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

    

June 28- Using the different formulas in finding the z-value

       Some of my classmates, including myself find it difficult when to use the different formulas in finding the z-value. But our teacher, Ms. Macatigos explained further the idea when to use this different formulas.
       
       Acccording to her,  the formula z = value- mean / standard deviation is used to gain information about an individual data value obtained from the population while the formula z = value- population mean / standard deviation/ square root of n is used to gain information about a sample mean.


Consider the problems below for more clarification:


     * The average number of pounds of rice a person consumes a year is 225 pounds. Assume that the standard deviation is 30 pounds and the distribution is normally distributed. Find the probability that a person selected at random consumes less than 215 pounds per year.


   Solution: z = 215-225/ 30
                    = -0.34
       P(x less than 215) = P(z less than -0.34)
                       = 0.5- 0.1331
                       =0.3669


     - we use the first formula because the question asks about an individual person.




       But if the question will be, " If a sample of 35 individuals is selected, find the probability that the mean of the sample will be less than 215 pounds per year.


Solution: z = 215-225/ 30 / square root of 35
                 = -10/ 5.71
                 = -1.75


  P(x less than 215 ) = P(z less than -1.75 )
                  = 0.5 - 0.4599
                  = 0.0401


- we use the second formula because the question concerns the mean of a sample with a size of 35.






       I hope it will become more easier for us to know what formula will we use in finding the z- value after the examples I had given.






         By: Dustin Joshua Esquia III- Gold

On MOndAy...

                                                                                                                                                June 25, 2012  

   Last week which is the 3rd week of June we III-Gold students were entangled to do a mind teasing quiz about our past lesson which is the Discrete Probability Distribution. Well, even though it was hard like all of our past lessons it became more easy for us to answer because of the cooperative learning, group study and mostly because each one of us studied hard to have high scores that we finally had on Monday when we Checked our Quiz #3 which is the The Discrete Probability Distribution.
     Our scores were pretty high this time which means almost all of us didn't have a below passing score which also means that almost the majority of the III-Gold's score's percentage were 80% and even 100% sharp.
     Triumphant to have high scores on our quiz, we III- Gold students will strive to maintain this kind of attitude  specifically being studious so that the next lessons or quizzes and assignments will be just an even smaller piece of cake for us. We will remain hungry. Hungrier than ever to have not just passing scores but to have scores equivalent to 100% grade.

By: Carl Joel Escanilla Palma III-Gold

What happened on Wednesday?

                    Last Wednesday morning, almost all my classmates including me are rushing along the quadrangle just to finish our assignment in Statistics. 'Copy here, copy there, copy everywhere', that's the motto of III-Gold during flag ceremony.
                   That day is a checked day for all of the assignments, activities and quizzes that we haven't checked yet. And then, 30 minutes was given to finish our Exercises on the Application of Normal Distribution. But almost all of us, did not finished our task, so mam Macatigos let us pass our assignment that  afternoon.
                  Honestly, I copied some of my answers from my classmates, but I also give an effort and try my very best to answer that Exercise. I slightly don't understand about the shaded part when the z value is less than or greater than and it is negative.
Example:  P( z < -1.52). What should be the shaded part? to the left or right?

                                                Which is correct? to the left or to the right?
Some of my classmates says that to the left is the shaded part because -2 is less than -1. But some answers to the right because it is just like in the positive value of z that when z is less than; to the 0 point and when greater than, to the tail point.

               That's the big question in my mind??? 

By: Tito Nuevacobita Jr. 

III-Gold

Binomial Distribution

Binomial Distribution

Binomial distribution is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. Such a success/failure experiment is also called a Bernoulli experiment or Bernoulli trial; when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.

·       Bernoulli distribution is the same as binomial distribution; it was named after the Swiss mathematician Jakob Bernoulli.

