Pakistan International School Riyadh
Class XII
1. How many possible permutations can be formed from the word ‘statistics’? 2. In how many ways can a team of 11 players be chosen from a total of 16 players? 3. State multiplicative theorem of probability for dependent events. 4. An aptitude test with 4 options. If a student marks the options of the questions randomly and independently, then find the probability of being correct to 4 questions. 5. A can solve 75% questions in a book and B can solve 60% questions in that book. Find the probability that randomly selected questions is solved by them? 6. What is the probability of getting at least 2 heads when 3 coins are tossed? 7. Find P(A ∩ B) if P(A) = ¼ and P(B) = 1/3 and p(A U B) = ½ . 8. State the laws of expectation.

9. Describe the properties of discrete probability distribution. 10. Given E(X) = 0.55, Var(X) = 1.55 and Y = 2X + 1. Find E(Y) and Var(Y). 11. Show that the mean of the binomial distribution (q + p)2 is 2p. 12. Show that mean is 2p and σ2 = 2pq for a binomial distribution in which n = 2. 13. A random variable X has a binomial distribution with E(X) = 2.4 and p = 0.3. Find the standard deviation of X. 14. Describe the normal distribution and write down its equation. 15. What is standard normal variable?

16. The value 2nd moment about mean in a normal distribution is 5. Find the 3rd and 4th moment about mean for this distribution. 17. Is every symmetrical distribution a normal distribution? 18. What is the relation between the binomial and normal distribution? 19. Define sampling units and frame.

20. What is the difference between sampling and non-sampling error? 21. Name the four techniques used in probability sampling. 22. Given P1 = 2/3, n1 = 2, P2 = 2, n2 = 2. Find µP1-P2 and σP1 – P2. 23. What is the difference between sampling with replacement and without replacement? 24. Find 90% confidence interval of µ from a sample...

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Contents
Question 1 3
Question 2a 5
Question 2b 6
Question 2c 7
Question 3a 8
Question 3b 8
Question 3c 10
Question 3d 11
Question 4 12
Question 5 14
References 15
Question 1
The sampling method that Mr. Kwok is using is Stratified Random Sampling Method. In this case study, Mr Kwok collected a random sample of 1000 flights and proportions of three routes in the sample. He divides them into different sub-groups such as satisfaction, refreshments and departure time and then selects proportionally to highlight specific subgroup within the population. The reasons why Mr Kwok used this sampling method are that the cost per observation in the survey may be reduced and it also enables to increase the accuracy at a given cost.
TABLE 1: Data Summaries of Three Routes
Route 1
Route 2
Route 3
Normal(88.532,5.07943)
Normal(97.1033,5.04488)
Normal(107.15,5.15367)
Summary Statistics
Mean
88.532
Std Dev
5.0794269
Std Err Mean
0.2271589
Upper 95% Mean
88.978306
Lower 95% Mean
88.085694
N
500
Sum
44266
Summary Statistics
Mean
97.103333
Std Dev
5.0448811
Std Err Mean
0.2912663
Upper 95% Mean
97.676525
Lower 95% Mean
96.530142
N
300
Sum
29131
Summary...

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NAME: SHU ZHAOHUI
ID: 17329164
Q5.
Descriptive Statistics |
| N | Minimum | Maximum | Mean | Std. Deviation | Skewness |
| Statistic | Statistic | Statistic | Statistic | Statistic | Statistic | Std. Error |
Gasolinescore | 1000 | 3.00 | 21.00 | 14.9090 | 4.83654 | -.493 | .077 |
Globalscore | 1000 | 3.00 | 21.00 | 17.0490 | 3.78774 | -1.073 | .077 |
Valid N (listwise) | 1000 | | | | | | |
The mean in the gaslinescore and globalscore stand for the average the respondents choose is 14.9090 and 17.0490, the respondents choose is concentrate around 14.9090 and 17.0490.
The standard deviation of gasolinescore refer to most of the respondents answer is rang from (14.9090-4.83654) to (14.9090+4.83654), and the standard deviation of globalscore refer to most of the respondents answer is rang from (17.0490-3.78774) to (17.0490+3.78774), and the globalscore’s float is smaller than gasolinescore’s float.
The data for Gasolinescore and Globalscore are not the normally distributed. Because the normally distributed situation means the median equal to mean. but in the table of above, the gaslinescore’s median(12) is not equal to the mean(14.9090) and the globalscore’s median(12) also not equal to mean(17.0490), they are all negative skew. So the data of each variable is not a normally distributed.
If the two variables were normally distributed, the median should be equal...

...QUESTION 21
The finishing process on new furniture leaves slight blemishes. The table below displays a manager's probability assessment of the number of blemishes on one piece of new furniture.
Number of Blemishes
0
1
2
3
4
5
Probability
0.34
0.25
0.19
0.11
0.07
0.04
1. On average, how many blemishes do we expect on one piece of new furniture?
2. What is the variance of blemishes on one piece of new furniture? (round to the nearest hundredth) QUESTION 22
The probability that a person catches a cold during the cold-and-flu season is 0.4. Assume that 10 people are chosen at random.
On average, how many of these ten people would you expect to catch a cold?
What is the standard deviation of the number of people who catch a cold? (round to the nearest hundredth)
QUESTION 23
The number of nails in a five-pound box is normally distributed with a mean of 566 and a standard deviation of 33.
What is the probability that there are less than 500 nails in a randomly-selected five-pound box of nails? (express as a decimal, not a percentage)
The probability is 0.99 that a randomly-selected five-pound box of nails contains at least how many nails approximately?
QUESTION 24
You are the owner of a small casino in Las Vegas and you would like to reward the high-rollers who come to your casino. In particular, you want to give free accommodations to no more than 10% of your patrons....

...cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value.
38. A recent article in the Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30- year rates (in percent):
4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6
At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value.
Chapter 11
27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below.
Statistic Men Women
Sample mean 24.51 22.69
Population standard deviation 4.48 3.86
Sample size 35 40
At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value?
46. Grand Strand Family Medical centre is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, on in the Little River area, and the other in Murrells Inlet. The Quality Assurance Department wishes to compare mean waiting times for patients in two locations. Samples of waiting times, reported in minutes follow:
Location Waiting time
Little River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49...

...Practice Problems 1-KEY 1. The closing stock price of Ahmadi, Inc. for a sample of 10 trading days is shown below. Day Stock Price 1 84 2 87 3 84 4 88 5 85 6 90 7 91 8 83 9 82 10 86
For the above sample, compute the following measures. a. b. c. The mean = ∑X/n = 860/10 = 86 The median = (85+86)/2 = 85.5 The variance = ∑ X - X 2/ n-1 = {(84-86)2 + (87-86)2 + (84-86)2 + (88-86)2 + (85-86)2 + (90-86)2 + (91-86)2 + (83The standard deviation = √8.89 = 2.98 The coefficient of variation = 2.98/86 * 100% = 3.47%
86)2 + (82-86)2 + (86-86)2 } / (10 -1) = 8.89 d. f.
2. In 2008, the average age of students at GUST was 22 with a standard deviation of 3.96. In 2009, the average age was 24 with a standard deviation of 4.08. In which year do the ages show a more dispersed distribution? Show your complete work and support your answer. CV2008 = 3.96/22 * 100% = 18% CV2009 = 4.08/24 * 100% = 17% So, 2008 shows more dispersed distribution 3. A local university administers a comprehensive examination to the recipients of a B.S. degree in Business Administration. A sample of examinations are selected at random and scored. The results are shown below. Grade For the above data, determine a. The mean = ∑X/n = 664/8 = 83 b. c. The median = (85+87)/2 = 86 The standard deviation = √variance ariance = ∑ X - X 2/ n-1 = {(93-83)2 + (65-83)2 + (80-83)2 + (97-83)2 + (85-83)2 + (87-83)2 + (97-83)2 + (60 - 83)2 } / (8 -1) = 196.29 S0, standard deviation = √196.29 = 14.01 d. The coefficient of...

...balances by branch. Does it appear that account balances are related to the branch?
|Branch |Smallest third |Middle third |Largest third |
|Ohio |8 |6 |2 |
|Georgia |1 |6 |10 |
|Kentucky |5 |5 |4 |
|Pensylvania |6 |3 |4 |
Ho: Account balances are independent of branch.
H1: Account balances are independent of branch
Test Statistic:
Chi square is given by (Oi – Ei)^2 / Ei ~ χ2 with 2*3= 6 degrees of freedom
The calculated value of chi-square is 11.89625
Critical value for 5% level of significance is 12.5916
Since the calculated value is less than critical value we accept null hypothesis and conclude that account balances are not related to the branch
g. Cite some examples and comment on your findings.
If we wish to know if any distinction is made in appointment on the basis of sex for the following data
| |Employed |Not Employed |Total |
|Male |1480 |5720 |7200 |
|Female |120...

...undergraduates at your college or university are more susceptible to hypnosis than the University of Tennessee undergraduates.
a. Set up and for the test.
b. Describe a Type I error for the test.
c. Describe a Type II error for the test.
8.24 (2 points) A random sample of 64 observations produced the following summary statistics: = .323 and = .034
a. Test the null hypothesis that= .36 against the alternative hypothesis that < .36, using = .10.
b. Test the null hypothesis that = .36 against the alternative hypothesis that .36, using = 10, interpret the result.
8.26 (3 points) Latex allergy in health-care workers. Refer to the Current Allergy & Clinical Immunology (March 2004) study of = 46 hospital employees who were diagnosed with a latex allergy from exposure to the powder on latex gloves used per week by the sampled workers is summarized as follow: = 19.3 and = 11.9. Let represent the mean number of latex gloves used per week by all hospital employees. Consider testing : = 20 against : < 20.
a. Give the rejection region for the test at a significance level of = .01.
b. Calculate the value of the test statistics.
c. Use the results from parts a and b to draw the appropriate conclusion.
8.29 (2 points) Post-traumatic stress of POWS. Psychological Assessment (March 1995) published the results of a study of World War II aviators captured by German forces after having been shot...

...1. Introduction
This report is about the case study of PAR, INC. From the following book: Statistics for Business an Economics, 8th edition by D.R. Anderson, D.J. Sweeney and Th.A. Williams, publisher: Dave Shaut. The case is described at page 416, chapter 10.
2. Problem statement
Par, Inc. has produced a new type of golf ball. The company wants to know if this new type of golf ball is comparable to the old ones. Therefore they did a test, which consists out of 40 trials with the current and 40 trials with the new golf balls. The testing was performed with a mechanical fitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the design. The outcomes are given in the table of appendix 1.
3. Hypothesis testing
The first thing to do is to formulate and present the rationale for a hypothesis test that Par, Inc. could use to compare the driving distance of the current and new golf balls. By formulation of these hypothesis there is assumed that the new and current golf balls show no significant difference to each other. The hypothesis and alternative hypothesis are formulated as follow:
Question 1
H0 : µ1 - µ2 = 0 (they are the same)
Ha : µ1 - µ2 ≠ 0 (the are not the same)
4. P-value
Secondly; analyze the data to provide the hypothesis testing conclusion. The p-value for the test is:
Question 2
Note: the statistical data is provide in §...