Chapter Six Normal Curves and Sampling Probability Distributions
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Transcript of Chapter Six Normal Curves and Sampling Probability Distributions
The Central Limit Theorem is The Central Limit Theorem is applied in the following ways:applied in the following ways:
A given population has fixed parameters µ and σ .For each population, there are very many samples of size n, which can be taken. Each of these samples has a sample mean x. The x statistic varies from sample to sample. The Central Limit Theorem tells us what to expect about the sample means.
If x is normally distributed for any size sample, or if the sample
size is 30 or more, the sample means will be normally
distributed.
When working with any normal distribution, we have to
know the mean of the distribution and the standard
deviation. The Central Limit Theorem relates the mean
and the standard deviation of the original x to the mean
and standard deviation of x .
When working with normal distribution probabilities, we
must convert all values to z-scores.
The Central Limit Theorem is used to investigate the probability
of a sample mean being in a given interval.
Since we will be working with normal
distributions, the x values in the given
interval must be converted to z-scores.
There are three formulas, all which are
equivalent, which may be used to
convert from an x value to a z-score.
The simplest formula is
The deviation of x value from the mean is divided by the
standard deviation of the distribution σ x. To use this
version, first we calculate σ xby dividing σ by n. Then substitute.
You may prefer to directly substitute the fraction σn for
σ x in the denominator and use either of the two formulas that follow. You could use ...
z =x−μσ x
To find the probability that a sample To find the probability that a sample mean is within a given interval:mean is within a given interval:
First, check to see if the distribution of x (from which the
sample mean was taken) is either a normal distribution, or
if not, the sample size is 30 or more. If either of these are
true, the sample distribution is normal.
Find the mean and the standard deviation for the sample
distribution of x using the formulas for the Central Limit
Theorem.
Convert each endpoint of the given interval to the standard
normal z-score.
Rewrite the problem with the z-score interval.
Sketch a standard normal distribution curve and shade the
area you wish to calculate.
Use the table for standard normal probability distribution
to calculate the area, and thus, the probability.
Case 1Case 1When a variable x has a mean of μx, and a standard deviation of σ x, and is normally distributed. For a random sample of any size n, the following statements about the sampling distribution of x are true.
1. The distribution of x is normal.
2. The mean of the sample means is equal to the
mean of the population. That is: μx =μ
3. The standard deviation of the distributions of the sample means is called the standard error of the mean and is smaller than the
distribution of x by a factor of 1n.
That is: σ x =σn.
CASE 2CASE 2When a variable x comes from any
type of distribution, no matter how
unusual, as long as the random sample
has at least 30 members, then the
following statements about the
distribution of the sample size are true.
1. The distribution of x is normal.
2. The mean of the sample means is equal to the
mean of the population. That is: μx =μ
3. The standard deviation of the distributions of the sample means is called the standard error of the mean and is smaller than the
distribution of x by a factor of 1n.
That is: σ x =σn.
Sampling DistributionSampling Distribution
A probability distribution
for the sample statistic
we are using.
Example of a Sampling Example of a Sampling DistributionDistribution
Select samples with two elements
each (in sequence with
replacement) from the set
{1, 2, 3, 4, 5, 6}.
Constructing a Sampling Distribution Constructing a Sampling Distribution of the Mean for Samples of Size n = 2of the Mean for Samples of Size n = 2
List all samples and compute the mean of each sample.sample: mean: sample: mean{1,1} 1.0 {1,6} 3.5{1,2} 1.5 {2,1} 1.5{1,3} 2.0 {2,2} 2.0{1,4} 2.5 … ...{1,5} 3.0
How many different samples are there?How many different samples are there? 3636
Sampling Distribution of the MeanSampling Distribution of the Mean
p1.0 1/361.5 2/362.0 3/362.5 4/363.0 5/363.5 6/364.0 5/364.5 4/365.0 3/365.5 2/36 6.0 1/36
x
Let x be a random variable with a
normal distribution with mean μ and standard deviation σ . Let x be the sample mean corresponding to random samples of size n taken from the distribution.
The following are true:The following are true:
1. The x distribution is a normal distribution.
2. The mean of the x distribution is μ (the same mean as the original distribution).
3. The standard deviation of the x distribution
is σn (the standard deviation of the original
distribution, divided by the square root of the sample size).
We can use this theorem to draw conclusions about means of samples taken from normal distributions.
If the original distribution is normal, then the sampling distribution will be normal.
The Standard Deviation of the The Standard Deviation of the Sampling DistributionSampling Distribution
σ x
The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square
root of the sample size.
σ x =σ
n
As the sample size continue to
increase closer and closer
to the population size the
following statements are true.
3. The standard deviation will take on the
intermediate value of the population
standard deviation divided by the
square root of the sample size. However,
because of the increasing sample size
this value will approach zero.
limn→ N
σ x =limn→ N
σn=0⎛
⎝⎜⎞⎠⎟
4. The probability of an interval that
contains the population mean
will approach 1.
limn→ N
P x1 < μ < x2( ) =1( )
5. The probability of an interval
that does NOT contain the
population mean will approach 0.
limn→ N
P x1 < μ( ) =0( )
or
limn→ N
P x> μ( ) =0( )
***Note***
As the sample size increases to the population size
Graph 1 → Graph 2→ Graph 3the graphs get closer and closer to the population mean.
The Central Limit Theorem states that
if the probability interval contains the
population mean and the sample size
continues to increase closer and closer
to the population size then the probability
will continue to increase and get closer
and closer to 1.
or
The Central Limit Theorem states that
if the probability interval does not
contain the population mean and the
sample size continues increase closer
and closer to the population size then
the probability will continue to decrease
and get closer and closer to 0.
