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Business Statistics

Home | Whitty's MBA Home | Introduction To Statistics | Descriptive Statistics and Graphical Analysis | Measures of Central Tendency | Probability and the Normal Distribution | A Statistical Review | Confidence Intervals and Sample Size | Hypothesis Testing | More Hypothesis Testing | Confidence and Testing Review | Stats Websites Links
Probability and the Normal Distribution

Applied Business Probability and Statistics

Probability and the Normal Distribution

          Recognize basic probability concepts

          Discern the basics of the Normal Probability Distribution

          Apply the Normal Distribution to solve probability problems

          Solve probability problems using the Central Limit Theorem when the distribution is not Normal

Probability and the Normal Distribution

Probability is an everyday part of our lives. We use it to measure uncertainty and to help us make choices. For example, when the weatherman says that there is a 90% chance of rain, do you take an umbrella that day? Probability is NOT the same as betting odds (payoffs); Kentucky Derby and Pro Sports sets odds to insure they make money and the odds depend on the trend in betting. Lower odds do not mean one is more likely to win, it just means that more people think something is true, or that an event will happen. Vegas tables set payoffs lower than true odds so they always come out ahead in the money.

 

Mathematically, probability equals the number of events meeting the specified condition divided by the number of possibilities. With a deck of playing cards, there are 52 possible events. Thus, the probability of drawing an ace from a deck of cards is equal to the number of cards meeting the condition of being an ace (i.e., 4) divided by the number of possibilities (i.e., 52) = 4 / 52.

 

An event is just an uncertain outcome, the result of an experiment; an experiment is the process of making an observation. If one flips a coin (experiment), there are two possible events - heads or tails. If one flips a coin several times, one can expect 1/2 of the event to be Heads and 1/2 of the events to be Tails. If one flips a coin 10 times, the expected outcome is 5 Heads and 5 Tails. However, what if the outcome is 10 Tails in a row? Has the probability changed? In a fair coin, the probability of Heads or Tails is 50% because each coin toss and its outcome is completely independent of any other coin toss. What happened in the past has nothing to do with the future outcomes of independent events. Probability falls in a range of 0 and 1 (0% - 100%) inclusive. The probability that something will occur is NEVER less than 0% and NEVER greater than 100%.

Solving Probability Problems

 

Probability problems can easily be solved with tables. Setting up the table is usually the tricky part, but once it is done, the rest is easy. With a deck of cards, one would make a table with the values serving as the columns (A, 2, 3, …, J, Q, K) and the suits serving as the rows (diamonds, hearts, clubs, spades). We employ rules of Set Theory in which the objects in question are circled and those items that are relevant in the circle are counted.

 

Suppose 100 people were asked about their political affiliation, and the following shows their responses:

 

 

Dem

Rep

Ind

Total

Male

10

25

5

40

Female

30

20

10

60

Total

40

45

15

100

 

Now if one were asked the following, simple rules can be used to solve them.

1. P[Female]

2. P[Republican]

3. P[Female and Democrat]

4. P[Female or Democrat]

5. P[Female | Democrat] (i.e., probability of female given Democrat)

6. P[Democrat | Female] (i.e., probability of Democrat given female)

 

Let us take them one at a time.

1. P[Female] = total females / total people = 60/100

 

Dem

Rep

Ind

Total

Male

10

25

5

40

Female

30

20

10

60

Total

40

45

15

100

2. P[Republican] = total republicans / total people = 45/100

 

Dem

Rep

Ind

Total

Male

10

25

5

40

Female

30

20

10

60

Total

40

45

15

100

3.      P[Female and Dem] = intersection of females and democrats / total people

= 30/100

 

Dem

Rep

Ind

Total

Male

10

25

5

40

Female

30

20

10

60

Total

40

45

15

100

4.      P[Female or Dem] = union of females and dem / total people =

(10+30+20+10)/100 =70/100

 

Dem

Rep

Ind

Total

Male

10

25

5

40

Female

30

20

10

60

Total

40

45

15

100

5.      P[Female/Dem] = percentage of democrats who are female (i.e., prob of

female given Dem)

 

= # of female democrats / total democrats = 30/40

 

 

Dem

 

 

 

Male

10

 

 

 

Female

30

 

 

 

Total

40

 

 

 

6.      P[Dem/Female] = percentage of females who are democrat (i.e., prob of

Dem given female)

 

= # of female democrats / total females = 30/60

 

 

Dem

Rep

Ind

Total

 

 

 

 

 

Female

30

20

10

60

 

 

 

 

 

 

Excel can perform these computations, but it is important to understand how to derive the answers.

