Thursday, June 27, 2013

How to Work with Normal Distributions


One of the most widely used and important continuous probability distributions seen in statistics is the normal distribution. The normal distribution is characterized by a bell shaped curve with the highest point above the mean. The distribution is symmetrical and approaches the horizontal axis but never touches it. But how do we work with normal distributions? At the end of this article, you should be able to calculate probabilities for a random variable that has a normal distribution.

Suppose a random variable x has a normal distribution with a mean m and standard deviation s. To find areas and probabilities for such a random variable, convert x values to z using the formula z = (x - m)/s. Then use a table for the area of a standard normal distribution, found in most statistics textbooks, to find the corresponding areas and probabilities.

Consider the following example:

Let x have a normal distribution with m = 15 and s = 4. Find the probability that an x value chosen at random from this distribution is between 8 and 19.

In symbols, we write this problem as P(8 ≤ x ≤ 19).

Since the probabilities correspond with areas under the normal distribution curve, we can find the area under the curve from x = 8 to x = 19. To do this, we convert the x values to z values.

For x = 8, z = (8 - 15)/4, therefore z = -1.75. For x = 19, z = (19 - 15)/4, therefore z = 1. Writing this in symbols, we get P(-1.75 ≤ z ≤ 1), This is the same as thinking of the area to the left of z = 1 minus the area to the left of z = -1.75. These values can be found using a table for the area of a standard normal distribution. Using the table, we get 0.8413 - 0.0401 = 0.8012.

Suppose we want to know the probability that a value chosen at random is greater than a certain value? Using the same value for mean and standard deviation, suppose we want the probability that x > 12?

This is the same as one minus the probability of x less than or equal to 12. In symbols, this is 1 - P(x ≤ 12). Converting the x value to z value, we get z - (12 - 15)/4 = -0.75. The z value corresponding to -0.75 is 0.2266. Therefore, P(z > -0.75) = 1- .2266 = .7734.

This guide should give a student a basic understanding of how to calculate probabilities for a random variable that has a normal distribution.

Monday, June 24, 2013

Change Raw Scores to Z Scores to Compare Normal Distributions


The normal distribution is a distribution that is symmetrical about a vertical line through the mean. The curve of the distribution is bell-shaped with the highest point over the mean. The curve will never cross or touch the horizontal axis. The differences in normal distributions makes calculating the area under the curve in a specific interval difficult. Therefore, there is a method to make this easier by changing raw scores into standardized scores known as "z scores".

The standardization of scores must be done in such a way that we can use one table for all of the normal distributions. This is done by considering how many standard deviations a data value lies from the mean. This makes it to compare a value from one normal distribution to a value from another normal distribution.

For example, suppose Frank and Tom are taking a history course. Frank is in the 8 o'clock class and Tom is in the 10 o'clock class. Since each class has a large number of students, the scores on the final exam each follow a normal distribution. Frank's section had a class average of 75 and his score was a 83. In Tom's section, the average was a 68 and Tom scored a 76. When comparing scores they both felt good that they scored 8 points higher than the average. But which one did better with respect to the rest of the students in the class?

Suppose a graph of the data shows that Frank scored higher than most of the other students in his class, but Tom's score places him more in the middle of the upper half. Then you know that Frank's score is clearly much better with respect to the students in his class than Tom's score is. But how can you tell if the distributions are not graphically represented?

You have to change the raw scores to z scores to see how many standard deviations from the mean each score lies. Suppose the standard deviation of scores in Frank's section is 5 and the standard deviation of scores in Tom's section is 6. The z score is the sample value minus the mean value divided by the standard deviation. Therefore, the z score for Frank is 8/5 = 1.6, and the z score for Tom is 8/6 = 1.33. It's clear that Frank had a better score relative to the class since his score fell 1.6 standard deviations about the mean and Tom's score was only 1.33 standard deviations about the mean.

Note that when standardizing the scores, the mean of the original distribution is always zero, which is logical since the mean lies zero standard deviations from itself. Any value above the mean has a positive z score, and any value lower than the mean has a negative z score.

This guide should help anyone understand how to change a raw score to a z score, which makes comparing scores from different normal distributions easy.

Friday, June 21, 2013

Remember in a two column geometric proof, the statements go on the left side and the reasons on the right side. The reasons should be postulates and axioms.   The given statement is almost always first in this kind of proof.

Indirect proofs are done in paragraph form.  You first assume that the statement you are to prove is false.  Then you use deductive reasoning and come up with two statements that cannot both be true. There is the contradiction that shows the earlier assumption is false, therefore the statement that is to be proven is true.

Monday, June 17, 2013

A little math humor, not sure this angle is acute though haha


Sunday, June 16, 2013

Math tidbit of the day:

Note the graphs of f(x) = log(x) and f(x) = ln(x) are very similar

Both have an x intercept of (1, 0), both do not have a y intercept.

Both have a range of (0, infinity)

Both have a range of (-infinity, infinity)

Thursday, June 13, 2013

Math tidbit of the day.


When calculating square roots and cube roots, it's easier if you know the perfect squares from 1 to 25 and the perfect cubes from 1 to 10

perfect squares : 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625

perfect cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000

Tuesday, June 11, 2013


Math tidbit of the day.

The set of all points equidistant from a single point forms a circle.

Saturday, June 8, 2013

When graphing functions, it's important to know the basic parent functions. From there you can shift the graph up, down, left and right accordingly. Theses shifts are also known as translations.

f(x) = x^2,  f(x) = sqrt(x),  f(x) = x^3,  f(x)= |x| are some examples of parent functions

f(x) = (x - 2)^2 + 3 has the same graph as f(x) = x^2 except it is translated 2 units left and 3 units up.

f(x) = sqrt(x + 1) - 3 is the same as the graph of f(x) = sqrt(x) except it is translated 1 unit right and 3 units down.

f(x) = (x + 4)^3 - 1 is the same as the graph of f(x) = x^3 except it is translated 4 units right and 1 unit down.

f(x)= |x - 2| + 1 is the same as the graph of f(x)= |x| except it is translated 2 units right and 1 unit up.

Wednesday, June 5, 2013

When trying to identify angles on the unit circle with their corresponding sine and cosine value, there is an easy way to do this.

Remember that the angles start at 0, then 30, 45, 60, 90 in the first quadrant.  So they angle increases by 30, 15, 15 and 30.  Likewise they increase by the same amount in the 2nd , 3rd and 4th quadrants.

So the remaining angles on the unit circle are 120, 135, 150, 180, 210, 225, 240, 270, 300, 315, 330, 360.

The sine and cosine values in the first quadrant are 1/2, sqrt(2)/2, sqrt(3), 2   - Sine
                                                                            sqrt(3)/2, sqrt(2)/2, 1/2    - Cosine

These values remain the same in all quadrants with the exception of + or -.

Just remember the signs are follows

Quadrant,   Sine,   Cosine

       1            +           +
      II             +           -
      III           -            -
       IV          -           +

Sunday, June 2, 2013

Just for fun, enjoy! Lots of truth to this, especially in higher mathematics.