Discussion on Descriptive statistics and Inferential statistics

Statistical exam
Statistical exam, this is limited time exam (2 hours)the exam start at the 2th Dec 5pm-7pm(Sydney time )i will provide the midterm exam

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Topic 4 introduction to statistical inference
Descriptive statistics Inferential statistics
Recall：
Why we need statistical inference?
Population:
1. Parameters describe populations.
2. Parameters are almost always unknown.
3. We take a random sample from the population of interest to
obtain the necessary data.
4. We calculate one or more statistics from the sample data.
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To achieve the following five objectives:
1. Describe a single population
2. Compare two populations
3. Compare more than two populations
4. Study the relationship between two variables
5. Study the relationship amongst k + 1 variables
The sample is said to be independently and identically
distributed (IID):
(1) draws from the population are independent
(2) they have the same distribution.
Sampling distribution
A sampling distribution is a probability distribution of a statistic obtained through a
large number of samples drawn from a specific population. The sampling distribution
of a given population is the distribution of frequencies of a range of different
outcomes that could possibly occur for a statistic of a population.
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𝑺𝒂𝒎𝒑𝒍𝒊𝒏𝒈 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒂𝒎𝒑𝒍𝒆 𝒎𝒆𝒂𝒏 𝑿 ̅
The mean of X ̅ :
The expected value of the sampling distribution of the sample mean equals the
population mean :
A. when the population is normally distributed.
B. when the population is symmetric.
C. when the population size N > 30.
D. for all populations.
The variance of X ̅ :
The standard deviation of X ̅ :–standard error
𝜎̅𝑥
=
𝜎 𝑛√
If X1, X2,…,Xn is normal distribution X ̅ is normal distribution
X
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If X1, X2,…,Xn is non-normal distribution X ̅ is ?????
Central Limit Theorem, C.L.T.
Let X1, X2, …,Xn be a random sample of size n from any distribution with a
finite mean μ and a finite standard deviation σ. If nis large (n > 30), the
distribution of the sample mean can be approximated by:
The standardised random variable is:
☺ Practice:
Which of the following best describes the Central Limit Theorem?
A. If the distribution of a random variable is non-normal, the sampling distribution of the
sample mean will be approximately normal for samples n ≥ 30.
B. The sample mean is close to 0.50.
C. The underlying population is normal.
D. All of these choices are correct.

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Finite population:
𝑰𝒇
𝒏 𝑵
≥ 𝟏𝟎% 𝒕𝒉𝒆𝒏
The central limit theorem states that if a random sample of size n is drawn from a
population, then the sampling distribution of the sample mean :
A. is approximately normal if n > 30.
B. is approximately normal if n < 30.
C. is approximately normal if the underlying population is normal.
D. has the same variance as the population.
【】
1. A sample of n = 100 observations is drawn from a normal population, with µ
= 1000 and σ = 200. Find:
a) P(X ̅ >1050)
b) P(X ̅ < 960)
c) P(X ̅ >1100)
2. A sample of 50 observations is taken from a normal population, with µ = 100
and σ = 10. If the population is finite with N = 250. Find:
a) P(X ̅ >103)
b) P(98 < X ̅<101)
𝑺𝒂𝒎𝒑𝒍𝒊𝒏𝒈 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝑺𝒂𝒎𝒑𝒍𝒆 𝑷𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝒑 ̂
X
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Let X be the number of successes in n independent Bernoulli trials with
probability of success p.
𝒑 ̂ = 𝑿
𝒏
sample proportion of successes
By C.L.T., the sampling distribution of is approximately:
☺ Practice:
1. A commercial for a manufacturer of household appliances claims that only 3%
of all its products require a service call within the first year after purchase. A
consumer protection association wants to check the claim by surveying 400
households that recently purchased one of the company’s appliances. What is
the probability that more than 5% will require a service call within the first year?
What would you say about the commercial`s honesty if, in a random sample of
400 households, 5% report at least one service call?
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2. An accounting professor claims that no more than one-quarter of
a) What is the probability that in a random sample of 1200 undergraduate
business students, 336 or more will major in accounting?
 Point estimates–draws inferences about a population by
estimating the value of an unknown population parameter, using a
single value based on a random sample.
 Interval estimate–draws inferences about a population by
estimating the value of an unknown population parameter using an
interval that is likely to include the value of the population
parameter based on a random sample.
Which of the following best describes an interval estimator?
