高考一轮数学(理)复习课时作业70

课时作业70二项分布、正态分布及其应用

1．设X～N(μ1，σ21)，Y～N(μ2，σ22)，这两个正态分布密度曲线如图所示．下列结论中正确的是(C)

A．P(Y≥μ2)≥P(Y≥μ1)

B．P(X≤σ2)≤P(X≤σ1)

C．对任意正数t，P(X≤t)≥P(Y≤t)

D．对任意正数t，P(X≥t)≥P(Y≥t)

解析：由题图可知μ1<0<μ2，σ1<σ2，

∴P(Y≥μ2)

P(X≥σ2)>P(X≤σ1)，故B错；

当t 为任意正数时，由题图可知

P (X ≤t )≥P (Y ≤t )，

而P (X ≤t )＝1－P (X ≥t )，P (Y ≤t )＝1－P (Y ≥t )，

∴P (X ≥t )≤P (Y ≥t )，故C 正确，D 错．

2．(2019·福建厦门模拟)袋中装有2个红球，3个黄球，有放回地抽取3次，每次抽取1球，则3次中恰有2次抽到黄球的概率是( D )

A.25

B.35

C.18125

D.54125

解析：袋中装有2个红球，3个黄球，有放回地抽取3次，每次

抽取1球，每次取到黄球的概率P 1＝35，∴3次中恰有2次抽到黄球的

概率是P ＝C 23? ????352?

????1－35＝54125. 3．(2019·河北唐山模拟)甲乙等4人参加4×100米接力赛，在甲不跑第一棒的条件下，乙不跑第二棒的概率是( D )

A.29

B.49

C.23

D.79

解析：甲不跑第一棒共有A 13·A 33＝18种情况，甲不跑第一棒且乙

不跑第二棒共有两类：(1)乙跑第一棒，共有A 33＝6种情况；(2)乙不跑

第一棒，共有A 12·A 12·A 22＝8种情况，∴甲不跑第一棒的条件下，乙不

跑第二棒的概率为6＋818＝79.故选D.

4．(2019·山东淄博一模)设每天从甲地去乙地的旅客人数为随机变量X ，且X ～N (800,502)．则一天中从甲地去乙地的旅客人数不超过900的概率为( A )

(参考数据：若X ～N (μ，σ2)，有P (μ－σ

A ．0.977 2

B ．0.682 6

C ．0.997 4

D ．0.954 4

解析：∵X ～N (800,502)，

∴P (700≤X ≤900)＝0.954 4，

∴P (X >900)＝1－0.954 42

＝0.022 8， ∴P (X ≤900)＝1－0.022 8＝0.977 2.

故选A.

5．甲、乙两个小组各10名学生的英语口语测试成绩如下(单位：分)．

甲组：76,90,84,86,81,87,86,82,85,83

乙组：82,84,85,89,79,80,91,89,79,74

现从这20名学生中随机抽取一人，将“抽出的学生为甲组学生”记为事件A ；“抽出的学生的英语口语测试成绩不低于85分”记为事件B ，则P (AB )，P (A |B )的值分别是( A )

A.14，59

B.14，49

C.15，59

D.15，49

解析：由题意知，P (AB )＝1020×510＝14，根据条件概率的计算公式

得P (A |B )＝P (AB )P (B )

＝14920

＝59. 6．为向国际化大都市目标迈进，某市今年新建三大类重点工程，它们分别是30项基础设施类工程、20项民生类工程和10项产业建设类工程．现有3名民工相互独立地从这60个项目中任选一个项目参与建设，则这3名民工选择的项目所属类别互异的概率是( D )

A.12

B.13

C.14

D.16

解析：记第i 名民工选择的项目属于基础设施类、民生类、产业

建设类分别为事件A i ，B i ，C i ，i ＝1,2,3.由题意，事件A i ，B i ，C i (i ＝1,2,3)

相互独立，则P (A i )＝3060＝12，P (B i )＝2060＝13，P (C i )＝1060＝16，i ＝1,2,3，

故这3名民工选择的项目所属类别互异的概率是P ＝A 33P (A i B i C i )＝

6×12×13×16＝16.

7．位于坐标原点的一个质点P 按下述规则移动：质点每次移动一个单位，移动的方向为向上或向右，并且向上、向右移动的概率都

是12.质点P 移动五次后位于点(2,3)的概率是516.

解析：由于质点每次移动一个单位，移动的方向为向上或向右，移动五次后位于点(2,3)，所以质点P 必须向右移动两次，向上移动三次，故其概率为C 35? ????123·? ????122＝C 35? ????125＝C 25? ??

??125＝516. 8．(2019·江西南昌模拟)口袋中装有大小形状相同的红球2个，白球3个，黄球1个，甲从中不放回地逐一取球，已知第一次取得红球，

则第二次取得白球的概率为35.

