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....

...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...

...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 §...

...1. A radio station that plays classical music has a “By Request” program each Saturday night. The percentage of requests for composers on a particular night are listed below:
Composers Percentage of Requests
Bach 5
Beethoven 26
Brahms 9
Dvorak 2
Mendelssohn 3
Mozart 21
Schubert 12
Schumann 7
Tchaikovsky 14
Wagner 1
a. Does the data listed above comprise a valid probability distribution? Explain.
The individual probabilities are all between 0 & 1 and the sum = 100%
b. What is the probability that a randomly selected request is for one of the three B’s?
P(one of the B’s) = P(Bach) + P(Beethoven) + P(Brahms) = 5 + 26 + 9 = 40%
c. What is the probability that a randomly selected request is for a Mozart piece?
P(Mozart) = 21%
d. What is the probability that a randomly selected request is not for one of the two S’s?
P(not Schubert or Schumann) = 1 – P(Schubert or Schumann)
= 1 – (12 + 7)
= 81%
e. Neither Bach nor Wagner wrote any symphonies. What is the probability that a randomly selected request is for a composer who wrote at least one symphony?
P(Symphony) = 1 – P(Bach or Wagner)
= 1 – (5 + 1)
= 94%
f. What is the probability that a randomly selected request is for a composer other...

...Preliminary Observations
• The Mean of AA Fly (7.167) is lesser than TT Air (10.229). This means that the average number of hours of AA Fly is lesser than TT Air during each 24-hour period i.e. (10.229-7.167=3.062).
• The sample size of AA Fly (27) is lesser than TT Air (28). Even though AA Fly’s sample is 1 day shorter than TT Air, it does not significantly affect the output as the one day does not change much (unless there had been a market shock). However based on Central Limit Theorem, if the sample size is greater than or equal to 30 samples, the sample statistics reflect the time population parameters.
• The Median data does not show much in this case as the range (4.7 AA Fly and 6.8 TT Air) is small for both AA Fly and TT Air. This means that the mean is more applicable to be used.
• The Mode of AA Fly (7.3) is much nearer to the mean (7.167) compared to TT Air (9.2) which is much further (mean of TT Air, 10.229). This means that there is a high occurrence of AA Fly clustered around its mean.
• The Standard Deviation which compares the dispersion shows that AA Fly (1.2478) is clustered more closely compared to TT Air (1.6602).
• Since to the Standard Deviation for AA Fly is closer, the variance for AA Fly (1.557) as compared to TT Air (2.756) shows clearly the dispersion rate of both variables.
In conclusion, this could mean that TT Air has maximized their usage of aircrafts and is more efficient than AA Fly.
[Com: Good work on...