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Probability & Statistics
Review I
1.
2.
3.
Normal Distribution
Sampling Distribution
Inference
- Confidence Interval
1. Normal Distribution: N(μ,σ2)
 The normal probability density function is
• ‘Bell Shaped’, Symmetrical
• Location is characterized by the mean, μ
• Spread is characterized by the standard deviation, σ
Changing μ shifts the distribution
left or right.
Changing σ increases or
decreases the spread.
Standard Normal Distribution: N(0,12)
 The standard normal distribution has a mean of 0 and a standard
deviation of 1.
 A random variable X following Normal distribution N(μ , σ 2)
can be translated to the random variable Z following standard
normal distribution by subtracting the mean of X and dividing by
its standard deviation:
f(Z)
Z
X μ
σ
1
0
Z
Ex) If X is distributed normally with mean of 100 and standard deviation
of 50, the Z value for X = 200 is
Z
X  μ 200  100

 2.0
σ
50
Standard Normal Probability Table
 The Standardized Normal table in the textbook (Table I) gives the
probability less than a desired value for Z (i.e., from negative infinity to Z)
.9772
Example:
P(Z < 2.00) = .9772
0
Z
The row shows
the value of Z to
the first decimal
point
0.00
0.01
2.00
Z
0.02 …
The column gives the value of Z to
the second decimal point
0.0
0.1
.
.
.
2.0
.9772
The value within the table gives the probability
from Z =   up to the desired Z value.
Finding Normal Probability
To find P(a < X < b) when X is distributed normally:
f(X)
P(a ≤ X ≤ b)
a

b
Draw the normal curve for the problem in terms of X.
 Translate X-values to Z-values.
Z
 Use the Standardized Normal Table.
X μ
σ
Ex)
Suppose X is normal with mean 8.0 and standard
deviation 5.0. Find P(X < 8.6) and P(X > 8.6) .
μ=0
σ=1
μ=8
σ = 10
8
X
8.6
Z
0 0.12
P(X < 8.6) = P(Z < 0.12)
X  μ 8.6  8.0
Z

 0.12
σ
5.0
P(X < 8.6) = P(Z < 0.12) = 0.5478
P(X > 8.6) = 1.0 - .5478 = .4522
Z
.00
.01
.02
0.0
.5000 .5040 .5080
0.1
.5398 .5438 .5478
0.2
.5793 .5832 .5871
0.3
.6179 .6217 .6255
Ex) Suppose X is normal with mean 8.0 and standard
deviation 5.0. Find P(8 < X < 8.6)
8 8.6
P(8 < X < 8.6)= P(0 < Z < 0.12) = P(Z < 0.12) – P(Z ≤ 0)
Z
.00
.01
.02
0.0
.5000
.5040
.5080
0.1
.5398
.5438
.5478
0.2
.5793
.5832
.5871
0.3
.6179
.6217
.6255
= .5478 - .5000 = .0478
X
Standard Normal Distribution Table
Ex) Suppose X is normal with mean 20 and standard deviation 5 Find
the following:
1)
P( X > 30)
2)
P(X<15)
3)
P(10<X<25)
2. Sampling Distributions
POPULATION
A population consists of all the items or individuals about which you want to draw a
conclusion.
SAMPLE
A sample is the portion of a population selected for analysis.
PARAMETER
A parameter is a numerical measure that describes a characteristic of a population.
STATISTIC
A statistic is a numerical measure that describes a characteristic of a sample.
Population
Measures used to describe the
population are called parameters
Sample
Measures computed from sample
data are called statistics
Chap 1-10
• A sampling distribution is a distribution of all of the possible
values of a statistic for a given size sample selected from a
population.
Ex) Suppose you sample 50 students from your college regarding
their mean GPA. If you obtained many different samples of 50,
you will compute a different mean for each sample. We are
interested in the distribution of all potential mean GPA we
might calculate for any given sample of 50 students.
Chap 7-11
Probability and Statistics
Probability
Population
Sample
Statistics
12
We specify the population
and study the behavior of
samples selected from the
population
We make inferences in
populations based on the
information contained in a
sample
Statistics
Descriptive Statistics
Collecting, summarizing, and
describing data
 Collect data
ex. Survey
 Present data
ex. Tables and graphs
 Characterize data
ex. Sample mean = X i
n
Inferential Statistics
Drawing conclusions and/or making
decisions concerning a population
based only on sample data
 Estimation
-Point Estimation
ex) Estimate the population mean
weight using the sample mean
weight
- Interval Estimation
ex) Confidence Interval
 Hypothesis testing
ex. Test the claim that the
population mean weight is 120
pounds
Chap 1-13
Sampling Distributions: Sample Mean Example
•
•
•
•
Suppose your population is “brothers and sisters” in your family.
Population size N=4
Random variable, X, is age of individuals
Values of X: 18, 20, 22, 24 (years)
P(x)
•
X
x=18
x=20
x=22
x=24
P(X=x)
0.25
0.25
0.25
0.25
.3
.2
.1
0
18
20
22
24
Population has Uniform distribution with mean=21and variance=5
x
μ
N
i
18  20  22  24

