{"msg":"第二节 概率统计应用","code":200,"data":{"currentIndex":null,"examId":null,"examTime":null,"questionList":[{"id":"796239485191868417","title":"<p><strong>请作答第<span style=\"color: red\">1</span>个空。</strong></p>根据近几个月的数据统计，某车次火车到站晚点时间t (分钟）的概率分布密度函数可用函数<img alt=\"\" width=\"100\" height=\"17\" src=\"https://image.chaiding.com/ruankao/366f32c1661da9b5fd4e9b2c3535260a.jpg?x-oss-process=style/ruankaodaren\">)来描述，因此可以计算出其中的待定系数k= (54), 晚点超过5分钟的概率为（55）。","analyze":"本题考查应用数学（概率统汁）知识。<br>本题中，某次列车的晚点时间t是随机变量<img alt=\"\" width=\"50\" height=\"17\" src=\"https://image.chaiding.com/ruankao/cdbdb229c5e52a1566cc21a90cb788e3.jpg?x-oss-process=style/ruankaodaren\">，其分布密度函数f(t)意味着晚点时间在<img alt=\"\" width=\"45\" height=\"15\" src=\"https://image.chaiding.com/ruankao/79eee8f9237bc89ac86da2a9e519bd1c.jpg?x-oss-process=style/ruankaodaren\">这个时间段内的概率为<img alt=\"\" width=\"35\" height=\"17\" src=\"https://image.chaiding.com/ruankao/9402bc946d649ee5769148e44346eb15.jpg?x-oss-process=style/ruankaodaren\">。由于总概率为1，因此<br><img alt=\"\" width=\"150\" height=\"43\" src=\"https://image.chaiding.com/ruankao/e06f0a4a1ce921f1047613f85b9c44f4.jpg?x-oss-process=style/ruankaodaren\"><br>从而k=0.003。晚点时间超过5分钟的概率为<br><img alt=\"\" width=\"180\" height=\"30\" src=\"https://image.chaiding.com/ruankao/d68bdb87f0ee1a7fad2ce7d3ab3af85b.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"A","chooseItem":["796239486265610241"],"itemList":[{"id":"796239486265610241","questionId":"796239485191868417","content":" 0.003","answer":1,"chooseValue":"A"},{"id":"796239486282387457","questionId":"796239485191868417","content":" 0.03","answer":0,"chooseValue":"B"},{"id":"796239486303358977","questionId":"796239485191868417","content":" 0.3","answer":0,"chooseValue":"C"},{"id":"796239486320136193","questionId":"796239485191868417","content":" 3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239472374075393","title":"假设一个I/O系统只有一个磁盘，每秒可以接收50个I/O请求，磁盘对每个I/O请求服务的平均时间是10ms,则I/O请求队列的平均长度是（39）个请求。","analyze":"磁盘的I/O请求是一个随机过程，请求事件达到的时间间隔具有泊松分布的概率学特征。根据Little定律，平均队列长度=达到速率×平均等待时间。其中平均等待时间=平均服务时间X服务器利用率/(1-服务器利用率）<br>而服务器利用率=到达速率×平均服务时间，所以平均队列长度=服务器利用率×服务器利用率/(1-服务器利用率）<br>根据本题给出的相关数据，服务器利用率=1/50x0.01=0.5,因此平均队列长度等于0.5。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239473397485569"],"itemList":[{"id":"796239473363931137","questionId":"796239472374075393","content":" 0","answer":0,"chooseValue":"A"},{"id":"796239473397485569","questionId":"796239472374075393","content":" 0.5","answer":1,"chooseValue":"B"},{"id":"796239473426845697","questionId":"796239472374075393","content":" 1","answer":0,"chooseValue":"C"},{"id":"796239473456205825","questionId":"796239472374075393","content":" 2","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239488421482497","title":"<p><strong>请作答第<span style=\"color: red\">2</span>个空。