A. An interval estimator is the same as a point estimator.
B. An interval estimator is an interval that draws inferences about a population based on
a sample statistic.
C. An interval estimator can only be done for the population mean.
D. An interval estimator can only be done for the population proportion.
Properties of Point Estimate
☺ Unbiased
An unbiased estimator of the population parameter θ if its expected
value is equal to the parameter itself.
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The sample mean 𝑋 ̅ is an unbiased estimator of μ because:
The sample variance defined by is an unbiased
estimator
The sample proportion is an unbiased estimator of p ̂ because:
☺ Consistency
An estimator is said to be consistent if its value tends to become closer to
the population value θ as the sample size n becomes larger. That is, the
standard error of the estimator tends to zero.
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☺ Efficiency
For two unbiased estimators 𝜃
̂1
and 𝜃
̂2
of θ, the one with a smaller
variance (or standard error) is said to be relatively efficient (compared to
the other one).
Which of the following statements are correct?
A. If there are two unbiased estimators of a parameter, the one whose variance is larger is
said to be relatively efficient.
B. If there are two unbiased estimators of a parameter, the one whose mean is larger is
said to be relatively efficient.
C. If there are two unbiased estimators of a parameter, the one whose mean is smaller is
said to be relatively efficient.
D. If there are two unbiased estimators of a parameter, the one whose variance is smaller
is said to be relatively efficient.
Topic 5 Parameter estimation—confidence interval
We want to find an interval of values (LCL, UCL) for θ ̂ such that we are
confident to say that this interval will contain the true but unknown θ
with probability 100(1-α)%.
LCL = lower confidence limit
UCL = upper confidence limit
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The probability (1-α) is called the confidence level.
1. Interval estimating–μ
 when σ is known is:
 The 100(1-α)% confidence interval (CI) for μ:
☺ Practice:
A survey of 20 Australian companies indicated that the average annual
income of company secretaries was \$120,000. Assuming that the
population standard deviation is \$7,500 and that the annual incomes are
normally distributed, calculate the 90% confidence interval estimate of
the average annual income of all company secretaries.
 when σ is unknown is:
unknown μ + unknown σ
The Student-t distribution (t-distribution) is a symmetric probability
distribution. This distribution has a shape parameter, known as the
degrees of freedom (d.f.) ν (> 0) and therefore, it has more flexible
symmetric shapes than the normal distribution. If fact, the normal
distribution is a special case of the Student-t distribution.
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Eg,. find 1-α=0.95 df=30(=n-1)
 The 100(1-α)% confidence interval (CI) for μ
☺ Practice:
The degrees of freedom used to find the t/2 for a confidence interval for the population
mean?
A. The degrees of freedom are n − 2.
B. The degrees of freedom are n − 3.
C. The degrees of freedom are n − 1.
D. The degrees of freedom are n.
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its repair operators are the loneliest in the world because its washing
machines require the smallest number of service calls. To examine this
claim, researchers drew a random sample of 100 owners of five-year old
washing machines of that manufacturer. The mean and the standard
deviation of the number of services calls in the five-year period were 4.3
and 1.8 respectively. Find the 90% confidence interval estimate for the
mean number of services calls for all five-year-old washing machines
produced by that manufacturer.
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2. Interval estimating—P
 The 100(1-α)% confidence interval (CI) for p
☺ Practice:
Which of the following statistical distributions are used to find a confidence interval for the
population proportion?
A. Student t distribution
B. Standard normal distribution
C. Chi-square distribution
D. None of these choices are correct.
A statistics practitioner working for the Australian Cricket Board wants to
supply radio and television commentators with interesting statistics. He
observed several hundred games and counted the number of times a
batsman was run out off a no ball. He found there were 373 such attempts
of which 259 were successful. Estimate with 95% confidence the
proportion of times a batsman was run out off a no ball.
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3. Sample size determination
✓ Estimation of population proportion μ
Error bound
Note: You must round up the value of n to the next integer.
If the population standard deviation σ is unknown:
✓ Estimation of population proportion p
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☺ Practice:
1. Determine the sample size required to estimate μ to within 10 units,
with 99% confidence. We know that the range of the population
observations is 200 units.
