Lecture 5

# Lecture 5 - Example(p126 Lengths of Fish 1 Let us denote...

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Probability and Statistics in the Life Sciences (Fall 2010) AMS 110.02 Lecture 5 (chap4) Donghyung Lee

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Section 4.3 The Normal Curves Example (p132 ex. 4.7) Consider a standard normal distribution, Z. Find .10 .25 .05 .01 ( ) ? ( ) ? ( ) ? ( ) ? a Z b Z c Z d Z = = = =
Section 4.3 The Normal Curves Example (p132 ex. 4.7) Consider a standard normal distribution, Z. Find .10 .25 .05 .01 ( ) 1.28? ( ) .67? ( ) 1.64? ( ) 2.33? a Z b Z c Z d Z = = = =

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Section 4.3 The Normal Curves Example (p126 Lengths of Fish) In a certain population of the herring Pomolobus aestivalis , the lengths of the individual fish follow a normal distribution. The mean length of the fish is 54.0 mm, and the standard deviation is 4.5 mm. 1) Find the 70 th percentile of the fish length distribution. 2) Find the 20 th percentile of the fish length distribution.
Section 4.3 The Normal Curves

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Unformatted text preview: Example (p126 Lengths of Fish) 1) Let us denote the 70 th percentile by x*. By definition, x* is the value such that 70% of the fish lengths are less than x* and 30% are greater. Convert this Z value to the X scale. Thus, the 70 th percentile of the fish length distribution is 56.3 mm. .30 .52 Z = Q * 54 .52 4.5 * (.52)(4.5) 54 56.3 x x-= = = + = Section 4.3 The Normal Curves Example (p126 Lengths of Fish) 2) Let us denote the 20 th percentile by x*. By definition, x* is the value such that 20% of the fish lengths are less than x* and 80% are greater. Convert this Z value to the X scale. Thus, the 20 th percentile of the fish length distribution is 50.2 mm. .80 .84 Z = -Q * 54 .84 4.5 * ( .84)(4.5) 54 50.2 x x--= = = -+ =...
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Lecture 5 - Example(p126 Lengths of Fish 1 Let us denote...

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