</strong></p>根据近几个月的数据统计，某车次火车到站晚点时间t (分钟）的概率分布密度函数可用函数<img alt=\"\" width=\"100\" height=\"17\" src=\"https://image.chaiding.com/ruankao/7ecc2ad8d6596efa1e9202bec8d8a4a4.jpg?x-oss-process=style/ruankaodaren\">)来描述，因此可以计算出其中的待定系数k= (54), 晚点超过5分钟的概率为（55）。","analyze":"本题考查应用数学（概率统汁）知识。<br>本题中，某次列车的晚点时间t是随机变量<img alt=\"\" src=\"https://image.chaiding.com/ruankao/68c5e7056f28c1c9042e22d5bdd05930.jpg?x-oss-process=style/ruankaodaren\" height=\"15\" width=\"42\">，其分布密度函数f(t)意味着晚点时间在<img alt=\"\" width=\"40\" height=\"13\" src=\"https://image.chaiding.com/ruankao/47452bd7bb2c2f7b5fbda3064ddec271.jpg?x-oss-process=style/ruankaodaren\">个时间段内的概率为<img alt=\"\" width=\"40\" height=\"20\" src=\"https://image.chaiding.com/ruankao/97e7d59ba051ff854fe82a4bd9505472.jpg?x-oss-process=style/ruankaodaren\">。由于总概率为1，因此<br><img alt=\"\" width=\"150\" height=\"43\" src=\"https://image.chaiding.com/ruankao/63ce8db8f41c01391235c29b6a4c0071.jpg?x-oss-process=style/ruankaodaren\"><br>从而k=0.003。晚点时间超过5分钟的概率为<br><img alt=\"\" width=\"180\" height=\"30\" src=\"https://image.chaiding.com/ruankao/4bc1809516f6b6c0e7c7168838e8618e.jpg?x-oss-process=style/ruankaodaren\">","multi":0,"questionType":1,"answer":"C","chooseItem":["796239489419726849"],"itemList":[{"id":"796239489344229377","questionId":"796239488421482497","content":" 1/32","answer":0,"chooseValue":"A"},{"id":"796239489381978113","questionId":"796239488421482497","content":" 1/16","answer":0,"chooseValue":"B"},{"id":"796239489419726849","questionId":"796239488421482497","content":" 1/8","answer":1,"chooseValue":"C"},{"id":"796239489453281281","questionId":"796239488421482497","content":" 1/4","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239491395244033","title":"设甲乙丙三人独立解决某个问题的概率分别为0.45、0.55、0.6，则三人一起解决该问题的概率约为（53）。","analyze":"本题考查数学应用（概率）能力。<br>根据题意，三人一起无法解决该问题的概率为（1-0.45) x (1-0.55) x (1-0.6)=0.099。所以，三人一起能解决该问题的概率为1-0.099=0.901。<br>另一种解题思路是：甲解决了该问题的0.45部分，余下0.55部分没有解决。此时，乙能解决其中的0.55部分，即乙能解决总体的0.55x0.55=0.3025部分。甲乙共解决了45+0.3025=0.7525部分，余下0.2475部分没有解决。丙在其中解决了0.6,即丙解决了总体的0.2475x0.6=0.1485部分。