chapter 7 section 2
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Chapter 7 Section 2. The Standard Normal Distribution. 1. 2. 3. Chapter 7 – Section 2. Learning objectives Find the area under the standard normal curve Find Z -scores for a given area Interpret the area under the standard normal curve as a probability. 1. 2. 3. - PowerPoint PPT PresentationTRANSCRIPT
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 1 of 32
Chapter 7Section 2
The StandardNormal Distribution
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 2 of 32
Chapter 7 – Section 2
● Learning objectives Find the area under the standard normal curve Find Z-scores for a given area Interpret the area under the standard normal curve
as a probability
1
2
3
![Page 3: Chapter 7 Section 2](https://reader036.vdocuments.us/reader036/viewer/2022081603/56813a30550346895da219c9/html5/thumbnails/3.jpg)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 3 of 32
Chapter 7 – Section 2
● Learning objectives Find the area under the standard normal curve Find Z-scores for a given area Interpret the area under the standard normal curve
as a probability
1
2
3
![Page 4: Chapter 7 Section 2](https://reader036.vdocuments.us/reader036/viewer/2022081603/56813a30550346895da219c9/html5/thumbnails/4.jpg)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 4 of 32
Chapter 7 – Section 2
● The standard normal curve is the one with mean μ = 0 and standard deviation σ = 1
● We have related the general normal random variable to the standard normal random variable through the Z-score
● In this section, we discuss how to compute with the standard normal random variable
X
Z
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 5 of 32
Chapter 7 – Section 2
● There are several ways to calculate the area under the standard normal curve What does not work – some kind of a simple formula We can use a table (such as Table IV on the inside
back cover) We can use technology (a calculator or software)
● Using technology is preferred
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 6 of 32
Chapter 7 – Section 2
● Three different area calculations Find the area to the left of Find the area to the right of Find the area between
● Three different area calculations Find the area to the left of Find the area to the right of Find the area between
● Three different methods shown here From a table Using Excel Using StatCrunch
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 7 of 32
Chapter 7 – Section 2
● "To the left of" – using a table● Calculate the area to the left of Z = 1.68
● "To the left of" – using a table● Calculate the area to the left of Z = 1.68
Break up 1.68 as 1.6 + .08
● "To the left of" – using a table● Calculate the area to the left of Z = 1.68
Break up 1.68 as 1.6 + .08 Find the row 1.6
Enter
● "To the left of" – using a table● Calculate the area to the left of Z = 1.68
Break up 1.68 as 1.6 + .08 Find the row 1.6 Find the column .08
Enter
Enter
● "To the left of" – using a table● Calculate the area to the left of Z = 1.68
Break up 1.68 as 1.6 + .08 Find the row 1.6 Find the column .08
● The probability is 0.9535 Read
Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 8 of 32
Chapter 7 – Section 2
● "To the left of" – using Excel● The function in Excel for the standard normal is
=NORMSDIST(Z-score) NORM (normal) S (standard) DIST (distribution)
Enter
Read
Read
Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 9 of 32
Chapter 7 – Section 2
● To the left of 1.68 – using StatCrunch● The function is Stat – Calculators – Normal
EnterRead
Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 10 of 32
Chapter 7 – Section 2
● "To the right of" – using a table● The area to the left of Z = 1.68 is 0.9535
Read
Enter
Enter
● "To the right of" – using a table● The area to the left of Z = 1.68 is 0.9535
● The right of … that’s the remaining amount● The two add up to 1, so the right of is
1 – 0.9535 = 0.0465
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 11 of 32
Chapter 7 – Section 2
● "To the right of" – using ExcelRead
Read
Enter
● "To the right of" – using Excel
● The right of … that’s the remaining amount of to the left of, subtract from 1
1 – 0.9535 = 0.0465
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 12 of 32
Chapter 7 – Section 2
● "To the right of" – using StatCrunch
ReadEnter
● "To the right of" – using StatCrunch● Change the <= to >=
● "To the right of" – using StatCrunch● Change the <= to >= (the picture and number
change)
ReadEnter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 13 of 32
Chapter 7 – Section 2
● “Between”● Between Z = – 0.51 and Z = 1.87● This is not a one step calculation
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 14 of 32
Chapter 7 – Section 2
● The left hand picture … to the left of 1.87 … includes too much
● It is too much by the right hand picture … to the left of -0.51
Includedtoo much
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 15 of 32
Chapter 7 – Section 2
● Between Z = – 0.51 and Z = 1.87
We want
We start out with,but it’s too much
We correct by
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 16 of 32
Chapter 7 – Section 2
● We can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87
● We can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of 1.87, or 0.9693 … minus
● We can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of 1.87, or 0.9693 … minus The area to the left of -0.51, or 0.3050 … which
equals
● We can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of 1.87, or 0.9693 … minus The area to the left of -0.51, or 0.3050 … which
equals The difference of 0.6643
● We can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of 1.87, or 0.9693 … minus The area to the left of -0.51, or 0.3050 … which
equals The difference of 0.6643
● Thus the area under the standard normal curve between -0.51 and 1.87 is 0.6643
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 17 of 32
Chapter 7 – Section 2
● A different way for “between”
We want
We delete theextra on the left
We delete theextra on the right
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 18 of 32
Chapter 7 – Section 2
● Again, we can use any of the three methods to compute the normal probabilities to get
● Again, we can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of -0.51, or 0.3050 … plus The area to the right of 1.87, or .0307 … which equals The total area to get rid of which equals 0.3357
● Again, we can use any of the three methods to compute the normal probabilities to get
