Module 6: Normal Distribution
Using the Normal Distribution
Barbara Illowsky & OpenStax et al.
Example
The shaded area in the following graph indicates the area to the left of
x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).
The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(X ≤ x) and P(X > x) is the same as P(X ≥ x) for continuous distributions.
Calculations of Probabilities
Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. Additionally, this link houses a tool that allows you to explore the normal distribution with varying means and standard deviations as well as associated probabilities. The following video explains how to use the tool.
Note
To calculate the probability without the use of technology, use the probability tables provided here. The tables include instructions for how to use them.
If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.
try it
If the area to the left of x is 0.012, then what is the area to the right?
1 − 0.012 = 0.988
Example
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.
- Find the probability that a randomly selected student scored more than 65 on the exam.
- Find the probability that a randomly selected student scored less than 85.
- Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).
- Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).
Solution:
- LetX = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5
- Draw a graph. Then, find P(x > 65). P(x > 65) = 0.3446
- Draw a graph. Then, find P(x > 65). P(x > 65) = 0.3446
The probability that any student selected at random scores more than 65 is 0.3446.
-
- Go into
2nd DISTR
. After pressing2nd DISTR
, press2:normalcdf
. The syntax for the instructions are as follows:normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10
99) by pressing1
, theEE
key (a 2nd key) and then99
. Or, you can enter10^99
instead. The number 1099 is way out in the right tail of the normal curve. We are calculating the area between 65 and 1099. In some instances, the lower number of the area might be –1E99 (= –1099). The number –1099 is way out in the left tail of the normal curve.[latex]displaystyle{z}=frac{{{65}-{63}}}{{5}}={0.4}[/latex]Area to the left is 0.6554.
P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446 - Calculate the z-score:*Press
2nd Distr
*Press
3:invNorm
(*Enter the area to the left of z followed by )*Press