甲乙丙三人共解决了问题0.7525+0.1485=0.901部分。","multi":0,"questionType":1,"answer":"D","chooseItem":["796239492364128257"],"itemList":[{"id":"796239492297019393","questionId":"796239491395244033","content":" 0.53","answer":0,"chooseValue":"A"},{"id":"796239492317990913","questionId":"796239491395244033","content":" 0.7","answer":0,"chooseValue":"B"},{"id":"796239492338962433","questionId":"796239491395244033","content":" 0.8","answer":0,"chooseValue":"C"},{"id":"796239492364128257","questionId":"796239491395244033","content":" 0.9","answer":1,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796239475406557185","title":"有一名患者胸部长了一个肿瘤，医院X光检查结果呈阳性。据统计，胸部肿瘤为良性的概率为99%。对良性肿瘤，X光检查的正确率（呈阴性的概率）为90%；对恶性肿瘤，X光检査的正确率（呈阳性的概率）为80%。因此，可推算出该患者患恶性肿瘤的概率是（54）。","analyze":"我们可以将胸部肿瘤的检查情况画出概率树如下：<br><img alt=\"\" width=\"519\" height=\"147\" src=\"https://image.chaiding.com/ruankao/14f57bedc477d6e9a009a3bb3492b3cb.jpg?x-oss-process=style/ruankaodaren\"><br>该树的根为“胸部肿瘤”，其性质99%的概率为良性的，1%的概率为恶性的。对于良性肿瘤，X光检查的结果，90%的概率为阴性，10%的概率为阳性；对于恶性肿瘤，X光检查的结果，80%的概率为阳性，20%的概率为阴性。<br>从“胸部肿瘤”到“X光检查结果呈阳性”的路径有以下两条：<br>胸部肿瘤→良性→X光检査结果呈阳性<br>胸部肿瘤→恶性→X光检查结果呈阳性<br>前一条路径的概率等于其各段概率之积，为99%×10%=0.099。<br>后一条路径的概率等于其各段概率之积，为1%×80%=0.008。<br>从全概率公式可知道，对于胸部肿瘤，X光检查结果呈阳性的总概率的等于所有各条路径的概率之和，所以为0.099+0.008=0.107=10.7%。<br>如果已经知道X光检查结果呈阳性，那么从前一条路径过来（属于良性）的概率为：<br>0.099/(0.099+0.008)≈0.925=92.5%<br>从后一条路径过来（属于恶性）的概率为：<br>0.008/(0.099+0.008)≈0.075=7.5%<br>这个问题的结论常出乎大家的意料，即使医生也非常惊讶。这是著名的“反问题错乱 ”（confhsion of the inverse)现象。<br>对于患某种重病的概率很低的情况，当患者检查结果偏离正常值时，这种结果在医学上称为假阳性，还需要采用其他手段才能确诊。","multi":0,"questionType":1,"answer":"B","chooseItem":["796239476446744577"],"itemList":[{"id":"796239476413190145","questionId":"796239475406557185","content":" 0.8%","answer":0,"chooseValue":"A"},{"id":"796239476446744577","questionId":"796239475406557185","content":" 7.5%","answer":1,"chooseValue":"B"},{"id":"796239476476104705","questionId":"796239475406557185","content":" 80%","answer":0,"chooseValue":"C"},{"id":"796239476505464833","questionId":"796239475406557185","content":" 75%","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235417354522625","title":"某学校希望通过问卷调查了解学生考试作弊的真实情况。若直接在问卷调查中问：“你作弊了吗？”，极少有入真实做答。为此，专家设计的问卷调查表中包括两个问题：①你是男生吗？②你作弊了吗？而每个学生需要利用给自己配发的电子随机选题器选择一题并回答“是”或“否”。学校按照学生实际的男女比例，随机选择了60名男生和40名女生参与匿名答题，而电子随机选题器选择题1和题2的概率相同。学生们认为，此次调查不但匿名，还不透露自己选择了哪题，因此都如实做答。最后，学校回收到35份回答“是”，65份回答“否”，因此计算出考试作弊的比例大致为（59）。","analyze":"本题考查应用数学基础知识（概率统计）。