Binomial Experiment- a probability experiment that satisfies the following four requirements:

1.        Each trial can have only two outcomes or outcomes that can be reduced to to two outcomes.

2.      There must be a fixed no of trials.

3.      The outcomes of each trial must be independent of each other.

4.      The probability of a success must be remain the same for each trial.

Notation for thee Binomial Distribution

            P (S) The symbol for the probability of success

            P (F) The symbol for the probability of failure

            p       The numerical probability of success

            q       The numerical probability of failure

            n       The number of trials

            X       The number of successes

Binomial Probability Formula

           

The binomial Distribution has the following properties:

• The mean of distribution-   μ  =n · p 
• The Variance- σ² = n ∙ p ∙ q
• The Standard Deviation-  

Examples:

Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?

Solution: This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is:

b(2; 5, 0.167) = ₅C₂ * (0.167)² * (0.833)³
b(2; 5, 0.167) = 0.161

Application of Binomial distribution in our daily life;

            Some of my classmates always ask, “andat ginatun-an gd ra ang mga distribution”? magamit ta ra sa kabuhi  ta hw? Anu bay ra pulos na kon hindi related sa math ang kurso nga bul-on namun?’

           

In business;

·       In the insurance field;

Group insurance, which gets cheaper as the group gets larger, is an example of the principle in application; actuarial abnormalities have less influence on total claims.

·       In the auto industry;

when you are  evaluating  the  no. of  cars  are  poorly  painted  on the  system.
The Poisson distribution applies when you are counting the number of objects in a certain volume or the number of events in a certain time period. You know the average number of counts, and wish to know the chance of actually observing various numbers of objects or events. 

·       In the call center management;

to monitor  the  no. of  calls  received / no.  of  telephone machines  breakdown.

·       In the traffic management at a signal;

no of  traffic  flow /  no.  of   traffic  accidents.

·       IN  THE  BUSINESS  OPERATION   FIELD
Applying the technique to calculate, for example, losses from business disruption is, in principle, very straightforward:
* Identify distributions for both the frequency and severity of losses;
* Generate a random value from the frequency distribution to represent the number of losses in a given period;
* Generate random values form the severity distribution for each loss and aggregate to give a total loss for the period;
* Repeat many (several thousand) times and plot an overall loss distribution.
First of all they examine how many outages they suffer per year and it turns out that the mean number of outages in a year is five. They then look at the cost of these outages including:
* Overtime to catch up with lost work;
* Lost customer orders; 

   * Mis-processed customer orders; and
   * Delays / mistakes in billing customers. 

In my own understanding, we can apply binomial probability:

·       WHEN CHOOSING A BUSSINESS TO RUN

For example; you want to run a business of computer shops and a survey from National Cyber Community found that 95% of students go to computer shops every day. If 5 students are selected at random, find the probability that all five will go to computer shops every day. Does your business will succeed?

 

Yes, your business will succeed because more than ½ of the students go to computer shops every day.  

·       WHEN TAKING AN EXAMINATION

For example; if you randomly guesses a 10 multiple-choice questions in an examination. Find the probability that you will get a score of at least 8 correct answers or pass the examination. Each question has five possible choices. Will you continue to randomly guessing? 

With these result I will not continue randomly guessing because the probability that I will pass the 10 multiple-choice examinations with random guessing is 0.0000779264.

(di ko sure kon sakto answer koh)

By: Tito Nuevacobita

June 20, 2012- Lesson on Poisson Distribution and Hypergeometric Distribution

Poisson Distribution

A Poisson distribution is the probability distribution that results from a Poisson experiment.

Attributes of a Poisson Experiment

A Poisson experiment is a statistical experiment that has the following properties:

• The experiment results in outcomes that can be classified as successes or failures.
• The average number of successes (μ) that occurs in a specified region is known.
• The probability that a success will occur is proportional to the size of the region.
• The probability that a success will occur in an extremely small region is virtually zero.

Note that the specified region could take many forms. For instance, it could be a length, an area, a volume, a period of time, etc.

Notation

The following notation is helpful, when we talk about the Poisson distribution.

