Lec5 - Determining areas for any normal distribution The...

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Unformatted text preview: Determining areas for any normal distribution The lengths of herring follow a normal distribution. The mean length is 54.0 mm and the standard deviation is 4.5 mm. What percentage of the fish are less than 60 mm long? 1. Convert the limits of the area from the Y scale to the Z scale: for y = 60, the value of z is z = ( y- μ )/ σ = (60 - 54)/4.5 = 1.33 equivalent ‘What percentage of fish are less than 60 mm long?’ ‘What is the area under the standard normal curve below z = 1.33’ 2. Look up the area in the Normal Curve Table: Area = ?? 0 1.33 Z 1. Convert the limits of the area from the Y scale to the Z scale: z = ( y- μ )/ σ = (60 - 54)/4.5 = 1.33 2. Look up the area in the Normal Curve Table: 0.9082 → 90.82% of the fish are less than 60 mm long! Inverse reading of Normal Curve Table … Inverse reading: Sometimes we start with an area and use the table to figure out the corresponding z value. Pr{ Z > Z α ) = α Z .025 =1.96 Percentiles: The values that divide a distribution into 100 equal parts. For example, the 70th percentile of a standard normal distribution is the number that divides the bottom 70% from the top 30%. ?? 1.96 Finding percentiles What is the 70th percentile of the fish length distribution? 1. Look up the corresponding Z value: Z .30 = 0.52 2. Convert the Z value to a Y value: Z = ( y- μ )/ σ 0.52 = ( y *-54)/4.5 y * = (.52)(4.5) + 54 = 56.3 Now you try … The lengths of herring follow a normal distribution. The mean length is 54.0 mm and the standard deviation is 4.5 mm. What is the 20th percentile of the fish distribution?...
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This note was uploaded on 11/04/2009 for the course BIO 50935 taught by Professor Bryant during the Fall '09 term at University of Texas.

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Lec5 - Determining areas for any normal distribution The...

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