<br>根据题意画出概率图如下（设作弊的比例为X）:<br><img alt=\"\" width=\"338\" height=\"127\" src=\"https://image.chaiding.com/ruankao/a096916fb2fe3f56a99df04ea0521eb5.jpg?x-oss-process=style/ruankaodaren\"><br>则回答“是”的比例等于0.5*0.6+0.5x=0.35，因此x=0.35*2-0.6=0.1。","multi":0,"questionType":1,"answer":"A","chooseItem":["796235418361155585"],"itemList":[{"id":"796235418361155585","questionId":"796235417354522625","content":" 10%","answer":1,"chooseValue":"A"},{"id":"796235418390515713","questionId":"796235417354522625","content":" 15%","answer":0,"chooseValue":"B"},{"id":"796235418419875841","questionId":"796235417354522625","content":" 20%","answer":0,"chooseValue":"C"},{"id":"796235418440847361","questionId":"796235417354522625","content":" 25%","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796235395086962689","title":"某市场上某种零件由甲、乙、丙、丁四厂供货，供货数量之比为4:3:2:1。各厂产品的合格率分别为99%、98%、97.5%和95%。某抽检员发现了一件次品，它属于（52）厂的概率最大。","analyze":"本题考查应用数学基础知识（概率）。<br>先根据题意画出概率图如下：<br><img alt=\"\" width=\"399\" height=\"207\" src=\"https://image.chaiding.com/ruankao/22af92c37fb0904a47673db0d30d4c88.jpg?x-oss-process=style/ruankaodaren\"><br>总次品率=0.4*1%+0.3*2%+0.2*2.5%+0.1*5%<br>=0.004+0.006+0.005+0.005=0.02<br>该次品属于甲厂的概率=0.004/0.02=20%<br>该次品属于乙厂的概率=0.006/0.02=30%<br>该次品属于丙厂的概率=0.005/0.02=25%<br>该次品属于丁厂的概率=0.005/0.02=25%","multi":0,"questionType":1,"answer":"B","chooseItem":["796235396034875393"],"itemList":[{"id":"796235395997126657","questionId":"796235395086962689","content":" 甲","answer":0,"chooseValue":"A"},{"id":"796235396034875393","questionId":"796235395086962689","content":" 乙","answer":1,"chooseValue":"B"},{"id":"796235396060041217","questionId":"796235395086962689","content":" 丙","answer":0,"chooseValue":"C"},{"id":"796235396089401345","questionId":"796235395086962689","content":" 丁","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null},{"id":"796233978213322753","title":"根据历史数据和理论推导可知，某应用中，随机变量s的分布密度函数为f（x）=3x<sup>2</sup>，（0&lt;x&lt;1）。这意味着，当Δx充分小时，随机变量s落在区间（x，x+Δx）内的概率约等于f（x）Δx。为此，开发该应用的仿真系统时，可用（）来模拟该随机变量，其中，r1.r2.r3...为计算机逐个产生的、均匀分布在（0，1）区间内的互相独立的伪随机数。","analyze":"（ 0 ，1 ）区间内的分布密度函数3x<sup>2</sup>。意味着随着x的增大出现的概率也增大。显然，对于min（r1 ,r2 , r3 ） ，出现较小的数值的概率更大些；r1 *r2*r3（两个小于1 的数相乘会变得更小）也会这样。对于随机变量（r1+r2+r3）/2 ， 出现中等大小数值的概率更大一些，出现较大的或较小值的概率会小一些，其分布密度函数会呈凸型。只有max（r1 ,r2 ,r3） ，出现较大数值的概率更大些。","multi":0,"questionType":1,"answer":"A","chooseItem":["796233979287064577"],"itemList":[{"id":"796233979287064577","questionId":"796233978213322753","content":" max（r1，r2，r3）","answer":1,"chooseValue":"A"},{"id":"796233979324813313","questionId":"796233978213322753","content":" min（r1，r2，r3）","answer":0,"chooseValue":"B"},{"id":"796233979358367745","questionId":"796233978213322753","content":" r1*r2*r3","answer":0,"chooseValue":"C"},{"id":"796233979396116481","questionId":"796233978213322753","content":" （r1+r2+r3）/3","answer":0,"chooseValue":"D"}],"userAnswer":null,"userChooseItem":null,"answerCorrect":null,"userCollect":null}]}}