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I. (Questions requiring a short answer, mostly concepts. 1 point each).
(1) The likelihood that an event will occur given that another event has already occurred. conditional probability
(2) A concept of probability (a method of assigning probability) based on the assumption that each of the all possible outcomes is euqally likely. Classical probability
(3) A collection of one or more outcomes of an experiment event
(4) A table used to classify observations according to two nominal (qualitative) characteristics ? Contingency table
(5) A quantity resulting from an experiment that, by chance, can assume different values. Random variable
(6) A characteristic of a normal probability distribution that its probability curve gets closer and closer to X-axis (to either (+) or (-) direction), but never touches it. asymptotic
(7) A member of the family of normal probability distributions, with its mean being 0 and standard deviation being 1. Standard normal (z) distribution
(8) A formula to count the number of possible arrangements of certain number (r) of objects from a single group of (n) objects, with the order of objects being not important. nCr (n Combination r)
(9) The sampling distribution of the sample means will approach a normal distribution (regardless of the population distribution), if the size of sample is sufficiently large. Central limit theorem
(10) A single value computed from a sample and used to estimate a population parameter, contrasted with interval estimate. Point estimate
(11) The difference between a sample statistic and the corresponding population parameter. Sampling error
(12) A sampling method in which population is ordered in a way, such as alphabetically, then a starting point is randomly selected & every kth item from it belongs to sample. Systematic random sampling
(13) What does the df shown in the first column of t-distribution table stand for? Degree of freedom
(14) A distribution often used to approximate binomial probility distribution when n is large and is small. Poisson distribution
(15) A measure of the (linear) relationship between two random variables, which can have values from (-) infinite to (+) infinite. Covariance
II. Show both the process of getting the answer(s) and the answer(s) for the following problems.
1. From the following stem-and-leaf display, find the minimum, the 1st quartile, the median, the 3rd quartile, and the maximum value. (Hint: use the percentile formula.)
1| 9
2| 0 1 2 2 6 9
3| 0 1 2 3 4 5 5 7 8 8 9
4| 2 6
5| 0 1 2 4 5 7 8 9
Min: 19, Max: 59
The percentile formula (for the Pth percentile) Lp = (n+1)P/100
The first quartile: L25 = (28+1)*25/100 = 7.25
7th (from the minimum value): 29 & 8th : 30
So the the first quartile (the 7.25th) will be 29.25
The median (50th percentile) is 36 (=35+37/2)
The third quartile: 50.75 (obtain by applying the formula L75= 21.75)
2. A case of 24 bottles contain 1 bottle that is contaminated. 3 bottles are to be selected randomly for testing.
(1) How many different samples of 3 bottles could be selected? 24C3 = 2024
(2) What is the probability that the contaminated bottle is included in a randomly selected testing sample?
1 Pr(no contaminated bottle)= 1 (23/24)(22/23)(21/22)= 1/8= .125.
3. Routine physical examinations for the employees have discovered that 8% of the employees needed corrective shoes, 15% needed dental work, and 3% needed both corrective shoes and dental work. What is the probability that an employee selected at random will need either corrective shoes or dental work?
P(S or D) = P(S) + P(D) P(S and D)= .08 + .15 - .03 = .2
4. An insurance company classifies drivers as good, medium, or poor risks, with the proportions of 30%, 50%, and 20%, respectively. A good drivers probability of having accident is 0.01, medium drivers probability of having accident is 0.03, and the poor drivers probability of having accident is 0.10.
(1) What is the overall probability of a driver has accident?
P(A) = p(g)P(A/g) + P(m)P(A/m) + P(p)P(A/p)
= .003 + .015 + .02 = .038
(2) One person, who had accident, purchased insurance from this company. What is the probability that he is classified as a good driver?
(by Bayes Theorem)
P(g/A) = p(g)P(A/g) / p(g)P(A/g) + P(m)P(A/m) + P(p)P(A/p)
= .003 / .038 = .0787
5. The arrival of customers at a service desk follows a Poisson distribution. If they arrive at a rate of two every five minutes, what is the probability that no customers arrive in a five-minute period?
According to Poisson probability distribution P(0)= 20 e-2/0! = 0.1353
6. A credit card company has found out that customers charge between $100 and $1100 per month. Given that the monthly amount charged is uniformly distributed, answer the following questions.
(1) What is the average monthly amount charged? (100+1100)/2 = 600
(2) 75% of all monthly charges are greater than _____ ?
By solving (1100-x)*(1/1000)= .75 for x, we get x=350
7. The following table provide the information of the 20 large corporations on the salaries of the CEO and whether the company make profit or loss.
CEO paid more than $1 millionCEO paid less than $1 millionTotalProfit making21113Loss making437Total61420If a company is randomly selected from the list of the 20 company, what is the probability that
The CEO made more than $1 million. P(A)= 6/20 = .3
The CEO made more than $1 million or the company made loss.
P(A or B)= 6/20 + 7/20 -4/20(=P(A and B)) = .45
The CEO made more than $1 million given that the company made loss.
