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论文题目：题目未定 35道统计数据分析题目（用EXCEL和PHSTAT)+一篇论文
论文语言：English
论文专业：统计
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学校国家：USA
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论文用于：Master assignment 硕士课程作业
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论文要求老师还没说 可以先做35道题目 涉及的资料太大了 附件发不上去 我先发点题目 其他资料麻烦联系我我发你们
以下的习题 有的我已经做了 但是有的我没有回答 没有回答的习题 麻烦做下 有的习题我可能只做了其中的几个小题 还有几个小题需要补上 需要做的部分是：
第二题的bc小题 第三题的C 第四题的A 第七题 第八题 第九题的BC 第十题的DE
我贴上我做的部分是为了给您看下格式 之前发你的PDF和PPT格式的东西是讲义和例题，其中HAND ON为命名的是例题 所有涉及的知识点都在里面了 需要用phstat 做 数据分析 英文版
[1](10) A standard slot machine has three reels. The first reel has 3 walnuts, 7 cherries, 3 oranges, 5 lemons, 1 bell, and 1 jackpot bar. The second reel has 7 cherries, 6 oranges, 1 lemon, 3 bells, and 3 bars. The third reel has 4 walnuts, 7 oranges, 5 lemons, 3 bells and 1 bar. The payoffs are
Reel 1 Reel 2 Reel 3 Payoff in Coins
Bar Bar Bar 85
Bell Bell Bar 18
Bell Bell Bell 18
Lemon Lemon Bar 14
Lemon Lemon Lemon 14
Orange Orange Bar 10
Orange Orange Orange 10
Cherry Cherry Bell 5
Cherry Cherry Walnut 5
Cherry Cherry Any other 3
a) What is the expected payoff if each picture on each reel is equally likely to appear?
b) In the long run, what fraction of every dollar bet will this one-armed bandit keep?
Answer:
a).
P(X) 20 20 20 182
1 1 3 1 *85 0.03188
2 1 3 1 *18 0.00675
3 1 3 3 *18 0.02025
4 5 1 1 *14 0.00875
5 5 1 5 *14 0.04375
6 3 6 1 *10 0.0225
7 3 6 7 *10 0.1575
8 7 7 3 *5 0.09188
9 7 7 4 *5 0.1225
10 7 7 13 *3 0.23888
0.744625
So the expected payoff if each picture on each reel is equally likely to appear is 0.744625.
b). 1-0.744625=0.255375
So in the long run, the fraction of every dollar bet will this one-armed bandit keep is 0.255375.
[2](10) Maximum breadth of sample of male Egyptianhttp://www.ukassignment.org/daixieEssay/meiguoessaydaixie/ skulls from 4000 B.C. and 150 A.D. (based on data from Ancient Races of the Thebaid by Thompson and Randall-Maciver):#p#分页标题#e#
4000 B.C. 131 119 138 125 129 126 131 132 126 128 128 131
150 A.D. 136 130 126 126 139 141 137 138 133 131 134 129
a) Find the mean, median, and mode, range, variance, and the standard deviation for each of the two samples.
b) Compare and analyze the two sets using descriptive measures.
c) Changes in head size over time suggest interbreeding with people from other regions. Do the head sizes appear to have changed from 4000 B.C. to 150 A. D.? Explain.
Answer:
a).
4000 B.C.
Mean 128.6667
Standard Error 1.339003
Median 128.5
Mode 131
Standard Deviation 4.638443
Sample Variance 21.51515
Kurtosis 1.6906
Skewness -0.11247
Range 19
Minimum 119
Maximum 138
Sum 1544
Count 12
Largest(1) 138
Smallest(1) 119
150A.D.
Mean 133.3333
Standard Error 1.447743
Median 133.5
Mode 126
Standard Deviation 5.015129
Sample Variance 25.15152
Kurtosis -1.17336
Skewness -0.10128
Range 15
Minimum 126
Maximum 141
Sum 1600
Count 12
Largest(1) 141
Smallest(1) 126
[3](10) The Corner Convenience Store kept track of the number of paying customers it had during the noon hour each day for the last 150 days. The following are the resulting statistics rounded to the nearest integer:
mean = 94
standard deviation (s) = 12
median = 97
first quartile (Q1) = 85
mode (m) = 98 third quartile (Q3) = 107
range (R) = 56
a) The Corner Convenience Store served what number of paying customers during the noon hours more often than any other number? Explain how you determined your answer.
Answer: The Corner Convenience Store served 98 customers during the noon hours more often than any other number. Because mode means the most frequently number.
b) On how many days was there between 85 and 107 paying customers during the noon hour? Explain how you determined your answer.
Answer: Q1=85, Q2=107,
The comprises of 50% or (107-85)=28 of the data set. (28*2)=56 or the range of the data set
The smallest value of the data set= 69 the largest value=125.
(150 Days/56)=2.68 days.
Because 50% of the data lies between the first and third quartile
(.50)*56=28*2.68 Days=75 days.
So there was 75 days between 85 and 107 customers.#p#分页标题#e#
c) For how many of the 150 days was the number of paying customers within three standard deviations of the mean? Explain how you determined your answer.
[4](10) (Please do one of the following, but not both)
a) You are offered a choice of two envelopes and told that one envelop has twice as much as the other and that you can change your mind after opening one of the envelopes. You choose Envelope B and find $1000. Now will you always switch envelopes? Substantiate your decision.
b) Males who have never smoked average 14.8 days of restricted activity per year, while present smokers average 22.5 days and former smokers average 23.5 days. Do these data show that it is healthiest to never smoke, but once you start smoking, it is better not to stop?
