英语好的请进来帮个忙~

问题描述：

英语好的请进来帮个忙~
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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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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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应该是 C.P(X≤2)
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另外不好意思啊，ABCD是每个大题里的小题目，都是小题。

1个回答分类：英语 2014-10-27

问题解答：

我来补答

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

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