 

 

Is political party independent of gender? To be independent, P (A given B) = P (A). In other words, the probability of an event occurring is unchanged by the existence of the other event being known. The P [Female] = 60 / 100 = 60%. The P [Female given Democrat] = 30 / 40 = 75%. Knowing the person's political party alters the probability in regards to gender, so the two variables are dependent.

 

An example of independent variables would be the suit and value of a playing card. The probability of drawing an Ace from a deck of cards is 4 / 52. Given that the card drawn is known to be a diamond, the probability that the card is an Ace is now 1 / 13. Because 4 / 52 = 1 / 13, the two variables must be independent. Thus, the probability of a card being a specific value is independent of the suit (i.e., knowing the suit will not help determine the value of the card), AND the probability of a card being a specific suit is independent of its value (it works in both directions).

The Normal Distribution

If one were to look at a 5'8" tall man, he would probably be considered short. If one were to look at a 5'7" tall woman, she would probably be considered tall, despite the fact that she is shorter than the man is. Why? It is because we tend to compare each data point to its respective mean. To solve problems regarding heights, one would think we would need two separate formulas (one for men and one for women). However, this is not true. The Normal Probability Distribution allows us to put everything on the same scale. The Normal curve is a bell-shaped curve that peaks in the middle at the mean. The Standard Normal curve has a mean of zero. Units on the standard normal curve are measured in terms of the standard deviations; one standard deviation in both directions from the mean captures 68% of the data, two standard deviations in both directions captures 95% of the data, and three standard deviations in both directions captures 99.7% of the data. As expected, the bulk of the data is close to the mean.

 

The Standard Normal Distribution has values mainly between -3 and +3 as measured by a z-score, with the z-score being the number of standard deviations a value is from the mean. Thus, by converting the normal distribution of male height to a standard normal distribution, the height of 5'8" for a man is about one standard deviation below average (z = -1), and the height of 5'7" for a female is about two standard deviations above average (z = +2). The female would show up to the right of the male on this standard scale.

 

The SAT scores, for example, are computed by the comparing one's raw score with the mean raw score and dividing by the standard deviation (thus giving the z-score); then the z-score is converted to the SAT scale, which forces the mean to be 500 and the standard deviation to be 100. If the mean raw score on the math section was 40 and the standard deviation was 10, then a person who scored 60 would have scored 20 points higher than the mean, and since the standard deviation is 10, that person scored 2 standard deviations above average. On the SAT system, this translates to 200 points above the average of 500, so the SAT score would be 700. Because SAT scores are normally distributed, we know that about 68% of students score between 400 and 600, 95% score between 300 and 700, and the remaining 5% score below 300 or above 700.

 

One of the great features of using the Normal Distribution is that we can compute probabilities very easily. We only need the mean and standard deviation to define the curve completely. If one scored 400, and the mean is 500 with a standard deviation of 100, the computed score is 1 standard deviation below the mean.

 

The Central Limit Theorem

When a distribution is not Normal, we cannot compute probability directly. However, if we take samples from the data, the means of the samples will appear normally distributed if the sample size is large enough. Regardless of the look of the original distribution, larger samples will result in the curve starting to appear Normal. Generally, a sample of at least 30 guarantees that the distribution of sample means will be normally distributed. The mean of the new distribution is the same as the original mean, but the spread is cut down dramatically, and this dispersion in the distribution of samples is known as the standard error (which is computed from the standard deviation and the sample size).

 

When solving problems regarding the probability of a sample mean being in a specified range, only the mean, the standard deviation, and the sample size are needed. Excel will still do all of the work, but here is an example nonetheless.

 

If the mean age at GCU is 40 and the standard deviation is 12, then what is the probability of taking a sample of 36 students and finding the average age to be over 42? While 42 is only two years higher than 40, we are dealing with a sample mean here. The Central Limit Theorem will apply. The standard error = 12 / square root (36) = 12 / 6 = 2. So the z-score = (42 – 40) / 2 = 1.00. 16% of the values have a z-score greater than 1.00, so the probability of finding an average higher than 42 is 16%.

Conclusion

These core tools take the powers of probability to a new height. A measure of central tendency and a measure of dispersion can completely define a distribution, and now can allow one to solve for any probability in regards to that distribution. The only catch is that the data must be normally distributed, but if not, there are ways to bend the rules legally and make the data work

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