● The area between -0.51 and 1.87 The area to the left of -0.51, or 0.3050 … plus The area to the right of 1.87, or .0307 … which equals The total area to get rid of which equals 0.3357
● Thus the area under the standard normal curve between -0.51 and 1.87 is 1 – 0.3357 = 0.6643
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 19 of 32
Chapter 7 – Section 2
● Learning objectives Find the area under the standard normal curve Find Z-scores for a given area Interpret the area under the standard normal curve
as a probability
1
2
3
![Page 20: Chapter 7 Section 2](https://reader036.vdocuments.us/reader036/viewer/2022081603/56813a30550346895da219c9/html5/thumbnails/20.jpg)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 20 of 32
Chapter 7 – Section 2
● We did the problem:
Z-Score Area● Now we will do the reverse of that
Area Z-Score
● We did the problem:
Z-Score Area● Now we will do the reverse of that
Area Z-Score● This is finding the Z-score (value) that
corresponds to a specified area (percentile)● And … no surprise … we can do this with a
table, with Excel, with StatCrunch, with …
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 21 of 32
Chapter 7 – Section 2
● “To the left of” – using a table● Find the Z-score for which the area to the left of
it is 0.32
● “To the left of” – using a table● Find the Z-score for which the area to the left of
it is 0.32 Look in the middle of the table … find 0.32
Find
Read
Read
● “To the left of” – using a table● Find the Z-score for which the area to the left of
it is 0.32 Look in the middle of the table … find 0.32
The nearest to 0.32 is 0.3192 … a Z-Score of -.47
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 22 of 32
Chapter 7 – Section 2
● "To the left of" – using Excel● The function in Excel for the standard normal is
=NORMSINV(probability) NORM (normal) S (standard) INV (inverse)
Read
Read
Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 23 of 32
Chapter 7 – Section 2
● "To the left of" – using StatCrunch● The function is Stat – Calculators – Normal
Read Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 24 of 32
Chapter 7 – Section 2
● "To the right of" – using a table● Find the Z-score for which the area to the right of
it is 0.4332● Right of it is .4332 … left of it would be .5668● A value of .17
Enter
Read
Read
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 25 of 32
Chapter 7 – Section 2
● "To the right of" – using Excel● To the right is .4332 … to the left would be .5668
(the same as for the table)
Read
Read
Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 26 of 32
Chapter 7 – Section 2
● "To the right of" – using StatCrunch● Change the <= to >= (watch the picture change)
Read Enter
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 27 of 32
Chapter 7 – Section 2
● We will often want to find a middle range, to find the middle 90% or the middle 95% or the middle 99%, of the standard normal
● The middle 90% would be
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 28 of 32
Chapter 7 – Section 2
● 90% in the middle is 10% outside the middle, i.e. 5% off each end
● These problems can be solved in either of two equivalent ways
● We could find The number for which 5% is to the left, or The number for which 5% is to the right
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 29 of 32
Chapter 7 – Section 2
● The two possible ways The number for which 5% is to the left, or The number for which 5% is to the right
5% is to the left 5% is to the right
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 30 of 32
Chapter 7 – Section 2
● The number zα is the Z-score such that the area to the right of zα is α
● The number zα is the Z-score such that the area to the right of zα is α
● Some useful values are z.10 = 1.28, the area between -1.28 and 1.28 is 0.80
z.05 = 1.64, the area between -1.64 and 1.64 is 0.90
z.025 = 1.96, the area between -1.96 and 1.96 is 0.95
z.01 = 2.33, the area between -2.33 and 2.33 is 0.98
z.005 = 2.58, the area between -2.58 and 2.58 is 0.99
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 31 of 32
Chapter 7 – Section 2
● Learning objectives Find the area under the standard normal curve Find Z-scores for a given area Interpret the area under the standard normal curve
as a probability
1
2
3
![Page 32: Chapter 7 Section 2](https://reader036.vdocuments.us/reader036/viewer/2022081603/56813a30550346895da219c9/html5/thumbnails/32.jpg)
Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 32 of 32
Chapter 7 – Section 2
● The area under a normal curve can be interpreted as a probability
● The standard normal curve can be interpreted as a probability density function
● The area under a normal curve can be interpreted as a probability
● The standard normal curve can be interpreted as a probability density function
● We will use Z to represent a standard normal random variable, so it has probabilities such as P(a < Z < b) P(Z < a) P(Z > a)
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 33 of 32
Summary: Chapter 7 – Section 2
● Calculations for the standard normal curve can be done using tables or using technology
● One can calculate the area under the standard normal curve, to the left of or to the right of each Z-score
● One can calculate the Z-score so that the area to the left of it or to the right of it is a certain value
● Areas and probabilities are two different representations of the same concept
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 34 of 32
Example: Chapter 7 – Section 2
● Determine the area under the standard normal curve that lies
a. to the left of Z = –2.31. (0.0104)
b. to the right of Z = –1.47. (0.9292)
c. between Z = –2.31 and Z = 0. (0.4896)
d. between Z = –2.31 and Z = –1.47. (0.0603)
e. between Z = 1.47 and Z = 2.31. (0.0603)
f. between Z = –2.31 and Z = 1.47. (0.9188)
g. to the left of Z = –2.31 or to the right of Z = 1.47. (0.0812)
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Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 7 Section 2 – Slide 35 of 32
Example: Chapter 7 – Section 2
● The Graduate Record Examination (GRE) is a test required for admission to many U.S. graduate schools. The Department of Molecular Genetics at Ohio State University requires a GRE score no less than the 60th percentile. (Source: www.biosci.ohio-state.edu/~molgen/html/admission_criteria.html.)
a. Find the Z-score corresponding to the 60th percentile. In other words, find the Z-score such that the area under the standard normal curve to the left is 0.60. (0.25)
b. How many standard deviations above the mean is the 60th percentile? (0.25)