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4. Interval estimating— μ1-μ2
（ independent population）
 when σ1 and σ2 is known is:
=
 The 100(1-α)% confidence interval (CI) for is:
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Pooled variance
 when σ1 and σ2 is unknown is:
Case 1: σ12 and σ22 are unknown and σ12 = σ22 = σ2 equal variance
The 100(1-α)% confidence interval (CI) for is:
Case 2: σ12 and σ22 are unknown and σ12 ≠ σ22 unequal variance
The 100(1-α)% confidence interval (CI) for is:
where
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☺ Practice:
Two random samples from two normal populations produced the
following statistics:
n1 = 51, X 1 = 35, s1 = 28.
n2 = 40, X 2 = 28, s2 = 10.
Assume that the two population variances are different. Estimate with
95% confidence the difference between the two population means.
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5. Interval estimating— μ1-μ2
（ NOT independent population）
Matched pairs experiment
☺ Practice:
A simple random sample of ten firms was asked how much money
(in thousands of dollars) they spent on employee training programs
this year and how much they plan to spend on these programs next
year. The data are shown below.
Firm 1 2 3 4 5 6 7 8 9 10
This year 25 31 12 15 21 36 18 5 9 17
Next year21 30 18 20 22 36 20 10 8 15
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Assume that the populations of amount spent on employee training
programs are normally distributed.
a. Estimate with 95% confidence the mean difference.
6. Interval estimating—p1-p2
Assumptions:
1. The populations are binomial, Bin(n1,p1) and Bin(n2,p2).
2. The populations are independent.
 The 100(1 – α)% CI for p1 – p2 is
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Topic 4
1.
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2.
3.
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4.
Topic 5
5.
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6.
7.
8.
9.
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10.
a. -1.1 ± (2.262) *(1.0588) = (–3.495, 1.295).
11. (补充)
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Topic 2 Probability
Practice:
1. The results of three flips of a fair coin are observed. Consider the following events:
A: At least two heads are observed.
B: Exactly one tail is observed.
C: Exactly two tails are observed.
a Define an appropriate sample space S for this experiment.
b Assign probabilities to the outcomes in S.
c Find P(A) by summing the probabilities of the outcomes in A.
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d Find P
（ 𝐵 ̅ ).
e Find P(B C).
f Find P(A B).
g Find P(B|A).
h Is there a pair of independent events among A, B and C?
i Is there a pair of mutually exclusive events among A, B and C?
DWJWLXS 2020 S1 Qbus5001 2. A consumer goods company recruits several graduating students from universities each year. Concerned about the high cost of training new employees, the company instituted a review of attrition among new recruits. Over five years, 30% of new recruits came from a local university, and the balance came from more distant universities. Of the new recruits, 20% of those who were students from a local university resigned within two years, while 45% of other students resigned. Given that a student resigned within two years, what is the probability that she was hired from a) a local university? b a more distant university?
DWJWLXS 2020 S1 Qbus5001 3. Casino Windsor conducts surveys to determine the opinions of its customers. Among other questions, respondents are asked the question ‘What is your overall impression of Casino Windsor?’. The possible responses are: Excellent, Good, Average and Poor. Additionally, the gender of the respondent is also noted. After analysing the results, the following table of joint probabilities was produced. a What proportion of customers rate Casino Windsor as excellent? b Determine the probability that a male customer rates Casino Windsor as excellent. c Find the probability that a customer who rates Casino Windsor as excellent is a man. d Are gender and rating independent? Explain your answer.
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Topic 3 Distribution 3.1 Expected value and variance The population mean is the weighted average of all of its possible values. The population variance is the weighted average of the squared deviations from the mean.
Shortcut calculation for population variance Refer to practice 3.

Expected value E(x)
The same
𝜇 𝜎
DWJWLXS 2020 S1 Qbus5001 4. The quarterly rates of return for four shares are recorded.
a Calculate the mean and variance of each of the four shares.
b If you wish to construct a portfolio that maximises the expected return, what should you do?
c If you wish to construct a portfolio with two of these shares that minimises the risk, what should you do?
3.2 Binomial distribution binom.dist(x,n,p,0) Compute PMF of the Bin(n,p) distribution binom.dist(x,n,p,1) Compute CDF of the Bin(n,p) distribution
DWJWLXS 2020 S1 Qbus5001 5. A study of drivers reveals that, when lost, 45% will stop and ask for directions, 30% will consult a map and 25% will continue driving until the location has been determined. Suppose that a sample of 200 drivers was asked to report what they do when lost. Find the following probabilities.
a At least 100 stop and ask directions.
b At most 55 continue driving.
c Between 50 and 75 (inclusive) consult a map.
d. of all outcomes of consults a map which has the greatest marginal probability.
e. what is the average value for 200 drivers who consults a map? What is the standard deviation?
