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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 4: Variables mean, sd, leftTailArea have been initialized. Working methods normalcdf and invNorm are available.
The syntax for methods normalcdf and invNorm are shown in the examples 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, you can enter your e-mail address
Universal Computational Math Methods:
pow(5,2) returns 25.0
abs(-3.0) returns 3
sqrt(49.0) returns 7.0
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