June 26, 2012-Tuesday/Lesson: Normal Distribution Curve as a probability Distribution Curve

Thanks to our teacher, stat is getting better within the minds of the III-Gold students, yet their are still so much of learning that needs to take over in order for us to be true geniuses in the field of Statistics.


According to our lesson last Tuesday with Ms. Macatigos, the Normal Distribution Curve can be used as a probability curve for normally distributed variables. The area under the curve is more important than the frequencies. This area corresponds to a probability.

For example, find the probability for P(0<z<2.32).

P (0<z<2.32) means to find the area under the normal distribution curve between 0 and 2.32. So we should look up the area in Table E (The Standard Normal Distribution) corresponding to z=2.32, which means that it is 0.4898 or 48.98%.

We also had the Applications of the Normal Distribution wherein we knew that the standard normal distribution curve can be use to solve a wide variety of practical problems and the only requirement is that the variable be normally or approximately normally distributed. The formula is:

We can use the Normal Distribution in solving problems like this:

If the scores for the test have a mean of

100 and a standard deviation of 15, find

the percentage of the scores that will fall                                    

 below 112.

Solution:

First step: Draw the figure and represent

 the area.

Second step: Find the z value

corresponding to a score of 112.

Third step: Find the area using Table E.

The area between z=0 and z=0.8 is

0.2881. Since the area under the curve

To the left of z=0.8 is desired, add 0.5000

To 0.2881 (0.5000+0.2.881=0.7881).

Therefore, 78.81% of the scores fall below

112.

Finding Data Values Given Specific Probabilities. 
The normal distribution can also be used to find specific data values for given percentages.

Example: In order to qualify for a police academy,                             

candidates must score in the top 10% on a general

abilities test. The test has a mean of 200 and a

standard deviation of 20. Find the lowest possible

score to qualify. Assume the test scores are normally 

distributed.

Solution:

Since the test scores are normally distributed, the test

value (x) that cuts off the upper 10% of the area under

the normal distributed is desired. Work back to solve

the problem

first step: Subtract 0.1000 from 0.500 to get the area

under the normal distribution between 200 and x.

Second Step: Find the z value that corresponds to

an area of 0.4000 by looking up 0.4000 in the area

portion of the table e. If the specific value cannot be

found, use the closest value. If the area falls exactly

halfway between the 2 z values, use the larger of the

Two z values.

Third  Step: Substitute in the formula.            

The score of 226 should be used as cutoff. Anybody scoring 226 or higher qualifies.                 

From: Kent Spencer Manalo Mendez
          III-Gold