些原始样本数据不可查得, 但已知按照分层随机抽样原则抽取了样本, 其中男生$21$名, 身高样本平均数为$174.8 \\text{cm}$, 方差为$20.8$; 女生$29$名, 身高样本平均数为$163.1 \\text{cm}$, 方差为$18.4$. 试用这些已知的数据求该$50$名高三年级学生身高的样本平均数和方差, 并估计高三年级学生身高的总体方差. (结果精确到$0.01$)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题46", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017532": { + "id": "017532", + "content": "某校举行了一次数学竞赛, 为了了解本次竞赛学生的成绩情况, 从中抽取了部分学生的分数 (得分取正整数, 满分为$100$分) 作为样本 (样本容量为$n$) 进行统计, 按照$[50,60)$, $[60,70)$, $[70,80)$, $[80,90)$, $[90,100]$的分组作出频率分布直方图, 已知得分在$[50,60)$, $[90,100]$的频数分别为$16$, $4$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.1, yscale = 60]\n\\draw [->] (35,0) -- (110,0) node [below] {成绩(分)};\n\\draw [->] (35,0) -- (35,0.055) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (35,0) node [below left] {$O$};\n\\foreach \\i/\\j in {50/0.016,60/0.030,70/0.040,80/0.010,90/0.004}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (10,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {50/0.016,60/0.030/a,70/0.040,80/0.010,90/0.004/b}\n{\\draw [dashed] (\\i,\\j) -- (35,\\j) node [left] {$\\k$};};\n\\draw (100,0) node [below] {$100$};\n\\end{tikzpicture}\n\\end{center}\n(1) 求样本容量$n$和频率分布直方图中的$a, b$的值;\\\\\n(2) 估计本次竞赛学生成绩的平均数 (同一组中的数据用该组区间的中点值代表);\\\\\n(3) 在选取的样本中, 若男生和女生人数相同, 我们规定成绩在$70$分或以上称为``优秀'', $70$分以下称为``不优秀'', 其中男女生中成绩优秀的分别有$24$人和$30$人, 请完成列联表, 并判断是否有$90 \\%$的把控认为``学生的成绩优秀与否与性别有关''?\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|}\n\\hline & 男生 & 女生 & 总计 \\\\\n\\hline 优秀 & & & \\\\\n\\hline 不优秀 & & & \\\\\n\\hline 总计 & & & \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline$P(\\chi^2 \\geq k)$& 0.10 & 0.05 & 0.010 & 0.005 & 0.001 \\\\\n\\hline$k$& 2706 & 3.841 & 6.635 & 7.879 & 10.828 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n附: $\\chi^2=\\dfrac{\\dot{n}(a d-b c)^2}{(a+b)(c+d)(a+c)(b+d)}$, 其中$n=a+b+c+d$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题47", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017533": { + "id": "017533", + "content": "由于疫情, 学生在家经过了几个月的线上学习, 某高中学校为了了解学生在家学习情况, 复学后进行了复学摸底考试, 并对学生进行了问卷调查, 如表 (单位: 人) 是对高二年级数学成绩及``认为自己在家学习态度是否端正''的问卷调查的统计结果, 其中成绩不低于$120$分为优秀, 成绩不低于$90$分且小于$120$分的为及格, 成绩小于$90$分的为不及格.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|}\n\\hline & 优秀 & 及格 & 不及格 \\\\\n\\hline 学习态度端正 & 91 & 300 &$a$\\\\\n\\hline 学习态度不端正 & 9 & 200 & 322 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n按成绩用分层抽样的方法在高二年级中抽取$50$人, 其中优秀的人数为$5$.\\\\\n(1) 求$a$的值;\\\\\n(2) 用分层抽样的方法在及格的学生中抽取一个容量为$5$的样本. 