Answer：
[5](10) A passenger ferry boat is rated safe for carrying 130 people. Suppose that a load of more than 19,500 pounds is unsafe. If the average weight of people is 144.9 pounds with a standard deviation of 23.7, what is the probability that the boat will have an unsafe load when 130 people are aboard?
Answer:
X value: 19500/130=150
SD=23.7/ROOT(130)=2.0786
Normal Probabilities
Common Data
Mean 144.9
Standard Deviation 2.0786
Probability for X >
X Value 150
Z Value 2.4535745
P(X>150) 0.0071
Z row(2.4), and Z column(.05) is0.9929,
The probability is 1-0.9929=0.0071
So the probability that the boat will have an unsafe load when 130 people are aboard is 0.71%.
[6](10) A week before the final exam, Professor Gottesman gives his students ten questions. At the beginning of the final examination, a student will reach into a wastebasket and blindly select three of these ten questions, which then become the final exam. You are well prepared to answer six of the ten questions and poorly prepared to answer the other four.
a) What is the probability that the three selected questions will all be ones that you are well prepared to answer?
b) What is the probability that the three selected questions will include two that you are well prepared for and one that you are poorly prepared for?
c) What is the probability that the three selected questions will all be ones that you are poorly prepared to answer?
Answer:
a). 6C3/10C3=20/120=1/6
So the probability that the three selected questions will all be ones that you are well prepared to answer is 1/6.
b). 6C2*4C1/10C3=60/120=1/2
So the probability that the three selected questions will include two that you are well prepared for and one that you are poorly prepared for is 50%
c). 4C1/10C3=4/120=1/30
So the probability that the three selected questions will all be ones that you are poorly prepared to answer is 1/30.
[7](10) Suppose that the large international corporation wants to assess whether the mean, , of all one-day travel expenses in Moscow exceeds $500 provided that the standard deviation is known to be $41. The mean of a random sample of 35 one-day travel expenses is $535.#p#分页标题#e#
a) Assuming that equals $500 and the sample size is 35, what is the probability of observing a sample mean that is greater than or equal to $538?
b) Based on your answer in (a), do you think that the mean of all one-day travel expenses in Moscow exceeds $500? Explain.
Answer:
.
[8](10) The January 1986 mission of the Space Shuttle Challenger was the 25th such shuttle mission. It was unsuccessful due to an explosion caused by an O-ring seal failure.
a) According to NASA, the probability of such a failure in a single mission was 1/60000. Using this value of p and assuming all missions are independent, calculate the probability of no mission failures in 25 attempts.
b) In another report, it only states that such a failure occurs at one in 60,000 missions (of course we haven’t seen this many missions yet). Assuming all missions are independent, calculate the probability of no mission failures in 25 missions.
c) According to a study conducted for the Air Force, the probability of such a failure in a single mission is 1/35. Recalculate the probability of no mission failures in 25 mission attempts.
d) How small must the probability p be made in order to ensure that the probability of no mission failures in 25 attempts is 0.999?
Answer:
[9](10) A recent Gallup poll consisted of 1012 randomly selected adults who were asked whether “cloning of humans should or should not be allowed.” Results showed that 89% of those surveyed indicated that cloning should not be allowed.
a) Among the 1012 adults surveyed, how many said that cloning should not be allowed?
b) If we assume that people are indifferent so that 50% believe that cloning of humans should not be allowed, find the mean and standard deviation for the numbers of people in groups of 1012 that can be expected to believe that such cloning should not be allowed.
c) Based on the preceding results, does the 89% result for the Gallup poll appear to be unusually higher than the assumed rate of 50%? Does it appear that an overwhelming majority of adults believe that cloning of humans should not be allowed?
Answer:
a). 89%*1012=901
So 901 people from the sample believe that cloning should not be allowed.
[10](10) According to the U.S. Census Bureau, the mean household income in the United States in 2000 was $57,045 and the median household income was $42,148 (U.S. Census Bureau, “Money Income in the United States: 2000,” www.census.gov, September 2001). The variability of household income is quite large, with the 90th percentile (the top 10% of all households) approximately equal to $111,600, and an overall standard deviation of approximately $25,000. Suppose random samples of 225 households were selected.
a) What proportion of the sample means would be below $55,000?#p#分页标题#e#
b) What proportion of the sample means would be above $60,000?
c) What proportion of the sample means would be above $111,600?
d) Why is the probability you calculated in (c) so much lower than 0.10, even though the top 10% of individual households have incomes above $111,600?
e) If random samples of size 20 were selected, can you use the methods discussed in this chapter to calculate the probabilities requested in (a)--(c)? Explain.
a).
Proportion of the sample means would be below $55,000
Common Data
Mean 57045
Standard Deviation 1667
Probability for X <=
X Value 55000
Z Value -1.226755
P(X<=55000) 0.1099574
So the proportion of the sample means below $55,000 is 10.9%
b).
Proportion of the sample means would be above $60,000
Common Data
Mean 57045
Standard Deviation 1667
Probability for X >
X Value 60000
Z Value 1.7726455
P(X>60000) 0.0381
So the proportion of the sample means above $60000 would be 3.8.%
c).
Proportion of the sample means would be above $111,600
Common Data
Mean 57045
Standard Deviation 1667
Probability for X >
X Value 111600
Z Value 32.726455
P(X>111600) 0.0000
So the proportion of the sample means above $111,600 would be 0.%.
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