Conditional probability P(A/B)= P(A and B)/P(B)= (4/20)/7/20)= 4/7
8. Two random variables are described by the following probibility table.
Y X123P(Y)0.2.10.31.1.1.2.420.1.2.3P(X).3.3.41.0Find the mean of the variables X and Y.
E(X)= 1(.3) + 2(.3) + 3(.4) = 2.1
E(Y)= 0(.3) + 1(.4) + 2(.3) = 1
Find the variance of the variables X and Y.
Var(X) = (1-2.1)2(.3) + (2-2.1)2(.3) + (3-2.1)2(.4) = .69
Var(Y) = (0-1)2(.3) + (1-1)2(.4) + (2-1)2(.3) = .6
Find the covariance of X and Y.
Cov(X,Y) = (1-2.1)(0-1)(.2) + (2-2.1)(0-1)(.1) + (3-2.1)(0-1)(0)
+ (1-2.1)(1-1)(.1) + (2-2.1)(1-1)(.1) + (3-2.1)(1-1)(.2)
+ (1-2.1)(2-1)(0) + (2-2.1)(2-1)(.1) + (3-2.1)(2-1)(.2) = .4
Find the variance of a random variable Z, which is the sum of X and Y (Z=X+Y).
Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,y)= .69+ .6 + .8 = 2.09
9. A new medicine is claimed to be 80% effective for a disease, meaning that 80 out of 100 patients taking the medicine experience significant improvement. Suppose a group of 15 patients take the medicine. What is the probability that:
(1) All 15 will show significant improvement.
Using binomial probability formula (or using binomial table with n=15, =.8) P(15) = .035
(2) 12 or more patients will show significant improvement.
P(12)+P(13)+P(14)+P(15)= .648 (can get easily from the binomial table)
10. A company fills 100 orders a day with a 2% error rate in filling the order satisfactorily. Assume the number of errors to be a binomial distribution.
(1) What is the mean and the standard deviation of the error rate?
Binomial distribution with n=100, =.02, so that = 100 (.02) = 2
2 = 100 (.02) (.98) = 1.96, so that = 1.4
(2) What is the probability that there will be more than 5 order errors in a given day? Using binomial distribution 1 - {P(0)+P(1)+P(2)+P(3)+P(4)+P(5)}
Or use Poisson distribution with =2, P(X>5) = .0165 (From the Poisson table).
11. A study of a company's practice regarding the payment of invoices revealed that an invoice was paid an average of 20 days after it was received. The standard deviation was 5 days. Assuming that the distribution is normal, what percent of the invoices were paid within 15 days of receipt?
N(20,5) X=15 Corresponding Z = (15-20)/5 = -1
P(Z<-1) = .5-.3413 = .1587 (i.e., 15.87%)
12. A test scores are normally distributed with a mean of 500 and a standard deviation of 50. A school will admit those applicants who scores in the upper 6 percent of the distribution. What is the lowest score an applicant must earn to be admitted?
Find the z value corresponding to 6% tail of the Z distribution. It is 1.555
Find the X value following N(500,50), corresponding to Z value of 1.555.
1.555 = (X-500)/50, resulting in X= 578.
13. The mean SAT score for all students is 947 with a standard deviation of 205. Suppose we select a random sample of 60 students, what is the probability the sample mean SAT score is below 900.
Z= (900-947)/205/60 1/2 = -1.78
P= 0.0375 (=0.5-0.4625)
14. The mean amount of a customer purchase at a supermarket is $25, with a standard deviation of $9 & unknown distribution. A sample of 49 customers is selected.
(1) What is the probability of the sample mean being at least $27?
According to central limit theorem, the sample mean follows a normal distribution with mean of 25, and standard deviation of 9/7.
The z value corresponding to the sample mean of 27 is 1.5556 (=(27-25)/(9/7))
P (Xbare"27) = P (Ze"1.5556) = .0594 (about 6%)
(2) Within what range (limits) will 90 percent of the sample means occur?
Find A and B, satisfying P(A<Xbar<B) = .9
The corresponding z value is -1.645 & 1.645 (i.e., P(-1.645<Z<1.645)=.9).
A= 25 - 1.645*(9/7)= 2345 >
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(1) Find the standard error of the sample mean.
n= 10, = 5, Xbar=20 so that the standard error = 5/(10)1/2 = 1.581
(2) Construct the 95% confidence interval for the population mean.
95% confidence interval will be (using z value of 1.96 with two tail 95%)
Between 16.901 (=20-3.099) and 23.099 (=20+3.099)
16. A restaurant manager wants to know how much people spend for eating out. A sample of 10 households was selected with the information on their expenditure on eating out as follows.
$107, $92, $97, $95, $105, $101, $91, $99, $95, $104
He wants to estimate the population mean (of the spending) based on the sample.
Compute the sample mean and sample standard deviation.
Xbar= 986/10 = 98.6
s = 5.54
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