3.3 Poisson distribution The Poisson random variable is the number of occurrences of events, which we’ll continue to call successes. The difference between the two random variables is that a binomial random variable is the number of successes in a set number of trials, whereas a Poisson random variable is the number of successes in an interval of time or specific region of space.
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poisson.dist(x,mu,0) Compute PMF of the Poi(μ) distribution
poisson.dist(x,mu,1) Compute CMF of the Poi(μ) distribution 6. During the summer months (December to February, inclusive), an average of five marriages per month take place in a small city. Assuming that these marriages occur randomly and independently of one another, find the probability of the following occurring:
a Fewer than four marriages will occur in December.
b At least 14 but not more than 18 marriages will occur during the entire three months of summer.
c Exactly 10 marriages will occur during the two months of January and February.
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3.4 Uniform distribution 7. The amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 30 and 60 minutes. One student is selected at random. Find the probability of the following events.
a The student requires more than 55 minutes to complete the quiz.
b The student completes the quiz in a time between 30 and 40 minutes.
c The student completes the quiz in exactly 37.23 minutes.
d The professor would like to track (and possibly help) students who are in the top 10% of completion times. What completion time should she use?
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3.5 Exponential distribution one-parameter distribution 8. A firm has monitored the duration of long-distance telephone calls placed by its employees, to help it decide which long-distance call package to purchase. The duration of calls was found to be exponentially distributed, with a mean of 5 minutes.
a What proportion of calls last more than 120 seconds?
b What proportion of calls last more than 5 minutes?
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c What proportion of calls are shorter than 10 minutes?
3.6 Normal distribution-two parameter distribution (miu+sigma)
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Standard normal distribution 9. Every day a bakery prepares its famous marble rye. The statistically savvy baker determined that daily demand is normally distributed with a mean of 850 and a standard deviation of 90. a. How many loaves should the bakery bake if the probability of running short on any day is to be no more than 30%? b. Any marble rye loaves that are unsold at the end of the day are marked down and sold for half price. How many loaves should the baker prepare daily so that the proportion of days on which there are unsold loaves is no more than 60%?
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 Normal Approximation to Binomial Probability
If n is large, the binomial distribution can be approximated by the normal
distribution.
Continuity correction
A correction factor that allows for the approximation of a discrete random
variable by a continuous random variable.
It’s as simple as adding or subtracting 0.5 to the discrete x-value. 10. Suppose that X is a binomial random variable with n = 100 and p = 0.20. Use the normal approximation to find the probability that X takes a value between 22 and 25 (inclusive).
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Module 4. Statistical inference and sampling distributions
 Point estimates–draws inferences about a population by estimating the value of an unknown population parameter, using a single value based on a random sample.
 Interval estimate–draws inferences about a population by estimating the value of an unknown population parameter using an interval that is likely to include the value of the population parameter based on a random sample.
Properties of Point Estimate
Unbiased An unbiased estimator of the population parameter θ if its expected value is equal to the parameter itself.
Consistency An estimator is said to be consistent if its value tends to become closer to the population value θ as the sample size n becomes larger. That is, the standard error of the estimator tends to zero.
Efficiency For two unbiased estimators 𝜃
̂1 and 𝜃
̂2 of θ, the one with a smaller variance (or standard error) is said to be relatively efficient (compared to the other one).
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DWJWLXS 2020 S1 Qbus5001 11. A commercial for a manufacturer of household appliances claims that only 3% of all its products require a service call within the first year after purchase. A consumer protection association wants to check the claim by surveying 400 households that recently purchased one of the company’s appliances. What is the probability that more than 5% will require a service call within the first year?
DWJWLXS 2020 S1 Qbus5001 12. Refer to Excel
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Confidence interval – interval estimation
confidence interval estimator– An interval estimator that we have a certain degree of confidence that it contains the value of the parameter.
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Q13 refers to Excel
Interpretation of the confidence interval:
the 95% confidence interval estimator of . implies that 95% of the values of X will create intervals
that will contain the true value of the population mean.
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14. Tina surveyed 545 people who watched TV on Saturday. She then used this survey to construct a confidence interval estimating the proportion of people who watched the AFL Grand Final. The resulting confidence interval is (0.16, 0.24). what confidence level did Tina use in constructing this interval?
