Question

急！英语数学。英语好的请进来帮个忙~

请帮帮忙，下面的4道题目！谢谢。
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Question 2:
A space agency has determined that, of all trainee astronauts eligible for space flight, 23% will be over 50 years old before they are selected in a crew. If 12 astronauts are selected at random, what is the probability that:
A. 3 are over 50?
B. At least 10 are over 50? Use your graphics calculator functions to solve this calculation. You must either describe the functions you are using, or include a screen shot of your calculator.
C. At most 2 are not over 50?

Question 3:
The Geometric Distribution is related to the Binomial Distribution in that:
 there are two outcomes for each trial – success and failure
 the outcomes for each trial are statistically independent, and
 all the trials are identical, i.e. they have the same probability of success.
However, where a Binomial random variable is the number of successes in trials, a Geometric random variable is the number of trials until the first failure. Thus, the total number of trials for a Geometric distribution is potentially infinite.
A discrete random variable is said to follow a Geometric Distribution if: P(X=x)=p^x (1-p)^(1-x)
Suppose that 20% of items produced by a manufacturing production line are faulty and that a quality inspector is checking randomly sampled items. What is the probability that:
A. The first item is faulty, i.e. P(X=1)?
B. P(4≤X≤7)
C. (X≤2)
D. P(X≥3 )

Question 4: *
Charlotte and Jonathon are playing a game of chance. They are drawing marbles from a bag, containing 5 pink marbles and 5 blue marbles. Each turn the bag is shaken up and a marble taken out. If a blue marble is drawn, then Jonathon wins a point. If a pink marble is drawn, then Charlotte will win the point. The marble is replaced after each turn. They are playing the best of 5 turns.
After 3 turns, Charlotte is ahead 2-1. Jonathon decides that he no longer wishes to play and says they should split the pot (money that was bet) evenly. Charlotte does not think this is fair.
How should the pot be split so that it is done so with the most fairness? Use calculations to support your decision.
State any assumptions you are making, and explain the effect that these will have on your calculations.
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各位好心人帮帮忙，能做几道就几道。拜托啦~
好的一定一定一定加悬赏分。
忘了一个
Question 5: (REFER TO THE DATA SETS GIVEN TO YOU BY YOUR CLASSROOM TEACHER.)
In order to improve test results, a nearby school has introduced a new study program for year 11 mathematics students. The Head of Department has asked you to evaluate the effectiveness of the study program based on one class’ pre- and post- program test results.
For each of the KPS sections you have been asked to:
A. Construct a frequency distribution table.
B. Calculate the mean, median and mode.
C. Calculate the range and IQR.
D. Calculate the standard deviation.
For both the KPS and CAJ sections, you have been asked to:
E. Draw an appropriate graph.
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问题3 中
C，应该是 C.P(X≤2)
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另外不好意思啊，这些都不是选择题。ABCD是每个大题里的小题目，都是小题。
越详细越好。。谢谢。或者给我解题思路也行