将该样本看成一个总体, 从中任取$2$人, 求至少有$1$人学习不端正的概率;\\\\\n(3) 在及格的学生中随机抽取了$10$人, 他们的分数如茎叶图所示, 已知这$10$名学生的平均分为$104.5$, 求$a>b$的概率.\n\\begin{center}\n\\begin{tabular}{c|cccc}\n9 & 2 & $a$ & $b$ \\\\\n10 & 0 & 5 & 6 & 8 \\\\\n11 & 3 & 6 & 7\n\\end{tabular}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题48", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017534": { + "id": "017534", + "content": "我国是世界上严重缺水的宝家之一, 城市缺水问题较为突出, 某市政府为了节约生活用水, 计划在本市试行居民生活用水定额管理, 即确定一个居民月用水墨标准$x$, 用水量不超过$x$的部分按平价收费, 超出$x$的部分按议价收费. 下面是居民去均用水量的抽样频率分布直方图.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 1.5, yscale = 7]\n\\draw [->] (0,0) -- (5.5,0) node [below] {月均用水量(吨)};\n\\draw [->] (0,0) -- (0,0.63) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\foreach \\i/\\j in {0/0.08,0.5/0.16,1/0.3,1.5/0.4,2/0.52,2.5/0.3,3/0.12,3.5/0.08,4/0.04}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (0.5,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {0.5/0.16,1.5/0.4,2/0.52,2.5/0.3/a,3/0.12,3.5/0.08,4/0.04}\n{\\draw [dashed] (\\i,\\j) -- (0,\\j) node [left] {$\\k$};};\n\\draw (4.5,0) node [below] {$4.5$};\n\\end{tikzpicture}\n\\end{center}\n(1) 求直方图中$a$的值;\\\\\n(2) 试估计该市居民月均用水量的众数, 平均数;\\\\\n(3) 设该市有$30$万居民, 估计全市居民中月均用水量不低于$3$吨的人数, 并说明理由;\\\\\n(4) 如果希望$85 \\%$的居民月均用水量不超过标准$x$, 那么标准$x$定为多少比较合理?", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题49", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017535": { + "id": "017535", + "content": "垃圾是人类日常生活和生产中产生的废弃物, 由于排出量大, 成分复杂多样, 且具有污染性, 所以需要无害化、减量化处理. 某市为调查产生的垃圾数量, 采用简单随机抽样的方法抽取$20$个县城进行了分析, 得到样本数据$(x_i, y_i)$($i=1,2, \\cdots, 20)$, 其中$x_i$和$y_i$分别表示第$i$个县城的人口 (单位: 万人) 和该县年垃圾产生总量 (单位: 吨), 并计算得$\\displaystyle\\sum_{i=1}^{20} x_i=80$, $\\displaystyle\\sum_{i=1}^{20} y_i=4000$, $\\displaystyle\\sum_{i=1}^{20}(x_i-\\overline {x})^2=80$, $\\displaystyle\\sum_{i=1}^{20}(y_i-\\overline {y})^2=8000$, $\\displaystyle\\sum_{i=1}^{20}(x_i-\\overline {x})(y_i-\\overline {y})=7000$.\\\\\n(1) 请用相关系数说明该组数据中$y$与$x$间的关系可用线性回归模型进行拟合;\\\\\n(2) 求$y$关于$x$的线性回归方程;\\\\\n(3) 某科研机构研发了两款垃圾处理机器, 如表是以往两款垃圾处理机器的使用年限 (整年) 统计表:\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline \\backslashbox{款式}{台数}{使用年限} & 1 年 & 2 年 & 3 年 & 4 年 & 5 年 \\\\\n\\hline 甲款 & 5 & 20 & 15 & 10 & 50 \\\\\n\\hline 乙款 & 15 & 20 & 10 & 5 & 50 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n某环保机构若考虑购买其中一款垃圾处理器, 以使用年限的频率估计概率. 根据以往经验估计, 该机构选择购买哪一款垃圾处理机器, 才能使用更长久?