15. Refer to Excel

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QBUS5001 Mid-term
Part 1
Topic 1 Sample vs population
Topic 2 Probabilities
Topic 3 Distributions
Topic 4 Sampling distributions
Topic 5 Estimation 主讲人： Sara 21/09/2020
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Topic 1. Sample vs population
1.1 Introduction to Statistics
Descriptive statistics： organise, summarise and present data using graphical methods
or numerical techniques
Inferential statistics： Inferential statistics uses data from a sample to make inference
Q1. Which of the following statements is true?
A Descriptive statistics never uses graphical techniques.
B Descriptive statistics never uses numerical techniques.
C Statistical inference is used to describe a sample of data.
D Statistical inference is used to draw conclusions about characteristics of populations based on
sample data.
Q2. Which of the following is the goal of descriptive statistics?
A To summarise data.
B To display aspects of the collected data.
C To summarise data and display aspects of the collected data.
D To estimate characteristics of the population.
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1.2 Types of Data
Data categorised by measurement:
Data categorised by collection:
Cross sectional data: at the same point of time, or without regard to differences in time.
Time series data: in time order.
1.3 Graphical & Tabular Methods
nominal data:
Nominal Data
•Categories (no order of direction)
Ordinal Data
•Ordered Categories
Interval Data
•Differenes between measurements but no true zero
Ratio Data
•Differenes between measurements, true zero exists
Qualitative
Quantitative
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The main criticism against pie charts is that it is very difficult for the human eye to estimate
the magnitude of angles, thus making visualization not very clear.
numerical data:
Scatter plot: the relation between two variables.
Histogram:
Absolute or relative?
Pdf or cdf?
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Line chart: Time series
Contingence table: cross-sectional
Q3. Which of the following statements is true?
A. A contingency table may also be called a cross classification table
B. A contingency table is used to describe two nominal variables.
C. A bar chart may be used as a graphical display of a contingency table.
D. All of these choices are correct.
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1.4 Data Collection
Census vs Sample
1. Simple Random Sampling
2. Stratified Random Sampling
3. Cluster Sampling
Sampling & Non-Sampling Errors
Q4. A professor intends to compare the performances of his first year economics subject
students at the four campuses of the university. In order to do so, he selects a random
sample of 40 first year economics students from each campus. What is this type of sampling
called?
A. Stratified random sampling.
B. Simple random sampling.
C. Cluster sampling.
D. None of these choices are correct.
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1.5 Descriptive Statistics
• Population Parameters VS Sample statistics
1.5.1 Central tendency
Means:
Median: position=(n+1)/2, not affected by the extreme value.
Mode:
Skewness:
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Q5. Which of the following statements describes when the median is a better measure of centre than the
mean?
A. When the distribution is symmetric.
B. When the distribution is positively skewed.
C. When the distribution Is negatively skewed.
D. When the distribution is positively skewed or negatively skewed.
Q6. What measure of central location is best used with a categorical variable?
A. The mean, median or the mode if the distribution is symmetric.
B. The median if the distribution is skewed.
C. The range.
D. The mode.
1.5.2 Variability(dispersion)
Range= MAX-MIN
Boxplot：
IQR=Q3-Q1
Lower fence: Q1-1.5*IQR Upper fence:Q3+1.5*IQR
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Coefficient of variation (CV)
Chebyshev’ s Theorem
Empirical rules:
Q7. According to Chebyshev’s theorem, which of the following is the percentage of measurements in a
data set that fall within three standard deviations of their mean?
A. 75%.
B. At least 75%.
C. 89%.
D. At least 89%.
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Q8. According to the empirical rule, which of the following is the approximate percentage of
measurements in a data set (provided that the data set has a bell-shaped distribution) that fall within
two standard deviations of their mean?
A. 68%.
B. 75%.
C. 95%.
D. 99%.
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Topic 2 Probabilities
2.1 validity of the probabilities
0<=P(Xi)<=1 Σ P(Xi)=1
2.2 Assigning Probability
1. Classical approach – probability is defined based on equally likely events.
2. Relative frequency approach – probability of an outcome is defined as a long term
relative frequency of occurrence of the outcome.
3. Subjective approach – probability is assigned based on personal judgement.
Q9. An approach of assigning probabilities that assumes that all outcomes of the experiment
are equally likely is referred to as the:
A. subjective approach.
B. objective approach.
C. classical approach.
D. relative frequency approach.
4.
