mth3003 pjj sem i 2015/2016. assignment :25% assignment 1 (10%) assignment 2 (15%) mid exam :30%...
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MTH3003 PJJSEM I 2015/2016
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ASSIGNMENT :25%
Assignment 1 (10%)
Assignment 2 (15%) Mid exam :30%
Part A (Objective)
Part B (Subjective) Final Exam: 40%
Part A (Objective)Part B (Subjective - Short)Part C (Subjective – Long)
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oDefinitionoGraphing
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MEASURES OF CENTER- Arithmetic Mean or Average- Median- Mode
Group and ungrouped data
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• Range• Interquartile Range• Variance• Standard Deviation
Group an ungrouped data
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interpretCalculateQ1, Q2 and Q3, IQR, Upper fence, lower fence, outlier
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• The lower and upper quartiles (Qlower and upper quartiles (Q1 1 and Qand Q33), ), can be calculated as follows:
• The position of Qposition of Q11 is
0.75(n + 1)
0.25(n + 1)
•The position of Qposition of Q33 is
once the measurements have been ordered. If the positions are not integers, find the quartiles by interpolation.
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The prices ($) of 18 brands of walking shoes:
40 60 65 65 65 68 68 70 70
70 70 70 70 74 75 75 90 95
Position of Q1 = 0.25(18 + 1) = 4.75
Position of Q3 = 0.75(18 + 1) = 14.25
Example
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• Basic concept• The probability of an event - how to find prob
• Counting rules• Calculate probabilities
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Event Relations: Union, Intersection, Complement Calculating Probabilities for Unions
The Additive Rule for UnionsThe Additive Rule for Unions
A Special Case – Mutually Exclusive
Complements
Intersections
Independent and Dependent EventsIndependent and Dependent Events
Conditional Probabilities
The Multiplicative Rule for Intersections
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Probability Distributions forDiscrete Random VariablesProperties for Discrete Random VariablesExpected Value and Variance
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The properties for a discrete probability function (PMF) are:
Cumulative Distribution Function (CDF)
1)(
1)(0
)()(
all
x
xp
xxp
xXPxp
1)(0)(
)()()(
)()(
FF
xpbXPbF
xXPxFb
y
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Toss a fair coin three times and define X = number of heads.
1/8
1/8
1/8
1/8
1/8
1/8
1/8
1/8
P(X = 0) = 1/8P(X = 1) = 3/8P(X = 2) = 3/8P(X = 3) = 1/8
P(X = 0) = 1/8P(X = 1) = 3/8P(X = 2) = 3/8P(X = 3) = 1/8
HHHHHH
HHTHHT
HTHHTH
THHTHH
HTTHTT
THTTHT
TTHTTH
TTTTTT
x
3
2
2
2
1
1
1
0
X p(x)
0 1/8
1 3/8
2 3/8
3 1/8
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Discrete distributions: The binomialbinomial distribution
The PoissonPoisson distribution
The hypergeometrichypergeometric distributionTo find probabilities
formulacumulative table
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I. The Binomial Random VariableI. The Binomial Random Variable1. Five characteristics: n identical independent trials, each resulting in either success S or failure F; probability of success is p and remains constant from trial to trial; and x is the number of successes in n trials.
2. Calculating binomial probabilities
a. Formula:b. Cumulative binomial tables
3. Mean of the binomial random variable: np 4. Variance and standard deviation: 2 npq and
knknk qpCkxP )(
npq
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A marksman hits a target 80% of the time. He fires five shots at the target. What is the probability that exactly 3 shots hit the target?
P(P(xx = 3) = 3) = P(x 3) – P(x 2)= .263 - .058= .205
P(P(xx = 3) = 3) = P(x 3) – P(x 2)= .263 - .058= .205
Check from formula:
P(x = 3) = .205
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II. The Poisson Random VariableII. The Poisson Random Variable 1. The number of events that occur in a period of time
or space, during which an average of such events are expected to occur. Examples:Examples:
• The number of calls received by a switchboard during a given period of time.