Answer 1

Question 2: 选C。过程——0.23*0.24=0.0552；0.0552*50=2.76<3。所以不能选A,只能选C!
Question 3:抱歉，我几何、离散型随机变量这一块不精通，毕竟我不是一个专业的外式数学迷，偶只是一个国际通用的会计。提醒——您学的是哪国的数学？毕竟各国的数学规则在符号、术语方面还有一点小小的差异（在国际化的大环境下）。
例：相关的几何分布的二项分布：有两种结果——试验成功的和失败的。
试验结果独立统计,所有的试验是相同的,也就是说,它们有相同的成功概率。
于是 在一个二项式随机变量的数目是成功的实验中,几何随机变量的数目是直到第一个失败。因此,总数量的试验为几何分布的潜在无限
一个离散型随机变量 遵循几何分布:P(X=x)=p^x (1-p)^(1-x)；
Question 4: 这个经典！题目的构想来自著名的“赌徒赌注理论”，这是概率论的开端啊！哈哈！简化答案：一个布袋里有10颗珠子（5粉5蓝），抽出一个蓝色珠子,乔纳森赢得一分；抽出一个粉红色的珠子,夏绿蒂将获得分数。一共抽五次，同色珠子多的人胜。
共进行了3轮，现在的比分是夏洛特2-1领先。乔纳森却说要回家了，剩下两轮不比了，赌注平分算了。夏洛特不服。然后让你来裁决到底赌注到底怎么分；然后做出解释！
哈哈！这是概率论的鼻祖基础的基础！您自己算吧，我就不班门弄斧了。
Question 5：
为了提高测试结果,附近的学校已经引进了一种新的研究项目（针对11年级数学的学生）。这个部门已经要求你有效性的评价研究计划（基于一个类的售前？和邮报？-项目的测试结果）
唉，毕竟不是专业的，许多术语都不知道（这就是字母文化的弊端——一个新词出来还需要专门学习；哪像象形文字中文——一个术语出来，可以联系想象一下！）
这道题是让你如何解决问题的——又是关于频率、偏差的了~~~大哥我实在弱项，汗颜！斗胆在这耍斧了。

Answer 2

第二题第一问：c（3,12）* 0.23^3*0.77^9

第三题设计几何分布和二项分布。

1.在每次试验中只有两种可能的结果,而且是互相对立的;
2.每次实验是独立的,与其它各次试验结果无关;
3.结果事件发生的概率在整个系列试验中保持不变,则这一系列试验称为伯努力试验.

几何分布就是：

1. 得到1次成功而进行，n次伯努利实验，n的概率分布，取值范围为『1，2，3，...』;

2. m = n-1次失败，第n次成功，m的概率分布，取值范围为『0，1，2，3，...』.

由两种不同情况而得出的期望和方差如下：

E(n) = 1/p, var(n) = (1-p)/p^2;

E(m) = (1-p)/p, var(m) = (1-p)/p^2。

概率为p的事件A，以X记A首次发生所进行的试验次数，则X的分布列：

P(X=k)=p*(1-p)^(k-1)，k=1,2,3,……

具有这种分布列的随机变量，称为服从参数p的几何分布。

大概就是Pre- program Test Results:
KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
CAJ:
B‐ B B+ B‐ C C‐ C+ D+ A‐ D‐ B‐ D+ C+ B‐ B
Post- program Test Results:
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
CAJ:
C A C+ C A B+ C‐ A‐ B‐ B‐ C B A C B+

A).
B). KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
Mean: 63.2
Median: 37
Mode: 85
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
Mean: 78.3
Median: 58
Mode: 100, 50, 96, 79, 58, 83
C).
D). KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
standard deviation: 18.53
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
standard deviation: 17.40

第五题，按照3/4和1/4去分钱。

第六题，

Pre- program Test Results:
KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
CAJ:
B‐ B B+ B‐ C C‐ C+ D+ A‐ D‐ B‐ D+ C+ B‐ B
Post- program Test Results:
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
CAJ:
C A C+ C A B+ C‐ A‐ B‐ B‐ C B A C B+

A).
B). KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
Mean: 63.2
Median: 37
Mode: 85
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
Mean: 78.3
Median: 58
Mode: 100, 50, 96, 79, 58, 83
C).
D). KPS:
72 85 47 65 37 53 55 37 75 62 85 85 62 90 38
standard deviation: 18.53
KPS:
81 100 50 50 71 96 79 58 58 96 100 83 79 83 90
standard deviation: 17.40

Answer 3

第一个题 说是有一些宇航员中23%的超过50岁 现在随即挑选出12个宇航员 也就是12*23%=2.76 所以选A

Answer 4

这几个题的意思还能懂，但是都涉及概率统计的专业计算了。对不起，哪些东西只在大学学过一点，现在都不记得了。爱莫能助。

Answer 5

Q2：貌似都是no
其他正在做。。。。