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http://www.youtube.com/watch?v=xDIyAOBa_yUPowerPoint Presentation

http://www.youtube.com/watch?v=xDIyAOBa_yU

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The Standard Normal Distribution

http://www.youtube.com/watch?v=xDIyAOBa_yU

Notice that the x axis is standard scores, also called z scores. This means that the distribution has a population mean of zero, and a population standard deviation of 1.

The most significant thing about the normal distribution is that predictable proportions of cases occur in specific regions of the curve.

50%

50%

Notice that: (1) 34.13% of the scores lie between the mean and 1 sd above the mean, (2) 13.6% of the scores lie between 1 sd above the mean and 2 sds above the mean, (3) 2.14% of the scores lie between 2 sds above the mean and 3 sds above the mean, and (4) 0.1% of the scores in the entire region above 3 sds.

The curve is symmetrical, so the area betw 0 and -1 = 34.13%, -1 to -2 = 13.6%, etc. Also, 50% of the cases are above 0, 50% below.

IQ=115, z = 1.0 and 1

IQ Percentile Problem 1:

IQ: m=100, s=15. Convert an IQ score of 115 into a percentile, using the standard normal distribution.

Step 1: Convert IQ=115 into a z score:

z = xi-m/s = (115-100)/15 = 15/15; z = +1.0

(1 sd above the mean)

IQ=115, z = +1.0 and 1

The part with the slanty lines repre-sents the portion of the distribution we’re looking for.

Step 2: Calculate the area under the curve for all scores below z=1 (percentile=% of scores falling belowa score).

Area under the curve below z=1.0: 34.13+50.00= 84.13. We get this by adding 34.13 (the area between the mean and z=1) to 50.00 (50% is the area under the curve for values less than zero; i.e., the entire left side of the bell curve). So, an IQ score of 115 (z=1.0) has a percentile score of 84.13.

IQ=85, z = -1.0 and 1

IQ Percentile Problem 2:

IQ: m=100, s=15. Convert an IQ score of 85 into a percentile, using the standard normal distribution.

Step 1: Convert IQ=85 into a z score:

z = xi-m/s = (85-100)/15 = -15/15; z = -1.0

(1 sdbelow the mean)

IQ=85, z = -1.0 and 1

Step 2: Calculate the area under the curve for all scores below z=-1.

Area under the curve values below z=-1.0: 50.00-34.13=15.87. We get this by subtracting 34.13 (the area between the mean and z=-1) from 50.00 (50% is the total area under the curve for values less than zero; i.e., the entire left side of the bell curve). So, an IQ score of 85 (z=-1.0) has a percentile score of 15.87.

z = x and 1 i-m/s = 85-100/15 = -15/15

z = -1.0

IQ=85, z=-1.0

Area under the curve for z=-1.0: 50.00-34.13=15.87.

We get this by subtracting 34.13 from 50.00 (50.00 is the total area under the curve for values less than zero; i.e., the entire left side of the bell curve.) We therefore want the 50% minus the area between zero and -1.0). So, an IQ score of 85 has a percentile score of 15.87.

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