• e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.)
• μ: The mean number of successes that occur in a specified region.
• x: The actual number of successes that occur in a specified region.
• P(x; μ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is μ.

Poisson Distribution

A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution.
Given the mean number of successes (μ) that occur in a specified region, we can compute the Poisson probability based on the following formula:

Poisson Formula. Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is μ. Then, the Poisson probability is:

P(x; μ) = (e^-μ) (μ^x) / x!

where x is the actual number of successes that result from the experiment, and e is approximately equal to 2.71828.

The Poisson distribution has the following properties:

• The mean of the distribution is equal to μ .
• The variance is also equal to μ .

Example 1

The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow?
Solution: This is a Poisson experiment in which we know the following:

• μ = 2; since 2 homes are sold per day, on average.
• x = 3; since we want to find the likelihood that 3 homes will be sold tomorrow.
• e = 2.71828; since e is a constant equal to approximately 2.71828.

We plug these values into the Poisson formula as follows:

P(x; μ) = (e^-μ) (μ^x) / x!
P(3; 2) = (2.71828^-2) (2³) / 3!
P(3; 2) = (0.13534) (8) / 6
P(3; 2) = 0.180

Thus, the probability of selling 3 homes tomorrow is 0.180 .

Poisson Calculator

Clearly, the Poisson formula requires many time-consuming computations. The Stat Trek Poisson Calculator can do this work for you - quickly, easily, and error-free. Use the Poisson Calculator to compute Poisson probabilities and cumulative Poisson probabilities. The calculator is free. It can be found under the Stat Tables tab, which appears in the header of every Stat Trek web page.

Poisson Calculator

Cumulative Poisson Probability

A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit.
Example 1

Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari?
Solution: This is a Poisson experiment in which we know the following:

• μ = 5; since 5 lions are seen per safari, on average.
• x = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions.
• e = 2.71828; since e is a constant equal to approximately 2.71828.

To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5). To compute this sum, we use the Poisson formula:

P(x < 3, 5) = P(0; 5) + P(1; 5) + P(2; 5) + P(3; 5)
P(x < 3, 5) = [ (e^-5)(5⁰) / 0! ] + [ (e^-5)(5¹) / 1! ] + [ (e^-5)(5²) / 2! ] + [ (e^-5)(5³) / 3! ]
P(x < 3, 5) = [ (0.006738)(1) / 1 ] + [ (0.006738)(5) / 1 ] + [ (0.006738)(25) / 2 ] + [ (0.006738)(125) / 6 ]
P(x < 3, 5) = [ 0.0067 ] + [ 0.03369 ] + [ 0.084224 ] + [ 0.140375 ]
P(x < 3, 5) = 0.2650

Thus, the probability of seeing at no more than 3 lions is 0.2650.

What happened on Thursday

                                                                                                                                              06-19-12    
       Our hearts beat faster and faster as we approach the room. Walking slowly, talking towards the statistics room of Ms. Macatigos where our assignments will be checked and a question lingers in our minds, will we have passing score? Zero?
       Then after we checked our assignments the result came before us and revealed that almost all of us had a very, very low score. Negative of what we expected, we the III- Gold went out of the room feeling hard and hungry of having a goal to study harder so that next time the tragedy will not happen ever again...

Dustin Joshua Esquia and Carl Joel Palma

06-22-12
          Random variables can either be discrete or continuous.  Discrete variables cannot assume all values between any two given values of the variables.  On the other hand, a continuous variable can assume  all values between any two given variables.
          Many continuous variables have distributions that are bell shape and are called approximately normally distributed variables.
       

Understand the Standard Normal distribution and its connection to all other Normal distributions

A value, x, from a normal distribution specified by a mean of m and a standard deviation of s can be converted to a corresponding value, z, in a standard normal distribution with the transformation z=(x-m)/s.  And, of course, in reverse, any value from a standard normal graph, say z, can be converted to a corresponding value on a normal distribution with a mean of m and a standard deviation of s by the formula x=m+z*s.  Remember that the standard normal distribution has a mean of 0 and a standard deviation of 1, i.e., m=0, s=1.