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 Event, A – A set of outcomes from a random experiment.
 Complement event 𝐴̅
  
Mutually exclusive events
 Exhaustive events
 Independent events: A and B are said to be independent events if
P(A∩B) = P(A)P(B)
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Q10 . If P(A) = 0.20, P(B) = 0.30 and P(A  B) = 0.00, then A and B are:
A. dependent events.
B. independent events.
C. mutually exclusive events.
D. complementary events.
Q11. If A and B are mutually exclusive events, with P(A) = 0.30 and P(B) = 0.40, then P(A  B) is:
A. 0.10.
B. 0.12.
C. 0.70.
D. None of these choices are correct.
Q12. 4. If A and B are independent events, with P(A) = 0.50 and P(B) = 0.70, then the probability that
A occurs or B occurs or both occur is:
A. 1.20.
B. 0.20.
C. 0.85.
D. 0.10.
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Q13 If each of two independent file servers has a reliability of 93 percent and either alone
can run the website, then the overall website availability is:
A. .9951.
B. .8649.
C. .9300.
D. .9522.
 Joint, Marginal & Conditional Probabilities
 Unconditional probability—marginal probability
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Q14. Two events A and B are said to mutually exclusive if:
A. P(A | B) = 1.
B. P(B | A) =0.
C. P(A B) =1.
D. P(A B) = 0.5
Q15. A statistics professor classifies his students according to their gender and the number of
hours of paid work they do a week. The following table gives the proportions of students falling
into the various categories. One student is selected at random.
Paid Work (hours/week)
Gender 0 1–8 9–16 Over 16
Male 0.06 0.20 0.15 0.06
Female 0.12 0.18 0.20 0.03
a. If the student selected is female, what is the probability that she works between 1 and 8
hours a week?
b. If the selected student works more than 16 hours a week, what is the probability that the
student is male?
c. What is the probability that the student selected is female or does do any paid work or
both?
d. Is gender independent of the number of hours of paid work done a week? Explain using
probabilities.
ANS:
a. 0.34.
b. 0.67.
c. 0.59.
d. No, since, for example, P(female  no paid work) = 0.59 ¹ P(female)×P(no paid work) =
0.53×0.18 = 0.0954.
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Probability Tree, or a simply a tree diagram is a way to express probabilities of
possible events in branches of a tree.
Bayes’ Theorem
EG .From the following tree, find the probability that a randomly chosen person will get the flu vaccine
and will also get the flu.
A. .10
B. .07
C. .19
D. .70
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Q16. A standard admissions test was given at three locations. One thousand students took the
test at location A, 600 students at location B, and 400 students at location C. The percentages
of students from locations A, B and C who passed the test were 70%, 68% and 77%,
respectively. One student is selected at random from among those who took the test.
a. What is the probability that the selected student passed the test?
b. If the selected student passed the test, what is the probability that the student took the
test at location B?
c. What is the probability that the selected student took the test at location C and failed?
ANS:
a. 0.708.
b. 0.288.
c. 0.046.
Q17. An ice cream vendor sells three flavours: chocolate, strawberry and vanilla. 45% of the sales
are chocolate, 30% are strawberry and the rest are vanilla. Sales are by the cone or the cup.
The percentages of cones sales for chocolate, strawberry and vanilla are 75%, 60% and 40%,
respectively. For a randomly selected sale, define the following events:
A1 = chocolate chosen.
A2 = strawberry chosen.
A3 = vanilla chosen.
B = ice cream in a cone.
B = ice cream in a cup.
Use this information to answer the following question(s).
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1) Find the probability that the ice cream was strawberry-flavoured, given that it was sold in
a cup.
ANS:
P( | B ) = P(  B ) / P( B ) = 0.12 / 0.3825 = 0.3138.
2) Find the probability that the ice cream was vanilla-flavoured, given that it was sold in a
cup.
ANS:
P( | B ) = P(  B ) / P( B ) = 0.15 / 0.3825 = 0.3922.
A2 A2
A3 A3
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 Expected value
Q18. The lottery commission has designed a new instant lottery game. Players pay \$1.00 to
scratch a ticket, where the prize won, X, (measured in \$) has the following discrete probability
distribution :
X P(X)
0 0.95
10 0.049
100 0.001
Which of the following best describes the expected value of X ?