\n参考公式: 相关系数$r=\\dfrac{\\displaystyle\\sum_{i=1}^n(x_i-\\overline {x})(y_i-\\overline {y})}{\\displaystyle\\sqrt{\\sum_{i=1}^n(x_i-\\overline {x})^2} \\sqrt{\\displaystyle\\sum_{i=1}^n(y_i-\\overline {y})^2}}$. 对于一组具有线性相关关系的数据$(x_i, y_i)$($i=1,2, \\cdots, n$), 其回归直线$y=\\hat{a} x+\\hat{b}$的斜率和截距\n的最小二乘估计分别为: $\\hat{a}=\\dfrac{\\displaystyle\\sum_{i=1}^n(x_i-\\overline {x})(y_i-\\overline {y})}{\\displaystyle\\sum_{i=1}^n(x_i-\\overline {x})^2}$, $\\hat{b}=\\overline {y}-\\hat{a} \\overline {x}$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题50", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017536": { + "id": "017536", + "content": "已知回归方程$y=5 x+1$, 则该方程在样本$(1, 4)$处的离差为\\bracket{20}.\n\\fourch{$-2$}{$1$}{$2$}{$5$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题51", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017537": { + "id": "017537", + "content": "某歌唱兴趣小组由$15$个编号为$01,02, \\cdots, 15$的学生个体组成, 现要从中选取$3$名学生参加合唱团, 选取方法是从随机数表的第$1$行的第$18$列开始由左往右依次选取两个数字, 则选出来的第$3$名同学的编号为\\bracket{20}.\n\\begin{center}\n\\begin{tabular}{cccccccccccccccccc}\n49 & 54 & 43 & 54 & 82 & 17 & 37 & 93 & 23 & 78 & 30 & 35 & 20 & 96 &23 & 84 & 26 & 34 \\\\\n91 & 64 & 50 & 25 & 83 & 92 & 12 & 06 & 76 & 57 & 23 & 55 & 06 & 88 & 77 & 04 & 74 & 47 \\\\\n67 & 21 & 76 & 33 & 50 & 25 & 83 & 92 & 12 & 06 & 76 & 49 & 54 & 43 & 54 & 82 & 74 & 47\n\\end{tabular}\n\\end{center}\n\\fourch{$02$}{$09$}{$12$}{$03$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题52", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017538": { + "id": "017538", + "content": "某校从高一年级学生中随机抽取部分学生, 将他们的模块测试成绩分为$6$组:\n$[40,50)$, $[50,60)$, $[60,70)$, $[70,80)$, $[80,90)$, $[90,100]$加以统计; 得到如图所示的频率分布直方图, 已知高一年级共有学生$600$名, 据此估计, 该模块测试成绩不少于$60$分的学生人数为\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.08, yscale = 100]\n\\draw [->] (30,0) -- (32,0) -- (33,0.002) -- (35,-0.002) -- (36,0) -- (115,0) node [below] {分数};\n\\draw [->] (30,0) -- (30,0.04) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (30,0) node [below left] {$O$};\n\\foreach \\i/\\j in {40/0.005,50/0.015,60/0.030,70/0.025,80/0.015,90/0.010}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (10,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {40/0.005,60/0.030,70/0.025,80/0.015,90/0.010}\n{\\draw [dashed] (\\i,\\j) -- (30,\\j) node [left] {$\\k$};};\n\\draw (100,0) node [below] {$100$};\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$588$}{$480$}{$450$}{$120$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题53", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017539": { + "id": "017539", + "content": "某校对甲、乙两个数学兴趣小组的同学进行了知识测试, 现从两兴趣小组的成员中各随机选取$15$人的测试成绩(单位: 分) 用茎叶图表示, 如图, 根据茎叶图, 对甲, 乙两兴趣小组的测试成绩作比较, 下列统计结论正确的有\\blank{50}.