Density curves and the normal distribution

Transcription

1 Density curves and the normal distribution - Imagine what would happen if we measured a variable X repeatedly and make a histogram of the values. What shape would emerge? A mathematical model of this shape is called a density curve. The connection between a density curve of a variable X and repeated measurements of X can be described in either of the following ways: The area of the density curve between a range of two values on the measurement axis gives both: a) the probability that a single measurement of X will produce a value withing that range. b) or, if X is measured repeatedly,the

2 proportion (or percentage) of values that lie withing that range. Basic fact: Since probabilities and percentages can only add up to one, density curves are always scaled so that the total area under the density curve is one. Example A: A random number generator utilizes a uniform or rectangular density. Lets say that it is uniform over the interval [0,3] a) What is the probability that, in a single try, the random number generator produces a value less than 1?

3 b) If the random number generator selects a large number of numbers, what percentage of these numbers do we expect to see above 2.5? Answers: 1/3, 1/6 Normal Distributions A variable X is said to be normally distributed or to have a normal distribution if, after many many repeated measurements of X, the values display a dispersion pattern that is symmetric and bell shaped, according to a very specific density curve. All normal distribution are characterized simply by two parameters, one that indicates the location of the central value, and another

4 to indicate who spread out the values are around the mean. Borrowing terminology already used, we call these parameters the mean and the standard deviation. We use the Greek letters for the mean and for the standard deviation. Notation: (, ) N means normal distribution with mean and standard deviation Basic fact ( rule)

5 1. The portion within one standard deviation of the mean comprises 68% of the area. 2. The portion within two standard deviations of the mean comprises 95% of the area. 3. The portion within three standard deviations comprises 99.7% of the area. Lets say that typical college applicants take one of two possible tests. Let s also say that in a given year, the range of scores for each

6 test can be modeled as a normal distribution such that Test A: The scores are normally distributed with 400 and 100. Test B: The scores are normally distributed with 300 and 80. Karen takes Test A and scores a 550. Allison takes Test B and scores a 460. Who performed better relative to their populations? We would like to know who scored higher than a greater percentage of fellow test takers. Although Karen had a higher raw score, more of her fellow students also scored high as well, so it is not clear who

7 outperformed a greater percentage of classmates. We would like to compute both Karen s and Allison s scoring percentage, which is accomplished by converting to standardized values, or z-scores Standardized values (z-scores) 2 If x is a sample from a N (, ) distribution, then the equation z x relates x to its standardized score (zscore)

8 Thus, Karen s z-score is 1.5, whereas Allison s z-score is 2. This shows that Allison actually outperformed Karen, even though she had the smaller raw score. In fact, Allison scored better than 97.72% of her classmates, and Karen scored better than 93.32% of her classmates. See the left tail probability table. Tables of the Normal Distribution Probability Content from -oo to Z Z

9 Computing probabilities for a normal distribution The previous example shows the major steps needed to do normal probability calculations. Basically, calculating probabilities involve two steps:

10 1. Convert raw scores x to standardized scores z 2. Look up the left tail probabilities P( Z z) using a table 3. Compute the area using arithmetic: a) A right tail probability is computed by subtraction of a left tail probability from 1: P( Z z) 1 P( Z z) the one s complement rule b) A bounded region probability is computed by subtraction of left tail probabilities P( a Z b) P( Z b) P( Z a)

11 Notation: The symbol P( Z z) refers to the probability that a standardized score Z will fall below z. It also refers to the proportion of times one expects Z to fall below z. Thus a statement such as P ( Z 1.00) means the proportion of z-scores which fall below 1.00, is 84.13% Example: Ball bearing are manufactured to a target diameter of 1.00mm with a tolerance of mm above and below this value. Suppose in actuality the diameters are

12 normally distributed with mean 0.99 mm with a standard deviation of 0.02 mm a. What proportion of ball bearings have diameters within one standard deviation of the mean? (That means between 0.97 and 1.01mm?) b. What proportion of ball bearings have diameters within two standard deviations of the mean? (That means between 0.95 and 1.03mm?) c. What proportion of ball bearings have diameters at least 1.00mm? d. What proportion of ball bearings are within specifications? Answers:

13 a) Let X denote the diameter of a given ball bearing. We wish to calculate P (. 97 X 1.01). By the rule, this is approximately 68%. On the other hand, we can get a more accurate result from the tables. Since the z-score of.97 is z x and the z-score of 1.01 is 1.00 z x We convert the probability P(.97 X 1.01) to P( 1.00 Z 1.00). We can now use the standardized normal tables to compute P ( 1.00 Z 1.00). Since this is a bounded area, we subtract:

14 P( 1.00 Z 1.00) P( Z 1.00) P( Z 1.00) b) This time we wish to calculate P (. 95 X 1.03). By the rule, this is approximately 95%. On the other hand, we can get a more accurate result from the tables. Since the z-score of.95 is z x and the z-score of 1.03 is 2.00 z x We convert the probability P(.95 X 1.03) to P( 2.00 Z 2.00). We can now use the standardized normal tables to

15 compute P ( 2.00 Z 2.00). Since this is a bounded area, we subtract: P( 2.00 Z 2.00) P( Z 2.00) P( Z 2.00) c) We can t use the rule, so we have to use our strategy. Since the z-score of 1.00 is z x we convert the probability P ( X 1.00) to P( Z 0.50). This is a right tail area, so we can use the normal tables and the one s complement rule to obtain

16 P( Z 0.50) 1 P( Z 0.50) d) We wish to calculate (. 965 X 1.035) (The rule does not apply) Since the z-score of.96 is P. z x and the z-score of is 1.25 z x We convert the probability P(.965 X 1.035) to P( 1.25 Z 2.25). We can now use the standardized normal tables to compute P ( 1.25 Z 2.25). Since this is a bounded area, we subtract:

17 P( 1.00 Z 1.00) P( Z 2.25) P( Z 1.25) Working Backwards Example: In a national exam, the score distribution is approximately normal with mean 550 and standard deviation 110. Sondra scored at the 63 rd percentile, meaning she scored at least as well as 63% of the test takers. What was her score? Solution:

18 In this case, we have to first assign Sondra her correct z-score. We then have to find the raw score that corresponds to this z-score. Using the normal tables we find that the 63 rd percentile occurs at a z-score of 0.33 (approximately). So Sondra s z-score is 0.33 Now we can find her raw score using the relation : z x x in 550 and 110 plugging We know z, it is so now we can solve for x

19 x 110 ( 0.33)(110) x 550 x 550 (0.33)(110) 586 Example: Assume diastolic blood pressure readings are normally distributed with mean 88 and standard deviation 12. What range of blood pressures encompasses the middle 75% of readings? Solution: If X is the variable of blood pressure readings, we want to find numbers c and d such that P ( c X d) Because of symmetry, it is easy to find c once d is found. Now since the remainder area is

20 .2500, and this is split into two tails, it must be true that the area above d is Thus d must be the th percentile. So using the table: The z-score of d is Therefore: d 12 d 88 (12)(1.15) So d is 13.8 units above the central reading of 88. Therefore c is 13.8 units below 88. So c = 74.2 and d = 101.8