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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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- )/ = (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- )/ = (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 ) = 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- )/ 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?...
View Full Document

Page1 / 22

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online