University:
California State University, Northridge
Course:
MATH 340 | Introduction to Probability and Statistics
Academic year:

2011
Views:

62

Pages:

17
Author:

Kamden Bullock

Chapter 6: Probability P b biliti Probabilities, Proportions, P ti and d z-Scores S • • • • The unit normal table lists relationships between z-score locations and proportions in a normal distribution. For any z-score location, you can use the table to look up the corresponding proportions. Similarly, if you know the proportions, you can use the table to look up y p the specific z-score location. Because we have defined probability as equivalent to proportion, you can also use the h unit i normall table bl to look l k up probabilities for normal distributions. Probabilities,, Proportions, p , and z-Scores cont. • • • In the preceding section, we used the unit normal table to find probabilities and proportions corresponding to specific zscore values. In most situations situations, however however, it will be necessary to find probabilities for specific X values. Consider the following g example: p – It is known that IQ scores form a normal distribution with μ = 100 and σ = 15. – Given this information, what is the probability of randomly selecting an individual with an IQ score less than 130? Probabilities,, Proportions, p , and z-Scores cont. • Specifically, what is the probability of randomly selecting an individual with an IQ score less than 130? p(X (X < 130) = ? • • Restated in terms of proportions, we want to find the proportion of the IQ distribution that corresponds to scores less than 130. The distribution is drawn in Figure g 6.9,, and the portion we want has been shaded. Probabilities,, Proportions, p , and z-Scores cont. • The first step is to change the X values into z-scores. – In particular, the score of X = 130 is changed to • Thus, an IQ score off X = 130 corresponds Th d to a z-score of z =+2.00, and IQ scores less than 130 correspond to z-scores less than 2.00. Next, look up the z-score value in the unit normal table. • Fig. 6-9, p. 178 Probabilities,, Proportions, p , and z-Scores cont. • • • • • Because we want the proportion of the distribution in the body y to the left of X = 130 (see Figure 6.9), the answer will be found in column B. Consulting the table, we see that a z-score off 2.00 2 00 corresponds d tto a proportion ti off 0.9772. The probability of randomly selecting an individual with an IQ less than 130 is 0.9772: Finally, notice that we phrased this question in terms of a probability. Specifically, we asked, "What is the probability of selecting an individual with an IQ less than 130?" Probabilities,, Proportions, p , and z-Scores cont. • • However, the same question can be phrased in terms of a proportion: "What proportion of the individuals in the population have IQ scores less than 130?" Both versions ask exactly the same question and produce exactly the same answer. Finding g Proportions/Probabilities p / Located Between Two Scores • • • • • The highway department conducted a study measuring driving speeds on a local section of interstate highway. They found an average speed of μ. = 58 miles per hour with a standard deviation of σ = 10. The distribution was approximately normal. Given this information, what proportion of the cars are traveling between 55 and 65 miles per hour? Using probability notation, we can express the problem as Fig. 6-10, p. 179 Finding g Proportions/Probabilities p / Located Between Two Scores cont. • • Looking again at Figure 6.10, we see that the proportion we are seeking can be di id d into divided i t two t sections: ti (1) the th area left l ft of the mean, and (2) the area right of the mean. The first area is the proportion between the mean and z = -0.30 and the second is the proportion between the mean and z = +0.70. Finding g Proportions/Probabilities p / Located Between Two Scores cont. • • Using column D of the unit normal table, these two proportions are 0.1179 and 0.2580. The total proportion is obtained by adding these two sections: Finding g Scores Corresponding p g to Specific Proportions or Probabilities • • • • • Scores on the SAT form a normal distribution with μ = 500 and σ = 100. What is the minimum score necessary to be in the top 15% of the SAT distribution? (An alternative form of the same question is presented in Box 6.2.) This problem is shown graphically in Figure g 6.13. In this problem, we begin with a proportion (15% = 0.15), and we are looking for a score. According to the map in Figure 6.12, we can move from p (proportion) to X (score) via z-scores. Finding g Scores Corresponding p g to Specific p Proportions or Probabilities cont. • • • • The first step is to use the unit normal table to find the z-score that corresponds to a proportion of 0.15. Because the proportion is located beyond z in the tail of the distribution distribution, we will look in column C for a proportion of 0.1500. – Note that you may not find 0.1500 exactly, y, but locate the closest value possible. – In this case, the closest value in the table is 0.1492, and the z-score that corresponds d to this hi proportion i is i z = 1.04. 1 04 The next step is to determine whether the zscore is positive or negative. Remember that the table does not specify the sign of the z-score. Finding g Scores Corresponding p g to Specific p Proportions or Probabilities cont. • • • • Looking at the graph in Figure 6.13, you should realize that the score we want is above the mean, so the z-score is positive, z = + 1.04. Now you are ready for the last stage of the solution-that is, changing the z-score into an X value. By y definition,, a z-score of z = +1.04 corresponds to an X value that is located above the mean (+) by 1.04 standard deviations. O standard One d d deviation d i i is i σ = 100 points, i so 1.04 standard deviations is Finding g Scores Corresponding p g to Specific p Proportions or Probabilities cont. • • • Thus, our score is located 104 points above the mean. With a mean of μ = 500, the score is X = 500 + 104 = 604. Th conclusion The l i for f this thi example l is i that th t you must have an SAT score of at least 604 to be in the top 15% of the distribution. Fig. 6-13, p. 181

Probabilities, Proportions, and Z-Scores

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