Normal Random Variables In the continuous model there is no table. The distribution is described by a graph of a positive function and the probabilities are found using the areas under between that function and x-axis on a particular range. This function is called probability density function. Example 1. The height of 18-year-old male is distributed approximately according to the following density, This seems on average 70 inches with a standard deviation of 3 inches. We shall learn how to recognize this later. If we are interested in calculating the chance that a random 18-year-old would be anywhere between 72 and 76 inches tall we would need to calculate the following area:
*** The chance the height is between 65 and 70 inches is the following area:
This seems simple enough but we do not have knowledge to calculate these areas! Definition Random Variable with the density given by 2 x µ 2 2 1 σ f( x) = e for any real x (1.1) 2 2πσ where µ is real number and σ > 0 (positive real number), is called Normal Random 2 N µσ,. Variable with parameters µ and σ and often denoted by ( ) In case of ( 0,1) N we call the variable Standard Normal Random Variable and often denote with Z. Although this variable has no value as a particular model, it does serve for standardization and once we learn to calculate the areas under the density curve of Standard Normal Random Variable we will be able to calculate the areas under any normal density.
Example 2. What is the chance that Z is less than 1.3. We write that as P( Z < 1.3) following area:. This is the We use textbook tables (attend the class): P Z < 1.3 0.9032. ( ) Example 3. What is the chance that Z value is between 0 and 1? P 0< Z < 1. The area is shown below. We write that as ( ) We use the textbook tables (attend the lecture!) to determine that P 0 < Z < 1 0.8413 0.5 = 0.3413. ( )
Example 4. P Z > 0.45. The area is: Evaluate ( ) From the textbook tables again: P Z > 0.45 1 0.6736 = 0.3264. Evaluate ( 0.37 Z 0.82) ( ) Example 5. P < <. This is the following area: From the tables: P( Z ) 0.37 < < 0.82 0.7939 0.3557 = 0.4382.
Example 6. P 1.47 < Z < 0.28. Evaluate ( ) From the tables: P( Z ) 1.47 < < 0.28 0.3897 0.0708 = 0.3189. *** Notable features of the graph of pdf for Standard Normal RV are: 1. Obviously the total area under the curve is 1. This is forced by the meaning and it can be shown using calculus methods. 2. Bell shaped curve with maximum at x = 0 and symmetrical about that line. This is easy to see using calculus from the expression for the function. 3. Inflexions at x = ± 1. We know this from calculus as well. 4. Empirical rule 68.3 95.4 99.7. This will be explained in the class, attend the lecture! Homework: Check online
Standardization For general normal random variable the properties 1-4 from the previous section are analogous: 1. The total area under the curve is 1. 2. Bell shaped curve with maximum at x = µ (where µ is the mean of the variable) and symmetrical about that line. 3. Inflexions at x = ± σ where σ is the standard deviation of the variable. 4. Empirical rule 68.3 95.4 99.7 holds as well. The only difference between various normal distributions is the scale, check the following two images:
Therefore to calculate the areas under the curve of general normal density we need to rescale the numerical values to z-values according to the following formula: z value = x µ σ For example, in example 1, the z-value of height of 76 inches is x µ 76 70 z value = = = 2. σ 3 Example 7. What is the probability that a random 18-year-old male is anywhere between 70 and 76 inches tall? Solution: The question is formulated first in terms of original random variable, then re-scaled to z- values and then we use the tables, P ( 70 X 76) 70 70 76 70 = P Z 3 3 ( 0 Z 2) = P 0.4772. The answer is approximately 48%.
Example 8. The average length of adult garden snake is 7 inches while the variable (length of a random garden snake) has standard deviation of 1.5 inches. Assuming normal distribution, what is the chance a random garden snake would be longer than 8 inches? P ( 8 < X ) 8 7 = P < Z 1. 5 ( 0.67 Z) P < 0.2514. The answer is approximately 25%. Example 9. The average IQ of general population is 100 while the random variable (IQ of a random person) has a standard deviation of 15. The average IQ of 9-year-old children in Tokio is 109. What percentage of general population is less intelligent than an average 9-year-old in Tokio? P ( X < 109) 109 100 = P Z < 15 ( Z 0.6) = P < 0.7257. The answer is about 73%. Example 10. The average white blood cell count per cubic millimeter of whole blood is 7500. The standard deviation of white blood cell count per cubic millimeter is 1750. Assume normal distribution. What percentage of random lab tests shows the WCBC between 5,000 and 10,000?
P ( 5000 X < 10000) 5000 7500 10000 7500 = P Z 1750 1750 ( 1.43 Z 1.43) P < < 0.8472. Approximately 85%. Homework: Check online.