• The number of machine breakdowns in a day
2. Calculating Poisson probabilities
a. Formula:b. Cumulative Poisson tables
3. Mean of the Poisson random variable: E(x) 4. Variance and standard deviation: 2 and
( )!
keP x k
k
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III. The Hypergeometric Random VariableIII. The Hypergeometric Random Variable1. The number of successes in a sample of size n from a finite population containing M successes and N M failures2. Formula for the probability of k successes in n trials:
3. Mean of the hypergeometric random variable:
4. Variance and standard deviation:
N
Mn
N
Mn
12
N
nN
N
MN
N
Mn
12
N
nN
N
MN
N
Mn
Nn
NMkn
Mk
C
CCkxP
)( N
n
NMkn
Mk
C
CCkxP
)(
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A package of 8 AA batteries contains 2 batteries that are defective. A student randomly selects four batteries and replaces the batteries in his calculator. What is the probability that all four batteries work?
84
20
64)4(C
CCxP
Success = working battery
N = 8
M = 6
n = 470
15
)1)(2)(3(4/)5)(6)(7(8
)1(2/)5(6
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The Standard Normal DistributionThe Standard Normal Distribution1. The normal random variable z has mean 0 and standard deviation 1.2. Any normal random variable x can be transformed to a standard normal random variable using
3. Convert necessary values of x to z.4. Use Normal Table to compute standard normal probabilities.
x
z
x
z
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The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation 0.1. What is the probability that a randomly selected package weighs between 0.80 and 0.85 pounds?
)85.80(. xP
)5.12( zP
0440.0228.0668.
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We can calculate binomial probabilities usingThe binomial formulaThe cumulative binomial tables
When n is large, and p is not too close to zero or one, areas under the normal curve with mean np and variance npq can be used to approximate binomial probabilities.
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Make sure to include the entire rectangle for the values of x in the interval of interest. That is, correct the value of x by This is called the continuity correction.continuity correction. Standardize the values of x using
( 0.5)x npz
npq
( 0.5)x npz
npq
Make sure that np and nq are both greater than 5 to avoid inaccurate approximations!
0.50.5
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Suppose x is a binomial random variable with n = 30 and p = .4. Using the normal approximation to find P(x 10).
n = 30 p = .4 q = .6
np = 12 nq = 18
683.2)6)(.4(.30
12)4(.30
Calculate
npq
np
683.2)6)(.4(.30
12)4(.30
Calculate
npq
np
The normal approximation is ok!
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)683.2
125.10()10(
zPxP
2877.)56.( zP
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Sampling DistributionsSampling distribution of the sample
meanSampling distribution of a sample
proportion Finding Probabilities for the
Sample MeanSample Proportion
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A random sample of size n is selected from a population with mean and standard deviation
he sampling distribution of the sample mean will have mean and standard deviation .
If the original population is normalnormal, , the sampling distribution will be normal for any sample size.
If the original population is non normal, non normal, the sampling distribution will be normal when n is large.
The standard deviation of x-bar is sometimes called the STANDARD ERROR (SE).
xn/
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1587.8413.1)1(
)16/8
1012()12(
zP
zPxP
1587.8413.1)1(
)16/8
1012()12(
zP
zPxP
If the sampling distribution of is normal or approximately normal standardize or rescale the interval of interest in terms of
Find the appropriate area using Z Table.
If the sampling distribution of is normal or approximately normal standardize or rescale the interval of interest in terms of
Find the appropriate area using Z Table.
Example: Example: A random sample of size n = 16 from a normal distribution with = 10 and = 8.
x
/
xz
n
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The standard deviation of p-hat is sometimes called the STANDARD ERROR (SE) of p-hat.
A random sample of size n is selected from a binomial population with parameter p.
The sampling distribution of the sample proportion,
will have mean p and standard deviation
If n is large, and p is not too close to zero or one, the sampling distribution of will be approximately approximately normal.normal.
n
xp ˆ
n
pq
p̂
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0207.9793.1)04.2(
)
100)6(.4.
4.5.()5.ˆ(
zP
zPpP
0207.9793.1)04.2(
)
100)6(.4.
4.5.()5.ˆ(
zP
zPpPExample: Example: A random sample of size n = 100 from a binomial population with p = 0.4.
If the sampling distribution of is normal or approximately normal, standardize or rescale the interval of interest in terms of
Find the appropriate area using Z Table.
If the sampling distribution of is normal or approximately normal, standardize or rescale the interval of interest in terms of
Find the appropriate area using Z Table.
p̂
p̂ pz
pq
n
If both np > 5 and
np(1-p) > 5