Transcript Standard Normal Table

Chapter 3
The Normal Distributions
Essential Statistics
Chapter 3
1
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Z-Score Explained
http://www.youtube.com/watch?v=AT-HH0W_swA&feature=related
Basics of Using the Std Normal Table
http://www.youtube.com/watch?v=y6sbghmHwQA&feature=related
Normal Distribution & Z-score
http://www.youtube.com/watch?v=mai23vW8uFM&feature=related
Essential Statistics
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We’ll Learn The Topics
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Review Histogram
Density Curve
Normal Distribution
68 – 95 – 99.7 Rule
Z-score
Standard Normal Distribution
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Density Curves
Example: here is a
histogram of vocabulary
scores of 947 seventh
graders.
- We can describe the
histogram with a smooth
curve, a bell- shaped
curve.
- It corresponding to a
normal distribution
Model.
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Density Curves
Example: the areas of
the shaded bars in this
histogram represent the
proportion of scores that
are less than or equal to
6.0. This proportion in
the observed data is
equal to 0.303.
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Density Curves
■ now the area under the
smooth curve to the left
of 6.0 is shaded.
■ The scale is adjusted,
the total area under the
curve is exactly 1, this
curve is called a density
curve.
■ The proportion of the
area to the left of 6.0 is
now equal to 0.293.
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Density Curves
 Always
 Have
on or above the horizontal axis
area exactly 1 underneath curve
 Display
the bell-shaped pattern of a
distribution
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A histogram becomes a density curve if
the scale is adjusted so that the total
area of the bars is 1.
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Mean & Standard Deviation
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The mean and standard deviation computed from
actual observations (data) are denoted by and
s, respectively
 The
mean and standard deviation of the
distribution represented by the density curve are
denoted by µ (“mu”) and  (“sigma”), respectively.
 The
mean of a density curve is the "balance point"
of the curve.
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Bell-Shaped Curve:
The Normal Distribution
standard deviation
mean
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The Normal Distribution
■ Knowing the mean (µ) and standard deviation
() allows us to make various conclusions about
Normal distributions.
■ Notation: N(µ,).
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68-95-99.7 Rule for
Any Normal Curve
 68%
of the observations fall within one
standard deviation of the mean
 95% of the observations fall within two
standard deviations of the mean
 99.7% of the observations fall within
three standard deviations of the mean
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68-95-99.7 Rule for
Any Normal Curve
68%
-
95%
µ +
-2
µ
+2
99.7%
-3
Essential Statistics
µ
Chapter 3
+3
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68-95-99.7 Rule for
Any Normal Curve
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Health and Nutrition Examination
Study of 1976-1980
 Heights
of adult men, aged 18-24
– mean: 70.0 inches
– standard deviation: 2.8 inches
– heights follow a normal distribution, so we
have that heights of men are N(70, 2.8).
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Health and Nutrition Examination
Study of 1976-1980
 68-95-99.7
 68%
Rule for men’s heights
are between 67.2 and 72.8 inches
[ µ   = 70.0  2.8 ]
 95%
are between 64.4 and 75.6 inches
[ µ  2 = 70.0  2(2.8) = 70.0  5.6 ]
 99.7%
are between 61.6 and 78.4 inches
[ µ  3 = 70.0  3(2.8) = 70.0  8.4 ]
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Health and Nutrition Examination
Study of 1976-1980
 What
proportion of men are less than
72.8 inches tall? 68%
(by 68-95-99.7 Rule)
16%
?
-1
+1
? = 84%
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Chapter 3
72.8
(height values)
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Standard Normal Distribution
x
z
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Z – Score
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The standard Normal distribution N(0,1) is the Normal
distribution has a mean of zero and a standard
deviation of one
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Normal distributions can be transformed to standard
normal distributions by Z-score
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Health and Nutrition Examination
Study of 1976-1980
 What
proportion of men are less than
68 inches tall?
?
68 70
(height values)
How many standard deviations is 68 from 70?
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Standardized Scores