\n\\begin{center}\n\\begin{tabular}{ccccc|c|ccc}\n\\multicolumn{5}{r|}{甲} & & \\multicolumn{3}{l}{乙}\\\\\n&&&3&6&9&2&1\\\\\n&&2&5&8&8&4&6\\\\\n2&5&7&9&9&7&2&5&7\\\\\n&1&3&5&8&6&1&4&4\\\\\n&&&&9&5&2&3&6\\\\\n&&&&&4&6&9\n\\end{tabular}\n\\end{center}\n\\textcircled{1} 甲兴趣小组测试成绩的平均分高于乙兴趣小组测试成绩的平均;\\\\\\textcircled{2} 甲兴趣小组测试成绩较乙兴趣小组测试成绩更分散;\\\\ \\textcircled{3} 甲兴趣小组测试成绩的中位数大于乙兴趣小组测试成绩的中位数;\\\\ \\textcircled{4} 甲兴趣小组测试成绩的众数小于乙兴趣小组测试成绩的众数.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题54", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017540": { + "id": "017540", + "content": "已知$100$个数据的$75$百分位数是$9.3$, 则下列说法正确的是\\bracket{20}.\n\\onech{这$100$个数据中恰有$75$个数小于或等于$9.3$}{把这$100$个数据从小到大排列后, $9.3$是第$75$个数据}{把这$100$个数据从小到大排列后, $9.3$是第$75$个数据和第$76$个数据的平均数}{把这$100$个数据从小到大排列后, $9.3$是第$75$个数据和第$74$个数据的平均数}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题55", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017541": { + "id": "017541", + "content": "蟋蛒鸣叫可以说是大自然优美、和谐的音乐, 殊不知蟋蜶鸣叫的频率$x$(每分钟鸣叫的次数) 与气温$y$(单位: $^{\\circ} \\mathrm{C}$) 存在较强的线性相关关系. 某地观测人员根据如表的观测数据, 建立了$y$关于$x$的线性回归方程$y=0.25 x+k$, 则当蟋蟀每分钟鸣叫$56$次时, 该地当时的气温预报值为\\bracket{20}.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline$x$(次数/分钟) & 20 & 30 & 40 & 50 & 60 \\\\\n\\hline$y$($^\\circ \\mathrm{C}$)& 25 & 27.5 & 29 & 32.5 & 36 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\fourch{$33^{\\circ} \\mathrm{C}$}{$34^{\\circ} \\mathrm{C}$}{$35^{\\circ} \\mathrm{C}$}{$35.5^{\\circ} \\mathrm{C}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题56", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017542": { + "id": "017542", + "content": "张老师将某位高三学生$10$次选填题专测的成绩进行统计, 得到的统计结果如图所示, 但学习委员在将成绩登记在册的时候将$62$与$68$均登记成了$65$, 则两个成绩相比, 不变的数字特征是\\bracket{20}.\n\\begin{center}\n\\begin{tabular}{c|cccc}\n5 & 3 & 5 & 7\\\\\n6 & 2 & 3 & 5 & 8\\\\\n7 & 5 & 5 & 7\n\\end{tabular}\n\\end{center}\n\\fourch{众数}{中位数}{平均数}{方差}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题57", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017543": { + "id": "017543", + "content": "为了考察某校各班参加课外书法小组的人数, 在全校随机抽取$5$个班级, 把每个班级参加该小组的人数作为样本数据. 已知样本平均数为$7$, 样本方差为$4$, 且样本数据互不相同, 求样本数据中的最大值.