A. In the long run, the average prize won per \$1 played is \$0.
B. In the long run, the average prize won per \$1 played is \$0.41.
C. In the long run, the average prize won per \$1 played is \$0.59.
D. None of these choices are correct
Q19.The lottery commission has designed a new instant lottery game. Players pay \$1.00 to
scratch a ticket, where the prize won, X, (measured in \$) has the following discrete probability
distribution :
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X P(X)
0 0.95
10 0.049
100 0.001
Which of the following best describes the standard deviation of X ?
A. 14.552 in \$
B. 3.815 in \$
C. 0.348 in \$
D. None of these choices are correct
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 Covariance：
1. evaluates how the mean values of two variables move together.
2. the strength of linear relationship between the two variables.
 Coefficient of correlation：
Q20. An analysis of the stock market produces the following information about the returns of
two stocks:
Assume that the returns are positively correlated, with ρ12 = 0.80.
a. Find the mean and standard deviation of the return on a portfolio consisting of an equal
investment in each of the two stocks.
b. Suppose that you wish to invest \$1 million. Discuss whether you should invest your money
in stock 1, stock 2, or a portfolio composed of an equal amount of investments on both stocks.
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ANS:
a. The expected return on the portfolio is E(0.5X1 + 0.5X2) = 0.5E(X1) + 0.5E(X2)= 16.5%.
The variance of the portfolio’s return is (0.5*0.20)2 + (0.5*0.32)2 + 2(0.5)2 *0.80*0.20*0.32
= 0.612.
The standard deviation of the portfolio’s return when ρ12 = 0.80 is therefore 24.74%.
b. Your choice of investment in stock 1, the portfolio, or stock 2 depends on your desired
level of risk (variance of return). The higher the risk you choose, the higher will be the
expected return.
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Topic 3 Probability
 Binomial Distribution
A binomial experiment is the Bernoulli trial that is repeated independently for
a finite number of times. It has the following properties:
1. The binomial experiment consists of n (fixed) no. of trials.
2. Each trial has only two possible outcomes (Bernoulli trial).
3. The probabilities of success and failure are p and q = 1 – p.
4. The trials are independent, i.e. outcomes from the trials do not affect one
another.
Combination:
Bin~(n, p)
The mean and variance of X are given by:
p>0.5 left-skewness
p=0.5 symmetric
p<0.5 right-skewness
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Q21. Which of the following is not a characteristic of a binomial experiment? A. There is a sequence of identical trials.
B. Each trial results in two or more outcomes. C. The trials are independent of each other. D. Probability of success p is the same from one trial to another.
2. An official from the Australian Securities and Investments Commission
estimates that 75% of all investment bankers have profited from the use
of insider information. If 15 investment bankers are selected at random
from the Commission’s registry, find the probability that:
a. at most 10 have profited from insider information. b. at least 6 have profited from insider information. c. all 15 have profited from insider information.
ANS:
a. 0.314.
b. 0.999.
c. 0.013.
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 Poisson Distribution
A Poisson experiment is a random experiment that possesses the
following properties.
(1) The number of successes (or occurrences of an event) that occur in
any time interval is independent of the number of successes that occur in
any other interval.
(2) The probability of a success in an interval is the same for all equal-size
intervals.
(3) The probability of a success in an interval is proportional to the size of
the interval.
(4) The probability of more than one success in an interval approaches 0
as the interval becomes smaller.
Let X be the number of occurrences of an event in a given time interval
and μ be the average number of occurrences in this interval (or the rate
of occurrences of an event). The pmf of X is given by:
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Practice:
Q22. Which of the following cannot generate a Poisson distribution?
A. The number of children watching a movie.
B. The number of telephone calls received by a switchboard in a specified time period.
C. The number of customers arriving at a petrol station on Christmas day.
D. The number of bacteria found in a cubic yard of soil.
each afternoon between 2 and 4pm. The calls occur randomly and
independently of one another.
a. Find the probability that the executive will receive 13 calls between 2
and 4pm on a particular afternoon.
b. Find the probability that the executive will receive seven calls between
2 and 3pm on a particular afternoon.
c. Find the probability that the executive will receive at least five calls
between 2 and 4pm on a particular afternoon.
ANS:
a. μ = 10, P(X = 13) = 0.0729.
b. μ = 5, P(X = 7) = 0.105.
c. μ = 10, P(X 5) = 0.971.