 21
4
Population mean
σ
2
 (x

i
 μ)2
N
5
Population
variance
Chap 7-14
x
•
•
Suppose we are sampling of sample size n=2 from this
population with replacement.
16 possible samples => 16 sample means
Sample Sample
(X1, X2) Mean
Sample
(X1, X)
Sample
Mean
18, 18
18
22, 18
20
18, 20
19
22, 20
21
18, 22
20
22, 22
22
18, 24
21
22, 24
23
20, 18
19
24, 18
21
20, 20
20
24, 20
22
20, 22
21
24, 22
23
20, 24
22
24, 24
24
S ample Means
Distribution
P( )
.3
.2
.1
0
18
19
20
21
22
23
24
=> Not a Uniform Distribution
μX 
σX
2
X
N
i

 (X

i
18  19  21    24
 21
16
 μ X )2
N
(18 - 21) 2  (19 - 21) 2    (24 - 21) 2

 2.5
16
Chap 7-15
Sample Means Distribution
n=2
Population
N=4
μ  21 σ  2.236 σ 2  5
μ X  21
σ X  1.58 σ X  2.5
2
P( )
.3
P(X)
.3
.2
.2
.1
.1
0
18
20
22
24
X
0
18
19
20
21
22
23
24
Chap 7-16

A measure of the variability in the mean from sample to sample is given by
the Standard Error of the Mean:
σ
σX 
n

Note that the standard error of the mean decreases as the sample size
increases.
Chap 7-17
Sampling Distribution of Normal Populations
• If a population is normal with mean μ and standard deviation σ,
the sampling distribution of the mean is also normally distributed
with μ  μ and σ  σ
X
X
n
• That is,
•

X ~ N(μ,
σ2)
Nice
Property!
=>
Recall that, from the previous example, if X~ Unifrom =>
not follow a Uniform distiribution
does
Z-value for the sampling distribution of the sample mean:
Z
(X  μ X )
σX

(X  μ)
σ
n
~ N(0,12)
Chap 7-18
• As n increases,
σ x decreases
Larger
sample
size
Smaller
sample
size
μ
Statistics for Managers Using Microsoft Excel, 5e © 2008
x
Chap 7-19
Sampling Distributions of Non-Normal Populations
• The Central Limit Theorem states that as the sample size (that is,
the number of values in each sample) gets large enough, the
sampling distribution of the mean is approximately normally
distributed. This is true regardless of the distribution of the
individual values in the population.
σ
• Measures of the sampling distribution: μ x  μ , σ x 
n
Population
Distribution
Sampling Distribution
(becomes normal as n increases)
Smaller sample
size
μ
x
Larger
sample
size
μx
x
Chap 7-20
Sampling Distributions
• For most distributions, n > 30 will give a sampling distribution that is
nearly normal
• For fairly symmetric distributions, n > 15 will give a sampling
distribution that is nearly normal
• For normal population distributions, the sampling distribution of the
mean is always normally distributed
Chap 7-21
Example
• Suppose a population has mean μ = 8 and standard deviation σ = 3. Suppose a
random sample of size n = 36 is selected.
• What is the probability that the sample mean is between 7.75 and 8.25?
• Even if the population is not normally distributed, the central limit theorem
can be used (n > 30).
• So, the distribution of the sample mean is approximately normal with
μx  8
Z
7.75 - 8
 0.5
3
36
σx 
σ
3

 0.5
n
36
Z
8.25 - 8
 0.5
3
36
P(7.75  μ X  8.25)  P(-0.5  Z  0.5)  0.3830
Chap 7-22
Population
Distribution
= 2(.5000-.3085)
= 2(.1915)
μ8
X
= 0.3830
Sample
Sampling
Distribution
7.75
Standardized Normal
Distribution
μX  8
x
8.25
-0.5 μ  0 0.5
z
Z
Sampling Distributions : The Proportion

Let p be the proportion of the population having some characteristic.