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题58", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017544": { + "id": "017544", + "content": "由正整数组成的一组数据$x_1, x_2, x_3, x_4$, 其平均数和中位数都是$2$, 且标准差等于$1$, 则这组数据为\\blank{50}(从小到大排列).", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题59", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017545": { + "id": "017545", + "content": "在发生某公共卫生事件期间, 有专业机构认为该事件在一段时间没有发生在规模群体感染的标志为``连续$10$天, 每天新增颢似病例不超过$7$人''. 根据过去$10$天甲、乙、丙、丁四地新增疑似病例数据, 一定符合该标志的是\\bracket{20}.\n\\twoch{甲地: 总体均值为$3$, 中位数为$4$}{乙地: 总体均值为$1$, 总体方差大于$0$}{主地: 中位数为$2$, 众数为$3$}{丁地: 总体均值为$2$, 总体方差为$3$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题60", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017546": { + "id": "017546", + "content": "近年来, 随着互眹网技术的快速发展, 共享经济覆盖的范围迅速扩张, 继共享单车、共享汽车之后, 共享房屋以``民宿''、``农家乐''等形式开始在很多平台上线. 某创业者计划在某景区附近租赁一套农房发展成特色``农家乐'', 为了确定未来发展方向, 此创业者对该景区附近六家``农家乐''跟踪调查了$100$天. 得到的统计数据如下表, $x$为收费标准 (单位: 元/日), $t$为入住天数 (单位: 天), 以频率作为各自的``入住率'', 收费标准$x$与``入住率''$y$的散点图如图\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|c|}\n\\hline $x$& 50 & 100 & 150 & 200 & 300 & 400 \\\\\n\\hline $t$& 90 & 65 & 45 & 30 & 20 & 20 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (6,0) node [below] {收费标准};\n\\draw [->] (0,0) -- (0,3.3) node [left] {入住率};\n\\draw (0,0) node [below left] {$O$};\n\\foreach \\i in {100,200,300,400,500}\n{\\draw ({\\i/100},0.1) -- ({\\i/100},0) node [below] {$\\i$};};\n\\foreach \\i in {0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1}\n{\\draw [gray!50] (5,{\\i*3}) --++ (-5,0) node [left, black] {$\\i$};};\n\\foreach \\i/\\j in {50/0.9,100/0.65,150/0.45,200/0.3,300/0.2,400/0.2}\n{\\filldraw ({\\i/100},{\\j*3}) circle (0.03);};\n\\end{tikzpicture}\n\\end{center}\n(1) 若从以上六家``农家乐''中随机抽取两家深入调查, 记$\\xi$为``入住率''超过$0.6$的农家乐的个数, 求$\\xi$的概率分布列;\\\\\n(2) 令$z=\\ln x$, 由散点图判断$y=\\hat{a} x+\\hat{b}$与$y=\\hat{a} z+\\hat{b}$哪个更合适于此模型(给出判断即可, 不必说明理由)?, 并根据你的判断结果求回归方程; ($\\hat{a}$结果保留一位小数)\\\\\n(3) 若一年按$365$天计算, 试估计收费标准为多少时, 年销售额$L$最太? (年销售额$L=365\\times$入住率$\\times$收费标准$x$)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题61", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017547": { + "id": "017547", + "content": "某种疾病可分为I、II两种类型, 为了解该疾病类型与性别的关系, 在某地区随机抽取了患该疾病的病人进行调查, 其中男性人数为$z$, 女性人数为$2 z$, 男性患I型病的人数占男性病人的$\\dfrac{5}{6}$, 女性患I型病的人数占性病人的$\\dfrac{1}{3}$.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|}\n\\hline & I 型病 & II型病 & 合计 \\\\\n\\hline 男 & & & \\\\\n\\hline 女 & & & \\\\\n\\hline 合计 & & & \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n(1) 完成$2 \\times 2$联表; 若在犯错误的概率不超过$0.005$的前提下认为``所患疾病类型''与``性别''有关, 求男性患者至少有多少人?