Math 42 SIB

Normal Distribution and Z Scores
Equation of Normal Curve 
f ( x) 
1 2x
e
2
2
The normal probability distribution is a mathematical model which is generated by a
specific mathematical function. This mathematical model is useful because many
variables observed in nature closely approximate a normal distribution.
Normal distributions are symmetric and the tails are asymptotic, that is, the tails approach
the baseline but never quite reach it no matter how far away from the mean you are.
Obviously, our illustrations do not completely capture the nature of normal distributions.
The Standard Normal Distribution
All normal distributions can be converted to the standard normal distribution using the
standard score formula.
The standard normal distribution has a mean of zero (mean = 0) and a variance of one (s2
= 1) thus, of course, the standard deviation is one (s = 1).
Since most distributions are observed rather than generated mathematically, there are
features of normal distributions that can also be observed. In a normal distribution
68.26% of the distribution will fall between plus and minus one standard deviation
(±1.0s). Usually, this will be denoted using the standard normal (Z-score) distribution as
±1.0Z. Here are some other values form the standard normal distribution, 95.44%
between ±2.0Z and 99.74% between ±3.0Z.
Example Areas Under the Normal Curve
Here are some example areas under the standard normal curve.
The easy way to find the areas (probabilities) under the standard normal curve are to use
the tables found in every statistics book. These tables can be formatted several different
ways. We will use a method that displays the area between the mean (Z = 0.0) and a
given Z-score.
 Determining areas beneath normal curve using tables.
ex 1) area for z = 0 and z = 1.
ex 2) area between z = - 1.2 and z = 0
ex 3) area between z = - 1.63 and z = 2.22
ex 4) area of all z scores which are less than z = .43
ex 5) area of all z scores which are greater than z = 1.82 ?
Some Example Z-score Problems
820 8th grade students in a school district took a standardized social studies test that is
normally distributed and has a mean of 340 and a variance of 256. Here are the scores for
four of the students: John scored 364, Paul scored 356, George scored 344, and Ringo
scored 332.
Note: Graphs are designed to give a rough estimate and are not drawn to exact scale or
proportion.
1. In this sample, how many students would be expected to score above Paul?
First, compute the standard deviation: s = SQRT(variance) = SQRT(256) = 16
Convert Paul's raw score to a Z-score: Z = (356 - 340)/16 = +1.0
Find area from the mean to a Z-score of +1.0: .3413
Find area above: .5000 - .3413 = .1587
Find number of people: .1587 * 820 = 130.134 or 130 people
2. What proportion of the students would be expected to score above John?
Convert John's raw score to a Z-score: Z = (364 - 340)/16 = +1.5
Find area from the mean to a Z-score of +1.5: .4332
Find area above: .5000 - .4332 = .0668
3. What percent of students would be expected to score above Ringo?
Convert Ringo's raw score to a Z-score: Z = (332 - 340)/16 = -0.5
Find area from the mean to a Z-score of -0.5: .1915
Find area above: .5000 + .1915 = .6915
Convert to percent: 69.15%
4. How many students would be expected to score below Ringo?
Ringo's Z-score: Z = -0.5
Find area from the mean to a Z-score of -0.5: .1915
Find area below: .5000 - .1915 = .3085 or using information from #3: 1.0 - .6915
= .3085
Find number of people: .3085 * 820 = 252.97 or 253 people
5. What percentage of students would be expected to score below George?
Convert George's raw score to a Z-score: Z = (344 - 340)/16 = +0.25
Find area from the mean to a Z-score of +0.25: .0987
Find area below: .5000 + .0987 = .5987
Convert to percent: 59.87%
6. What proportion of students would be expected to score below John?
John's Z-score: Z = +1.5
Find area from the mean to a Z-score of +1.5: .4332
Find area below: .5000 + .4332 = .9332
7. How many students would be expected to score between George and Ringo?
George's Z-score: Z = +0.25
Ringo's Z-score: Z = -0.5
Find area from the mean to a Z-score of +0.25: .0987
Find area from the mean to a Z-score of -0.5: .1915
Find area between: .0987 + .1915 = .2902
Find number of people: .2902 * 820 = 237.964 or 238 people
8. What proportion of students would be expected to score between John and Paul?
John's Z-score: Z = +1.5
Paul's Z-score: Z = +1.0
Find area from the mean to a Z-score of +1.5: .4332
Find area from the mean to a Z-score of +1.0: .3413
Find area between: .4332 - .3413 = .0919
9. What percentage of students would be expected to score between George and Paul?
George's Z-score: Z = +0.25
Paul's Z-score: Z = +1.0
Find area from the mean to a Z-score of +0.25: .0987
Find area from the mean to a Z-score of +1.0: .3413
Find area between: .3413 - .0987 = .2426
Convert to percent: 24.26%
10. 225 students score above Linda, what is her test score?
Proportion above Linda: 225/820 = .2744
Proportion from mean to Linda: .5000 - .2744 = .2256
Z-score for Linda: +0.6
Find the raw test score: +0.6 = (X-340)/16, X = 0.6*16 + 340 = 349.6
11. What are the approximate test scores for Q1 and Q3?
Find area from the mean to Q1: .25
Find area from the mean to Q3: .25
Z-score for Q1: -0.67
Z-score for Q3: +0.67
Find the raw test score: -0.67 = (X-340)/16, X = -0.67*16 + 340 = 329.28
Find the raw test score: +0.67 = (X-340)/16, X = +0.67*16 + 340 = 350.72
12. A middle-school student took two standardized tests. On the Language Proficiency
test she scored 114.4 (mean = 100, sd = 16) and on Mathematical Reasoning she scored
61.7 (mean = 50, sd = 9). On which test did she do better?
Compute Z-score for test 1: Z = (114.4 - 100)/16 = .9
Compute Z-score for test 2: Z = ( 61.7 - 50)/ 9 = 1.3
Relatively speaking, the student did better on the Mathematical Reasoning test.