解析：口袋中装有大小形状相同的红球2个，白球3个，黄球1个，甲从中不放回地逐一取球，设事件A 表示“第一次取得红球”，

事件B 表示“第二次取得白球”，则P (A )＝26＝13，P (AB )＝26×35＝15，

∴第一次取得红球后，第二次取得白球的概率为P (B |A )＝P (AB )P (A )＝1513

＝35.

9.如图，四边形EFGH 是以O 为圆心，半径为1的圆的内接正方形．将一颗豆子随机地扔到该圆内，用A 表示事件“豆子落在正方形EFGH 内”，B 表示事件“豆子落在扇形OHE (阴影部分)内”，则P (B |A )

＝14.

解析：由题意可得，事件A 发生的概率

P (A )＝S 正方形EFGH S 圆O ＝2×2π×12

＝2π.事件AB 表示“豆子落在△EOH 内”，则P (AB )＝S △EOH S 圆O ＝12×12π×12

＝1

2π，

故P (B |A )＝P (AB )P (A )＝1

2π2π

＝14. 10．某一部件由三个电子元件按如图所示方式连接而成，元件1或元件2正常工作，且元件3正常工作，则部件正常工作．设三个电子元件的使用寿命(单位：小时)均服从正态分布N (1 000,502)，且各个元件能否正常工作相互独立，那么该部件的使用寿命超过1 000小时

的概率为38.

解析：设元件1,2,3的使用寿命超过1 000小时的事件分别记为A ，

B ，

C ，显然P (A )＝P (B )＝P (C )＝12，

∴该部件的使用寿命超过 1 000小时的事件为(A B ＋A B ＋AB )C ，

∴该部件的使用寿命超过1 000小时的概率

P ＝? ??

??12×12＋12×12＋12×12×12＝38. 11．(2014·新课标Ⅰ)从某企业生产的某种产品中抽取500件，测量这些产品的一项质量指标值，由测量结果得如下频率分布直方图：

(1)求这500件产品质量指标值的样本平均数x 和样本方差s 2(同一组中的数据用该组区间的中点值作代表)；

(2)由直方图可以认为，这种产品的质量指标值Z 服从正态分布N (μ，σ2)，其中μ近似为样本平均数x ，σ2近似为样本方差s 2.

(ⅰ)利用该正态分布，求P (187.8＜Z ＜212.2)；

(ⅱ)某用户从该企业购买了100件这种产品，记X 表示这100件产品中质量指标值位于区间(187.8，212.2)的产品件数．利用(ⅰ)的结果，求E (X )． 附：150≈12.2.

若Z ～N (μ，σ2)，则P (μ－σ＜Z ＜μ＋σ)＝0.682 6，

P (μ－2σ＜Z ＜μ＋2σ)＝0.954 4.

解：(1)抽取产品的质量指标值的样本平均数x和样本方差s2分别为

x＝170×0.02＋180×0.09＋190×0.22＋200×0.33＋210×0.24＋220×0.08＋230×0.02＝200，

s2＝(－30)2×0.02＋(－20)2×0.09＋(－10)2×0.22＋0×0.33＋102×0.24＋202×0.08＋302×0.02＝150.

(2)(ⅰ)由(1)知，Z～N(200,150)，

从而P(187.8

(ⅱ)由(ⅰ)知，一件产品的质量指标值位于区间(187.8,212.2)的概率为0.682 6，

依题意知X～B(100,0.682 6)，所以E(X)＝100×0.682 6＝68.26.

12．(2019·广东顺德一模)某市市民用水拟实行阶梯水价，每人月用水量不超过w立方米的部分按4元/立方米收费，超出w立方米的部分按10元/立方米收费，从该市随机调查了100位市民，获得了他们某月的用水量数据，整理得到如下频率分布直方图，并且前四组频数成等差数列．

(1)求a，b，c的值及居民月用水量在2～2.5内的频数；

(2)根据此次调查，为使80%以上居民月用水价格为4元/立方米，应将w定为多少？(精确到小数点后2位)

(3)若将频率视为概率，现从该市随机调查3名居民的月用水量，将月用水量不超过2.5立方米的人数记为X ，求其分布列及均值．

解：(1)∵前四组频数成等差数列，

∴所对应的频率组距

也成等差数列， 设a ＝0.2＋d ，b ＝0.2＋2d ，c ＝0.2＋3d ，

∴0.5(0.2＋0.2＋d ＋0.2＋2d ＋0.2＋3d ＋0.2＋d ＋0.1＋0.1＋0.1)＝1，

解得d ＝0.1，∴a ＝0.3，b ＝0.4，c ＝0.5.

居民月用水量在2～2.5内的频率为0.5×0.5＝0.25.

居民月用水量在2～2.5内的频数为0.25×100＝25.

(2)由题图及(1)可知，居民月用水量小于2.5的频率为0.7<0.8， ∴为使80%以上居民月用水价格为4元/立方米，

应规定w ＝2.5＋0.10.15×0.5≈2.83.