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 Uniform Distribution
P(x1≤X≤x2) =
𝑥2−𝑥1
𝑏−𝑎
Practice:
Q24. The probability density function f(x) of a random variable X that is
uniformly distributed between a and b is:
A. 1/(b a). B. 1/(a b). C. (b a)/2. D. (a b)/2.
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Q25 The time required to complete a particular assembly operation is
uniformly distributed between 12 and 18 minutes.
a. What is the probability density function for this uniform distribution?
b. What is the probability that the assembly operation will require more
than 16 minutes to complete?
c. Find the expected value and standard deviation for the assembly
time.
ANS:
a. f(x) = 1/6, 12  x  18.
b. P(16 X 18) = (18-16)(1/6)=1/3.
c. E(X) = 15 minutes, 𝜎 = 1.732 minutes.
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 Exponential Distribution
Q26. Which of the following is not true for an exponential distribution with
parameter  ? A.  = 1/  B.  =1/ C. The distribution is completely determined once the value of  is known.
D. The distribution is a two-parameter distribution, since the mean and standard deviation are equal.
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Q27. The time it takes a technician to fix a computer problem is exponentially
distributed, with a mean of 15 minutes. a. What is the probability that it will take a technician less than 10 minutes to fix a computer problem? b. What is the variance of the time it takes a technician to fix a computer problem? c. What is the probability that it will take a technician between 10 to 15 minutes to fix a computer problem?
ANS:
b. 0.4866.
c. 225.
d. 0.1455.
 Normal Distribution
This is the most important continuous distribution considered here.
Many random variables can be properly modelled as normally
distributed.
Many distributions can be approximated by a normal distribution.
Normal distribution is the cornerstone distribution of statistical
inference.
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Q28. Find (a) P(Z ≥ 1.47), (b) P(-2.25 < Z <1.85) and (c) P(0.65 ≤ Z ≤ 1.36).
Q29. A smaller standard deviation of a normal distribution indicates that
the distribution becomes:
Q30. Which of the following is a characteristic of a standard normal
distribution? A. It is a symmetrical distribution. B. The mean is always zero. C. The mean, median and mode are all equal.
D. All of these choices are correct.
Q31. Given that X is a normal variable, which of the following statements
is (are) true?
A. The variable X + 5 is also normally distributed.
B. The variable XQ32 – 5 is also normally distributed.
C. The variable 5X is also normally distributed.
D. All of these choices are correct. A. more skewed to the left. B. flatter and wider.
C. narrower and more peaked. D. symmetrical.
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Q32. A bank has determined that the monthly balances of the saving
accounts of its customers are normally distributed, with an average
balance of \$1200 and a standard deviation of \$250. What proportions
of the customers have monthly balances:
a. less than \$1000?
b. more than \$1125?
c. between \$950 and \$1075?
ANS:
a. 0.2119.
b. 0.6179.
c. 0.1498.
Q33. The recent average starting salary for new college graduates in
computer information systems is \$47 500. Assume that salaries are
normally distributed, with a standard deviation of \$4500. a. What is the probability of a new graduate receiving a salary between \$45 000 and \$50 000? b. What is the probability of a new graduate getting a starting salary in excess of \$55 000? c. What percentage of starting salaries are no more than \$42 250? d. What is the cut-off for the bottom 5% of the salaries? e. What is the cut-off for the top 3% of the salaries?
ANS:
a. 0.4246.
b. 0.0475.
c. 12.10%.
d. \$40 097.50.
e. \$55 960.
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 Normal Approximation to Binomial Probability
If n is large, the binomial distribution can be approximated by the normal
distribution.
Continuity correction
A correction factor that allows for the approximation of a discrete random
variable by a continuous random variable.
It’s as simple as adding or subtracting 0.5 to the discrete x-value.
Q34. Given a binomial distribution with n trials and probability p of a success
on any trial, a conventional rule of thumb is that the normal distribution will
provide an adequate approximation of the binomial distribution if:
A. np  5 and n(1–p)  5. B. np  5 and n(1–p)  5. C. np  5 and n(1–p)  5. D. np  5 and n(1–p)  5.
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Q35. Given that X is a binomial random variable, the binomial probability
P(X  x) is approximated by the area under a normal curve to the left of: A. x. B. –x. C. x + 0.5.
D. x – 0.5.
Q36. The publisher of a daily newspaper claims that 90% of its subscribers
are under the age of 30. Suppose that a sample of 300 subscribers is
selected at random. Assuming the claim is correct, calculate the probability
of finding at least 260 subscribers in the sample under the age of 30.
ANS:
0.9729.

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