\\\\\n(2) 某药品研发公司欲安排甲乙两个研发团队来研发此疾病的治疗药物, 两个团队各至多安排$2$个接种周期进行试验. 每人每次接种花费$m$($m>0$)元. 甲团队研发的药物每次接种后产生抗体的概率为$p$; 根据以往试验统计, 甲团队平均花费为$-2 m p^2+6 m$; 乙团队研发的药物每次接种后产生抗体的概率为$q$, 每个周期必须完成$3$次接种, 若一个周期内至少出现$2$次抗体, 则该周期结束后终止试验, 否则进入第三个接种周期. 假设两个研发团队每次接种后产生抗体与否均相互独立. 若$p=2 q$, 从两个团队试验的平均花费考虑, 该公司应选择哪个团队进行药品研发?\\\\\n附: $\\chi^2=\\dfrac{n(a d-b c)^2}{(a+b)(c+d)(a+c)(b+d)}$.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline $P(\\chi^2 \\geq k_0)$& 0.10 & 0.05 & 0.01 & 0.005 & 0.001 \\\\\n\\hline $k_0$& 2.706 & 3.841 & 6.635 & 7.879 & 10.828 \\\\\n\\hline\n\\end{tabular}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题62", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017548": { + "id": "017548", + "content": "为了解某地农村经济情况, 对该地农户家庭年收入进行抽样调查, 将农户家庭年收入的调查数据整理得到如下频率分布直方图. 根据此频率分布直方图, 下面结论中不正确的是\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, xscale = 0.8, yscale = 14]\n\\draw [->] (1.5,0) -- (16,0) node [below] {收入/万};\n\\draw [->] (1.5,0) -- (1.5,0.26) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw (1.5,0) node [below left] {$O$};\n\\foreach \\i/\\j in {2.5/0.02,3.5/0.04,4.5/0.10,5.5/0.14,6.5/0.20,7.5/0.20,8.5/0.10,9.5/0.10,10.5/0.04,11.5/0.02,12.5/0.02,13.5/0.02}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (1,0) --++ (0,-\\j);};\n\\foreach \\i/\\j/\\k in {5.5/0.14,6.5/0.20,8.5/0.10,10.5/0.04,11.5/0.02}\n{\\draw [dashed] (\\i,\\j) -- (1.5,\\j) node [left] {$\\k$};};\n\\draw (14.5,0) node [below] {$14.5$};\n\\end{tikzpicture}\n\\end{center}\n\\onech{该地农户家庭年收入低手 4.5 万元的农户比率估计为$6 \\%$}{该地农户家庭年收入不低于 10.5 万元的农户比率估计为$10 \\%$}{估计该地农户家庭年收入的平均值不超过 6.5 万元}{估计该地有一半以上的农户, 其家庭年收入介于 4.5 万元至 8.5 万元之间\\blank{50}}", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题63", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "017549": { + "id": "017549", + "content": "为了解甲、乙两种离子在小鼠体内的残留程度, 进行如下试验: 将$200$只小鼠随机分成 A、B 两组, 每组$100$只, 其中 A 组小鼠给服甲离子溶液, B 组小鼠给服乙离子溶液, 每组小鼠给服的溶液体积相同、摩尔浓度相同. 经过一段时间后用某种科学方法测算出残留在小鼠体内离子的百分比. 