MTH3003 PJJ SEM I 2015/2016. ASSIGNMENT :25% Assignment 1 (10%) Assignment 2 (15%) Mid exam :30%...
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Transcript of MTH3003 PJJ SEM I 2015/2016. ASSIGNMENT :25% Assignment 1 (10%) Assignment 2 (15%) Mid exam :30%...
MTH3003 PJJSEM I 2015/2016
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)
oDefinitionoGraphing
MEASURES OF CENTER- Arithmetic Mean or Average- Median- Mode
Group and ungrouped data
• Range• Interquartile Range• Variance• Standard Deviation
Group an ungrouped data
interpretCalculateQ1, Q2 and Q3, IQR, Upper fence, lower fence, outlier
• 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.
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
• Basic concept• The probability of an event - how to find prob
• Counting rules• Calculate probabilities
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
Probability Distributions forDiscrete Random VariablesProperties for Discrete Random VariablesExpected Value and Variance
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
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
Discrete distributions: The binomialbinomial distribution
The PoissonPoisson distribution
The hypergeometrichypergeometric distributionTo find probabilities
formulacumulative table
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
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
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
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
)(
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
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
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.
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.
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
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!
)683.2
125.10()10(
zPxP
2877.)56.( zP
Sampling DistributionsSampling distribution of the sample
meanSampling distribution of a sample
proportion Finding Probabilities for the
Sample MeanSample Proportion
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/
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
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
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