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381
Continuous Probability Distributions (The Normal Distribution-II)
QSCI 381 – Lecture 17(Larson and Farber, Sect 5.4-5.5)
381
Finding z-scores-I Yesterday we addressed the question:
What is the probability that a normal random variable, X, would lie between x1 and x2.
To address this question we found the probabilities P[X x1] and P[X x2] and calculated the difference between them.
Today we are going to address the inverse of this question. Find the z-score which corresponds to a
cumulative area under the standard normal curve of p.
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Finding z-scores-II
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
What value of x corresponds to an area of 0.8?
X?
Area=0.8
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Finding z-scores-II We can use a table of z-scores or the
EXCEL function NORMINV: NORMINV(p,,)
Once you have a z-score for a given cumulative probability, you can find x for any and using the formula:
x z
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Example-I The length distribution of the catch
of a given species is normally distributed with mean 500 mm and standard deviation 30 mm.
Find the maximum length of the smallest 5%, 50% and 75% of the catch.
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Example-II
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
-1.64 0
0.674Find the z-score for each level (5%, 50% and 75%)
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Example-III We now apply the formula:
so the maximum lengths are 450.7, 500 and 520.2 mm.
500 30x z
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Sampling and Sampling Distributions-I
So far we have been working on the assumption that we know the values for and . This is rarely the case and generally we need to estimate these quantities from a sample. The relationship between the population mean and the mean of a sample taken from the population is therefore of interest.
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Sampling and Sampling Distributions-II
A is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means.
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Sampling and Sampling Distributions-III
(Example)
Consider a population of fish in a lake. The mean and standard deviation of the lengths of these fish are 300 mm and 50 mm respectively.
We now take 100 random samples where each sample is of size 10, 20, or 100. What can we learn about the population mean?
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Sampling and Sampling Distributions-IV
(Example)
Sample Mean
240 260 280 300 320 340 360
0.0
0.0
10
0.0
20
0.0
30
Sample Mean
240 260 280 300 320 340 360
0.0
0.0
10
.02
0.0
3
Sample Mean
240 260 280 300 320 340 360
0.0
0.0
20
.04
0.0
6
N=10 N=20
N=100
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Properties of the Sampling Distribution for the Sample Mean The mean of the sample means is
equal to the population mean:
The standard deviation of the sample means is equal to the population standard deviation divided by the square root of n.
is often called the
x
/x n
x
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The Central Limit Theorem1. If samples of size n (where n 30) are
drawn from any population with a mean and a standard deviation , the sampling distribution of sample means approximates a normal distribution. The greater the sample size, the better the approximation.
2. If the population is itself normally distributed, the sampling distribution of the sample means is normally distributed for any sample size.
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The Central Limit Theorem(Example)
Fre
que
ncy
0.0 0.2 0.4 0.6 0.8 1.0
01000
2000
3000
0.0 0.2 0.4 0.6 0.8 1.0
010
20
30
40
Fre
quency
0.0 0.2 0.4 0.6 0.8 1.0
010
20
30
Fre
quency
0.0 0.2 0.4 0.6 0.8 1.0
010
20
30
40
Fre
quency
0.0 0.2 0.4 0.6 0.8 1.0
010
20
30
40
Fre
quency
0.0 0.2 0.4 0.6 0.8 1.0
010
20
30
40
Fre
que
ncy
0.15 0.25 0.35 0.45
050
100
150
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Probabilities and the Central Limit Theorem
The distribution of the heights of trees are not normally distributed. We sample 100 (of many) trees in a (very large) stand and calculate sample mean and sample standard deviation to be 12.5m and 2.3m respectively. What is the standard deviation of the mean? What is the probability that the population
mean is less than 12m?