The ability to carry out this transformation is very important since we can do all our analysis with the standard normal distribution and then apply the results toevery other normal distribution, including the one of interest.  For example, to draw a normal curve with a mean of 10 and a standard deviation of 2 (m=10, s=2), draw the standard normal distribution and just re-label the axis.  The first figure below is the standard normal curve and the next figure is the curve with (m=10,s=2).

Each value along the x-axis represents that many standard deviations from the mean.  The 1 (or -1) x-value is one standard deviation from the mean.  Similarly, the 3 (or -3) represents three standard deviations from the mean.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution. It is the distribution that occurs when a normal random variable has a mean of zero and a standard deviation of one.

Standard Normal Distribution Table

A standard normal distribution table shows a cumulative probability associated with a particular z-score. Table rows show the whole number and tenths place of the z-score. Table columns show the hundredths place. The cumulative probability (often from minus infinity to the z-score) appears in the cell of the table.

For example, a section of the standard normal table is reproduced below. To find the cumulative probability of a z-score equal to -1.31, cross-reference the row of the table containing -1.3 with the column containing 0.01. The table shows that the probability that a standard normal random variable will be less than -1.31 is 0.0951; that is, P(Z < -1.31) = 0.0951.

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
-3.0	0.0013	0.0013	0.0013	0.0012	0.0012	0.0011	0.0011	0.0011	0.0010	0.0010
...	...	...	...	...	...	...	...	...	...	...
-1.4	0.0808	0.0793	0.0778	0.0764	0.0749	0.0735	0.0722	0.0708	0.0694	0.0681
-1.3	0.0968	0.0951	0.0934	0.0918	0.0901	0.0885	0.0869	0.0853	0.0838	0.0823
-1.2	0.1151	0.1131	0.1112	0.1093	0.1075	0.1056	0.1038	0.1020	0.1003	0.0985
...	...	...	...	...	...	...	...	...	...	...
3.0	0.9987	0.9987	0.9987	0.9988	0.9988	0.9989	0.9989	0.9989	0.9990	0.9990

There are a few ways to find the area under a normal distribution curve for two z-scores on opposite sides of the mean using a z-table. Once you know how to read the table, finding the area only takes seconds!

If you are looking for other variations (finding the area for a value between 0 and any z-score, or between two z-scores on the same side, see this normal distribution curve index).

normal distribution with z-scores on opposite side of mean

Step 1: Look in the z-table for the given z-scores (you should have two) by finding the intersections. For example, if you are asked to find the area from z= -0.46 to z= +0.16, look up both 0.46* and 0.16. The table below illustrates the result for 0.46  (0.4 in the left hand colum and 0.06 in the top row. the intersection is .6772).

z	0.00	0.01	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
0.0	0.0000	0.0040	0.0080	0.0120	0.0160	0.0199	0.0239	0.0279	0.0319	0.0359
0.1	0.0398	0.0438	0.0478	0.0517	0.0557	0.0596	0.0636	0.0675	0.0714	0.0753
0.2	0.0793	0.0832	0.0871	0.0910	0.0948	0.0987	0.1026	0.1064	0.1103	0.1141
0.3	0.1179	0.1217	0.1255	0.1293	0.1331	0.1368	0.1406	0.1443	0.1480	0.1517
0.4	0.1554	0.1591	0.1628	0.1664	0.1700	0.1736	0.1772	0.1808	0.1844	0.1879
0.5	0.1915	0.1950	0.1985	0.2019	0.2054	0.2088	0.2123	0.2157	0.2190	0.2224

Step 2: Add both of the values you found in step 1 together.

*note. Because the graphs are symmetrical, you can ignore the negative z-values and just look up their positive counterparts. For example, if you are asked for the area of a tail on the left to -0.46, just look up 0.46.

 by: Jorey Mark A. Millamena