1.1.
Suppose that it is known that the time
spent by customers in the local coffee
shop is normally distributed with a
mean of 24 minutes and a standard
deviation of 6 minutes.
a.Find the probability that an individual
customer will spend more than 26
minutes in the coffee shop.
P x > 26( )
P z>26−24
6⎛⎝⎜
⎞⎠⎟
P z>26
⎛⎝⎜
⎞⎠⎟
P z> 0.33( )0.5000−0.1293
0.3707
0.33
b.Find the probability that a random sample
of 9 customers will have a mean stay of
more than 26 minutes in the coffee shop. P x > 26( )
P z>26−24( ) 9
6
⎛
⎝⎜⎞
⎠⎟
P z>2 3( )6
⎛⎝⎜
⎞⎠⎟
P z>66
⎛⎝⎜
⎞⎠⎟
P z>1.00( )0.5000−0.3413
0.1587
1.00
c.Find the probability that a random sample
of 64 customers will have a mean stay of
more than 26 minutes in the coffee shop. P x > 26( )
P z>26−24( ) 64
6
⎛
⎝⎜⎞
⎠⎟
P z>2 8( )6
⎛⎝⎜
⎞⎠⎟
P z>166
⎛⎝⎜
⎞⎠⎟
P z> 2.67( )0.5000−0.4962
0.0038
2.67
d.Find the probability that a random sample
of 100 customers will have a mean stay of
more than 26 minutes in the coffee shop. P x > 26( )
P z>26−24( ) 100
6
⎛
⎝⎜⎞
⎠⎟
P z>2 10( )6
⎛⎝⎜
⎞⎠⎟
P z>206
⎛⎝⎜
⎞⎠⎟
P z> 3.33( )0.5000−0.4996
0.0004
3.33
e.Find the probability that a random sample
of 144 customers will have a mean stay of
more than 26 minutes in the coffee shop. P x > 26( )
P z>26−24( ) 144
6
⎛
⎝⎜⎞
⎠⎟
P z>2 12( )6
⎛⎝⎜
⎞⎠⎟
P z>246
⎛⎝⎜
⎞⎠⎟
P z> 4.00( )0.5000−0.4999
0.0001
4.00
e. The Central Limit Theorem states that
if the probability interval does not
contain the population mean and the
sample size continues increase closer
and closer to the population size then
the probability will continue to decrease
and get closer and closer to 0.
2.2.
Suppose that it is known that the time
spent by customers in the local coffee
shop is normally distributed with a
mean of 24 minutes and a standard
deviation of 6 minutes.
a.Find the probability that an individual
customer will spend between 22 to 25
minutes in the coffee shop.P 22 ≤x≤25( )
P22−24
6≤z≤
25−246
⎛⎝⎜
⎞⎠⎟
P−26
≤z≤16
⎛⎝⎜
⎞⎠⎟
P −0.33≤z≤0.17( )0.1293+0.0675
0.1968
-0.33 0.17
b.
Find the probability that a random sample
of 9 customers will have a mean stay in
the coffee shop between 22 to 25 minutes.
P 22 ≤x≤25( )
P22−24 9( )
6≤z≤
25−24( ) 96
⎛
⎝⎜⎜
⎞
⎠⎟⎟
P−2( ) 3( )6
≤z≤1( ) 3( )6
⎛⎝⎜
⎞⎠⎟
P−66
≤z≤36
⎛⎝⎜
⎞⎠⎟
P −1.00 ≤z≤0.50( )0.3413+ 0.1915
0.5328
-1.00 0.50
c.
Find the probability that a random sample
of 64 customers will have a mean stay in
the coffee shop between 22 to 25 minutes.
P 22 ≤x≤25( )
P22−24 64( )
6≤z≤
25−24( ) 646
⎛
⎝⎜⎜
⎞
⎠⎟⎟
P−2( ) 8( )6
≤z≤1( ) 8( )6
⎛⎝⎜
⎞⎠⎟
P−166
≤z≤86
⎛⎝⎜
⎞⎠⎟
P −2.67 ≤z≤1.33( )0.4962 + 0.4082
0.9044
-2.67 1.33
d.
Find the probability that a random sample
of 100 customers will have a mean stay in
the coffee shop between 22 to 25 minutes.
P 22 ≤x≤25( )
P22−24 100( )
6≤z≤
25−24( ) 1006
⎛
⎝⎜⎜
⎞
⎠⎟⎟
P−2( ) 10( )
6≤z≤
1( ) 10( )6
⎛⎝⎜
⎞⎠⎟
P−206
≤z≤106
⎛⎝⎜
⎞⎠⎟
P −3.33≤z≤1.67( )0.4996 +0.4525
0.9521
-3.33 1.67
e.
Find the probability that a random sample
of 144 customers will have a mean stay in
the coffee shop between 22 to 25 minutes.
P 22 ≤x≤25( )
P22−24 144( )
6≤z≤
25−24( ) 1446
⎛
⎝⎜⎜
⎞
⎠⎟⎟
P−2( ) 12( )
6≤z≤
1( ) 12( )6
⎛⎝⎜
⎞⎠⎟
P−246
≤z≤126
⎛⎝⎜
⎞⎠⎟
P −4.00 ≤z≤2.00( )0.4999 + 0.4772
0.9771
-4.00 2.00
e. The Central Limit Theorem states that
if the probability interval contains the
population mean and the sample size
continues to increase closer and closer
to the population size then the probability
will continue to increase and get closer
and closer to 1.