(3)将频率视为概率，设A (单位：立方米)代表居民月用水量，

可知P (A ≤2.5)＝0.7，

由题意，X ～B (3,0.7)，

P (X ＝0)＝C 03×0.33＝0.027，

P (X ＝1)＝C 13×0.32×0.7＝0.189，

P (X ＝2)＝C 23×0.3×0.72＝0.441，

P (X ＝3)＝C 33×0.73＝0.343.

∴X 的分布列为

∵X ～B ∴E (X )＝np ＝2.1.

13．(2019·广东茂名一模)设X～N(1,1)，其正态分布密度曲线如图所示，那么向正方形ABCD中随机投掷10 000个点，则落入阴影部分的点的个数的估计值是(D)

(注：若X～N(μ，σ2)，则P(μ－σ

A．7 539 B．6 038

C．7 028 D．6 587

解析：∵X～N(1,1)，

∴μ＝1，σ＝1.

∵P(μ－σ

∴P(0

则P (1

∴阴影部分的面积为1－0.341 3＝0.658 7.∴向正方形ABCD 中随机投掷10 000个点，则落入阴影部分的点的个数的估计值是10 000×0.658 7＝6 587.故选D.

14．(2019·金华一中模拟)春节放假，甲回老家过节的概率为13，乙、

丙回老家过节的概率分别为14，15.假定三人的行动相互之间没有影响，

那么这段时间内至少1人回老家过节的概率为( B )

A.5960

B.35

C.12

D.160

解析：“甲、乙、丙回老家过节”分别记为事件A ，B ，C ，则P (A )＝13，P (B )＝14，P (C )＝15，所以P (A )＝23，P (B )＝34，P (C )＝45.由题知A ，B ，C 为相互独立事件，所以三人都不回老家过节的概率

P (A B C )＝P (A )P (B )P (C )＝23×34×45＝25，所以至少有一人

回老家过节的概率P ＝1－25＝35.

15．甲罐中有5个红球，2个白球和3个黑球，乙罐中有4个红球，3个白球和3个黑球．先从甲罐中随机取出一球放入乙罐，分别以A 1，A 2和A 3表示由甲罐取出的球是红球，白球和黑球的事件；再从乙罐中随机取出一球，以B 表示由乙罐取出的球是红球的事件，则下列结论中正确的是②④.(写出所有正确结论的序号)

①P (B )＝25；

②P (B |A 1)＝511；

③事件B 与事件A 1相互独立；

④A 1，A 2，A 3是两两互斥的事件；

⑤P (B )的值不能确定，它与A 1，A 2，A 3中哪一个发生都有关．

解析：由题意知A1，A2，A3是两两互斥的事件，

P(A1)＝5

10＝1

2，P(A2)＝

2

10＝

1

5，

P(A3)＝3

10，P(B|A1)＝1

2×

5

11

1

2

＝

5

11，

由此知，②正确；

P(B|A2)＝4

11，P(B|A3)＝4

11，

而P(B)＝P(A1B)＋P(A2B)＋P(A3B)

＝P(A1)P(B|A1)＋P(A2)P(B|A2)＋P(A3)P(B|A3)

＝1

2×

5

11＋

1

5×

4

11＋

3

10×

4

11＝

9

22.

由此知①③⑤不正确；A1，A2，A3是两两互斥事件，④正确，故答案为②④.

16．(2019·河北石家庄新华模拟)“过大年，吃水饺”是我国不少地方过春节的一大习俗.2018年春节前夕，A市某质检部门随机抽取了100包某种品牌的速冻水饺，检测其某项质量指标值，所得频率分布直方图如下：

(1)求所抽取的100包速冻水饺该项质量指标值的样本平均数x (同一组中的数据用该组区间的中点值作代表)；

(2)①由直方图可以认为，速冻水饺的该项质量指标值Z 服从正态分布N (μ，σ2)，利用该正态分布，求Z 落在(14.55,38.45)内的概率；

②将频率视为概率，若某人从某超市购买了4包这种品牌的速冻水饺，记这4包速冻水饺中这种质量指标值位于(10,30)内的包数为X ，求X 的分布列和数学期望．

附：计算得所抽查的这100包速冻水饺的质量指标值的标准差为σ＝142.75≈11.95；

若ξ～N (μ，σ2)，则P (μ－σ<ξ≤μ＋σ)＝0.682 6，P (μ－2σ<ξ≤μ＋2σ)＝0.954 4.

解：(1)所抽取的100包速冻水饺该项质量指标值的平均数x ＝5×0.1＋15×0.2＋25×0.3＋35×0.25＋45×0.15＝26.5.

(2)①∵Z 服从正态分布N (μ，σ2)，且μ＝26.5，σ≈11.95，

∴P (14.55

②根据题意得X ～B ?

????4，12， P (X ＝0)＝C 04? ??

??124＝116； P (X ＝1)＝C 14? ??

??124＝14； P (X ＝2)＝C 24? ????124＝38；

P (X ＝3)＝C 34? ??

??124＝14； P (X ＝4)＝C 44

? ????124＝116. ∴X 的分布列为

∴E (X )＝4×12＝2.

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