Sample proportion ( p̂) provides an estimate of p:
pˆ 

X
number of items in the sample having the characteri stic of interest

n
sample size
X ~ Binomial distribution, Bin(n, p) and
Yi ~ Bernoulli (p)

Thus, sample proportion p can be expressed as a sample mean from a
sample with Bernoulli distribution
σ p̂   Y 

Central Limit Theorem can be applied:
pˆ  p
Z


Z -value for the proportion:
σp
p(1  p)
n
pˆ  p
~ N(0,12)
p(1  p)
n
Chap 7-24
Example
•
If the true proportion of voters who support Proposition A is p = 0.4,
what is the probability that a sample of size 200 yields a sample proportion
between 0.40 and 0.45? In other words, if p =0 .4 and n = 200, what is
P(0.40 ≤ p̂ ≤ 0.45) ?
σ p̂   Y 
p(1  p)

n
0.4(1  0.4)
 0.03463
200
0.45  0.40 
 0.40  0.40
P(.40  pˆ  .45)  P
Z

0.03464 
 0.03464
 P(0  Z  1.44)  0.4251
Chap 7-25
3. Inference-Confidence Interval
•
A point estimate is a single number.
Ex) - For the population mean, a point estimate is the sample mean.
- For the population standard deviation, a point estimate is the
sample standard deviation.
•
A confidence interval provides additional information about variability.
Lower Confidence
Limit
Point Estimate
Width of
confidence interval
Upper Confidence
Limit
•
A confidence interval (C.I.) is stated in terms of confidence level
Ex) 95% confidence, 99% confidence
• Confidence level is a percentage denoting the confidence in which the
interval will contain the unknown population parameter.
•
Ex) Confidence level = 95%
( 1- =0.95)
 P(A specific C.I. will contain the true parameter) = 0.95
 95% of all the C.I. s will construct the true parameter
• The general formula for all confidence intervals is:
Point Estimate ± (Critical Value) (Standard Error)
Confidence Interval for μ with σ Known
Assumptions
– Population standard deviation σ is known
– Population is normally distributed
– If population is not normal, use large sample (CLT)
Confidence interval estimate:
or
X  z/2
σ
n
(, where Zα/2 is the standardized normal distribution critical value
for a probability of α/2 in each tail)
Chap 8-28
Critical Value: Zα/2
Consider a 95% confidence interval:
α
 .025
2
Z1- /2 = -1.96
1   .95
α
 .025
2
0
Zα/2 = 1.96
Commonly used confidence levels are 90%, 95%, and 99%
X units:
Chap 8-29
Example
A sample of 11 circuits from a normal population has a mean
resistance of 2.20 ohms. We know from past testing that the
population standard deviation is 0.35 ohms. Determine a 95%
confidence interval for the true mean resistance of the population.
X  Z / 2
σ
 2.20  1.96 (0.35/ 11)
n
 2.20  .2068
(1.9932, 2.4068)
Chap 8-30
Confidence Interval for μ with σ Unknown
•
If the population standard deviation σ is unknown, we can substitute
the sample standard deviation, S
• This introduces extra uncertainty, since S is variable from sample to
sample => Use the t distribution instead of the normal distribution
Assumptions
– Population standard deviation is unknown
– Population is normally distributed
– If population is not normal, use large sample
• Confidence Interval Estimate:
X  t /2, n -1
or
s
n
(,where t/2, n-1 is the critical value of the t distribution with n-1 d.f.
and an area of α/2 in each tail)
Chap 8-31
Student’s t Distribution
• T-distriburions are symmetric and bell shaped but have flatter tails
than normal
• The t value depends on degrees of freedom (d.f.)
• As d.f. goes infinity, t-distribution -> N(0,12)
Standard
Normal
(t with df = ∞)
t (df = 13)
t (df = 5)
0
t
Chap 8-32
Table of T-distiribution
Example
A random sample of n = 25 has the sample mean 50 and
the sample variance 8. Form a 95% confidence interval
for μ
– d.f. = n – 1 = 24, so
– The confidence interval is
S
8
X  t/2, n -1
 50  (2.0639)
n
25
(46.698 , 53.302)
Chap 8-34
Ex) The following is the fuel efficiency data (km per liter) of a new car
model;
17.2 16.9 17.6 18.0 17.4 16.3 15.8 17.2 17.3 16.0
Suppose that the km per liter follows a normal probability, calculate
the 95% confidence interval for the mean km per liter.
Confidence Intervals for the
Population Proportion, p
 Recall that the distribution of the sample proportion is approximately
normal if the sample size is large, with standard deviation
σ p̂ 
p(1  p)
n
Confidence interval:
p̂  Z/2
p̂(1  p̂)
n
Chap 8-36
Example
A random sample of 100 people shows that 25 wear glasses. Form a
95% confidence interval for the true proportion of the population
who wear glasses.
p̂  Z / 2 p̂(1  p̂)/n  25/100  1.96 0.25(0.75)/100
 0.25  1.96 (0.0433)
(0.1651 , 0.3349)
Note : We are 95% confident that the true percentage of people wearing
glasses in the population is between 16.51% and 33.49%. Although the
interval from .1651 to .3349 may or may not contain the true proportion,
95% of intervals formed from samples of size 100 in this manner will
contain the true proportion.
Chap 8-37