根据试验数据分别得到如下直方图:\n\\begin{center}\n\\begin{tikzpicture}[>=latex,xscale = 0.6, yscale = 8]\n\\draw [->] (0,0) -- (0,0.45) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw [->] (0,0) -- (0.1,0) -- (0.2,-0.02) -- (0.4,0.02) -- (0.6,-0.02) -- (0.8,0) -- (9.5,0) node [below] {百分比};\n\\draw (0,0) node [below left] {$O$};\n\\foreach \\i/\\j/\\k in {6.5/0.05/0.05,5.5/0.1/0.1,1.5/0.15/0.15,4.5/0.2/0.2,3.5/0.3/0.3}\n{\\draw [dashed] (\\i,\\j) -- (0,\\j) node [left] {$\\k$};};\n\\foreach \\i/\\j/\\k in {1.5/0.15/0.15,2.5/0.2/0.2,3.5/0.3/0.3,4.5/0.2/0.2,5.5/0.1/0.1,6.5/0.05/0.05}\n{\\draw (\\i,0) node [below] {$\\i$} --++ (0,\\j) --++ (1,0) --++ (0,-\\j);\n};\n\\draw (7.5,0) node [below] {$7.5$};\n\\draw (4.5,-0.1) node {甲离子残留百分比直方图};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex,xscale = 0.6, yscale = 8]\n\\draw [->] (0,0) -- (0,0.45) node [left] {$\\dfrac{\\text{频率}}{\\text{组距}}$};\n\\draw [->] (0,0) -- (0.1,0) -- (0.2,-0.02) -- (0.4,0.02) -- (0.6,-0.02) -- (0.8,0) -- (9.5,0) node [below] {百分比};\n\\draw (0,0) node [below left] {$O$};\n\\foreach \\i/\\j/\\k in {2.5/0.05/0.05,3.5/0.1/b,7.5/0.15/0.15,6.5/0.2/0.2,5.5/0.35/a}\n{\\draw [dashed] ({\\i-1},\\j) -- (0,\\j) node [left] {$\\k$};};\n\\foreach \\i/\\j/\\k in {2.5/0.05/0.05,3.5/0.1/0.1,4.5/0.15/0.15,5.5/0.35/0.35,6.5/0.2/0.2,7.5/0.15/0.15}\n{\\draw ({\\i-1},0) node [below] {$\\i$} --++ (0,\\j) --++ (1,0) --++ (0,-\\j);\n};\n\\draw (7.5,0) node [below] {$8.5$};\n\\draw (4.5,-0.1) node {乙离子残留百分比直方图};\n\\end{tikzpicture}\n\\end{center}\n记$C$为事件: ``乙离子残留在体内的百分比不低于$5.5$'', 根据直方图得到$P(C)$的估计值为$0.70$.\\\\\n(1) 求乙离子残留百分比直方图中$a, b$的值;\\\\\n(2) 分别估计甲、乙离子残留百分比的平均值(同一组中的数据用该组区间的中点值为代表).", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题64", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "017550": { + "id": "017550", + "content": "在校运动会上, 只有甲、乙、丙三名同学参加铅球比赛, 比赛成绩达到$9.50 \\text{m}$上 (含$9.50 \\text{m}$) 的同学将获得优秀奖. 为预测获得优秀奖的人数及冠军得主, 收集了甲、乙、丙以往的比赛成绩, 并整理得到如下数据 (单位: $\\text{m}$):\\\\\n甲: $9.80,9.70,9.55,9.54,9.48,9.42,9.40,9.35,9.30,9.25$;\\\\\n乙: $9.78,9.56,9.51,9.36,9.32,9.23$\\\\\n两: $9.85,9.65,9.20,9.16$.\\\\\n假设用频率估计概率, 且甲、乙、丙的比赛成绩相互独立.\\\\\n(1) 估计甲在校运动会铅球比赛中获得优秀奖的概率;\\\\\n(2) 设$X$是甲、乙、丙在校运动会铅球比赛中获得优秀奖的总人数, 估计$X$的数学期望$E[X]$;\\\\\n(3) 在校运动会铅球比赛中, 甲、乙、丙谁获得冠军的概率估计值最大? (结论不要求证明)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "杨高概率统计复习试题65", + "edit": [ + "20230602\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, "020001": { "id": "020001", "content": "判断下列各组对象能否组成集合, 若能组成集合, 指出是有限集还是无限集.\\\\\n(1) 上海市控江中学$2022$年入学的全体高一年级新生;\\\\\n(2) 中国现有各省的名称;\\\\\n(3) 太阳、$2$、上海市;\\\\\n(4) 大于$10$且小于$15$的有理数;\\\\\n(5) 末位是$3$的自然数;\\\\\n(6) 影响力比较大的中国数学家;\\\\\n(7) 方程$x^2+x-3=0$的所有实数解;\\\\ \n(8) 函数$y=\\dfrac 1x$图像上所有的点;\\\\ \n(9) 在平面直角坐标系中, 到定点$(0, 0)$的距离等于$1$的所有点;\\\\\n(10) 不等式$3x-10<0$的所有正整数解;\\\\\n(11) 所有的平面四边形.",