Project 9: Normal Distribution: Finding Raw Scores

A z-score is the number of standard deviations above/below the mean a certain value is.

Suppose IQ score are normally distributed with mean 100 and SD 15

A person who has an IQ of 85 has a z-score of -1.  

A person with an IQ of 145 has a z-score of 3.

A person with an IQ of 120 has a z-score of 20/15 = 1.333

normalcdf(lower, upper) returns the area (proportion) under the normal curve between z = lower and z = upper.

normalcdf(-1, 0) returns .3413

normalcdf(-1, 1) returns .6826

normalcdf(-1, 2) returns .8185

normalcdf(0, 10) returns .5

normalcdf(-100,0) returns .5

invNorm(leftTailArea) returns the z-score with left tail area given

invNorm(.0227) returns -2

invNorm(.5) returns 0

invNorm(.8412) returns 1

invNorm(.75) returns .6744

Example 1: Adult IQ scores are normally distributed with mean 100 and sd 15.  What proportion of adults have an IQ less than 90?

z = (90-100)/15 = -.666

normalcdf(-10, -.666) = .255 so about 25.5% of adults have an IQ score less than 90.

Example 2:  Adult IQ scores are normally distributed with mean 100 and sd 15.  What IQ score is at the top 10% of IQ scores?

z = invNorm(.9) because 90% of scores are less than this score (left-tail area)

z = 1.28 which means the adult has an IQ score 1.28 standard deviations above the mean 

1.28*15 + 100 is about 119 so an adult would need an IQ of about 119 to be at the top 10% of IQ scores.

Project 9:  Variables 'mean', 'sd', 'leftTailArea' have been initialized.    Working methods normalcdf and invNorm are available.

Methods normalcdf(lowerZ, upperZ) and invNorm(leftArea) work  as shown above.

Task: Appropriately assign the values of 'z' and 'x'

z represents how many standard deviations above/below the mean the value would be with left-tail area given.

x represents the value in the distribution with the given left-tail area.

**If your code works for 5 test cases in a row, you can enter your e-mail address.