standardized score (Z-score) =
(observed value minus mean) / (std dev)
[ = (68  70) / 2.8 = 0.71 ]
 The
value 68 is 0.71 standard
deviations below the mean 70.
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Health and Nutrition Examination
Study of 1976-1980
 What
proportion of men are less than
68 inches tall?
?
68 70
-0.71
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0
(height values)
(standardized values)
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Table A:
Standard Normal Probabilities
 See
pages 464-465 in text for Table A.
(the “Standard Normal Table”)
 Look
up the closest standardized score
(z) in the table.
 Find
the probability (area) to the left of the
standardized score.
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Table A:
Standard Normal Probabilities
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Table A:
Standard Normal Probabilities
z
.00
.01
.02
0.8
.2119
.2090
.2061
0.7
.2420
.2389
.2358
0.6
.2743
.2709
.2676
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Health and Nutrition Examination
Study of 1976-1980
 What
proportion of men are less than
68 inches tall?
.2389
68 70
-0.71
Essential Statistics
0
(height values)
(standardized values)
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Health and Nutrition Examination
Study of 1976-1980
 What
proportion of men are greater than
68 inches tall?
1.2389 =
.2389
.7611
68 70
-0.71
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0
(height values)
(standardized values)
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Health and Nutrition Examination
Study of 1976-1980
 How
tall must a man be to place in the
lower 10% for men aged 18 to 24?
.10
? 70
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(height values)
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Table A:
Standard Normal Probabilities
 See
pages 464-465 in text for Table A.
 Look
up the closest probability (to .10 here)
in the table.
 Find
the corresponding standardized score.
 The
value you seek is that many standard
deviations from the mean.
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Table A:
Standard Normal Probabilities
z
.07
.08
.09
1.3
.0853
.0838
.0823
1.2
.1020
.1003
.0985
1.1
.1210
.1190
.1170
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Health and Nutrition Examination
Study of 1976-1980
 How
tall must a man be to place in the
lower 10% for men aged 18 to 24?
.10
? 70
-1.28
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0
(height values)
(standardized values)
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Observed Value for a
Standardized Score
 Need
to “reverse” the z-score to find the
observed value (x) :
x
z

 observed
x    z
value =
mean plus [(standardized score)  (std dev)]
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Observed Value for a
Standardized Score
 observed
value =
mean plus [(standardized score)  (std dev)]
= 70 + [(1.28 )  (2.8)]
= 70 + (3.58) = 66.42
A
man would have to be approximately
66.42 inches tall or less to place in the
lower 10% of all men in the population.
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The Entry in Table A
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Using random variable z to get the entrance in Table A.
Variable z is z-score which follows the standard normal
distribution N(0, 1)
x
z
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Z-score:
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When search entry for a z value
♫ look up the most left column first, locate the most
close value to z value
♫ look up the top row to locate the 2th decimal place
for a z value
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The Entry in Table A
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Table A’s entry is an area underneath the curve, to
the left of z
Table A’s entry is a percent of the whole area, to the
left of z-score
Table A’s entry is a probability, corresponding to the
z-score value.
Math formula:
P (z ≤ z0) = 0.xxxx
P (z ≤ -0.71) = 0.2389
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Problem type I
z ≈ N(0, 1),
P (z ≤ z0) = ?
 By checking the Table A, find out the answer.
 For type I problem, check the table and get
the answer directly.
For example,
P (z ≤ -0.71) = 0. 2389
 If
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Problem type II
z ≈ N(0, 1),
P (z ≥ z0) = ?
 This type’s problem, cannot check the table
directly. Using the following operation.
 P (z ≥ z0) = 1 - (z ≤ z0)
 For example, p (z ≥ -0.71) = ?
 If
◙ P (z ≥ -0.71) = 1 - (z ≤ - 0.71) = 1 – 0.2389 = 0.7611
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Problem Type II
0.7611
0.2389
68 70
-0.71
Essential Statistics
0
(height values)
(standardized values)
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Problem Type III
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If z ≈ N(0, 1),
P ( z2 ≤ z ≤ z1 ) = ?
random variable z is between two numbers
Look up z1 → P1
Look up z2 → P2
The result is:
P ( z2 ≤ z ≤ z1) = P1 - P2
For example, P ( -1.4 ≤ z ≤ 1.3) = ?
look up 1.3
P1 = 0.9032
look up -1.4
P2 = 0.0808
P ( -1.4 ≤ z ≤ 1.3) = 0.9032 – 0.0808 = 0.8224
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Steps Summary
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Write down the normal distribution N(µ,) for
observation data set
Locate the specific observation value X0
Transform X0 to be Z0 by z-score formula
Check table A using random variable Z0 to find out
table entry P(z ≤ z0)
If is problem type I, the result is P(z ≤ z0)
If is problem type II, the result is:
P (z ≥ z0) =1- P(z ≤ z0)
If is problem type III, the result is:
P ( z 2 ≤ z ≤ z 1 ) = P1 - P 2
P1 = P(z ≤ z1),
P2 = P(z ≤ z2)
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