Transcript Ch 5
Normal Distribution The most commonly used distribution for continuous random variable f(x) μ x “Bell Shaped” Symmetrical Mean, Median and Mode are Equal BUS304 – Chapter 5 Normal Probability Theory 1 Examples of Normal Distribution Examples: Heights and weights of people Students’ test scores Amounts of rainfall in a year Average temperature in a year Characteristics: Most of the time, the value is around the mean; The higher (lower) a value is from the mean, the less likely it happens; The probability a value is higher than the mean is the same as the probability that the value is lower than the mean; It could be extremely higher or lower than the mean. BUS304 – Chapter 5 Normal Probability Theory 2 Understanding the Curve One striking assumption of continuous random variables is: The curve doesn’t represent a probability. The area below the curve does! We don’t study the probability that the random variable takes one value. We care about the range. f(x) E.g. we don’t care about the likelihood of a person’s income is $50,102 per year. We care the percentage a person’ income falls in the range of $50,000~$60,000 x a BUS304 – Chapter 5 Normal Probability Theory b 3 Characteristics of Normal Distribution f(x) σ μ μ Location is determined by the mean (expected value), μ Spread is determined by the standard deviation, σ The shape of the normal distribution is decided based on μ and σ In theory, the random variable has an infinite range: from - to + BUS304 – Chapter 5 Normal Probability Theory 4 Rules to assign probability for normal distribution Events: Typical events we care about: x<100 lbs, x>300K, 50<x<60, etc. Think: what are the complement events for the above events? Basic Rules: 1. P(-< x < +) = 1 ~ x is certainly between - and + f(x) P( x μ) 0.5 P(μ x ) 0.5 2. P(x = a) = 0 for any given a. (It is almost impossible to find that x equals to a specific value. E.g. it is impossible to find a person with a height exactly as 6.1334 feet.) 3. P(x > ) = P(x ) = P(x < ) = P(x ) = 0.5 – symmetry. 0 . μ 5 BUS304 – Chapter 5 Normal Probability Theory x 5 Determine the probability for normal distribution (2) More rules: 3. P(x<+b)= 0.5 + P(<x<+b) f(x) P ( x +b ) 0.5 +b x 4. P( - b<x<) = P( <x< +b) BUS304 – Chapter 5 Normal Probability Theory 6 Exercise Assume that =3, P(3<x< 4)=0.2, determine the following probabilities: f(x) 2 3 4 x P(x< 4)? P(2 <x< 3)? P(2<x< 4)? P(x<2)? BUS304 – Chapter 5 Normal Probability Theory 7 Empirical Rules The probability that x falls in to the range (μ-1σ, μ+1σ) is 0.6826 (or 68.26%) That is, the area under the curve is 0.6826 Derivation: A half of the area will be 0.6826/2 = 0.3413 f(x) P(μ-σ<x<μ)=0.6826 34.13% σ f(x) μ1σ σ σ σ μ μ+1σ x f(x) 68.26% 34.13% P(μ<x<μ+σ)=0.6826 μ1σ μ μ+1σ P(μ-σ<x<μ+σ)=0.6826 σ x μ1σ μ μ+1σ BUS304 – Chapter 5 Normal Probability Theory x 8 Empirical Rules More Derivation: Derivation: The tailed area is 0.5-0.3413 =0.1587 f(x) P(x<μ-σ)=0.1587 15.87% The complement of tailed area is 1-0.1587=0.8413 f(x) σ μ1σ μ P(x>μ-σ)=0.8413 84.13% σ μ+1σ x f(x) μ1σ σ μ μ+1σ x f(x) 15.87% 84.13% P(x>μ+σ)=0.1587 P(x<μ+σ)=0.8413 σ μ1σ μ σ μ+1σ x μ1σ μ μ+1σ BUS304 – Chapter 5 Normal Probability Theory x 9 Exercise A supermarket (Walmart) wants to reward its customers. Based on the past experience, the accumulative expenditure of each customer in a month follows a normal distribution with mean $700 and standard deviation $300. The criterion to reward the customer is that if a customer spend more than $1000 will receive a reward of free digital camera. If you randomly select a customer to check whether he/she receives a digital camera, what is the probability that you will get a confirmative answer? If all the customers whose accumulative expenditure in Oct exceeds $400 will receive a free burger from McDonalds, what is the probability that you meet a customer gets a burger? BUS304 – Chapter 5 Normal Probability Theory 10 More Empirical Rule μ ± 2σ covers about 95% of x’s 2σ 2σ μ μ ±3σ covers about 99.7% of x’s x 95.44% 3σ 3σ μ x 99.72% BUS304 – Chapter 5 Normal Probability Theory 11 Exercise Problem 5.40 (Page 210) BUS304 – Chapter 5 Normal Probability Theory 12 Use the normal table to compute the probability for any range: Concept 1: z score The z score of x is computed based on 1. the value of x, 2. the mean of the normal distribution , and 3. the standard deviation of the normal distribution . x μ Formula: z σ Z score represents how many standard deviation x is from the mean. • E.g. if x= , z =0. no deviation. if x = + , z = 1. one standard deviation above. if x = - , z = -1. one standard deviation below. BUS304 – Chapter 5 Normal Probability Theory 13 Use the table to compute the probability Standard normal table: (Page 595) Use the z score to figure out the probability. The z-score has to be positive. The table shows the probability between the mean to the value. f(x) The area is the probability P(<x<a). a x BUS304 – Chapter 5 Normal Probability Theory 14 How to use the table? Steps: 1. 2. 3. 4. 5. f(x) Calculate the z-score. (z = (4-3)/1.5=0.667) Round the z-score to two decimals (z =0.67) Find the integer and first decimal part from the row Find the 2nd decimal from the column Find the corresponding value the probability. =1.5 3 4 z … 0.06 0.07 0.08 … … … … … … … 0.5 … 0.2123 0.2157 0.2190 … 0.6 … 0.2454 0.2486 0.2517 … 0.7 … 0.2764 0.2794 0.2823 … x BUS304 – Chapter 5 Normal Probability Theory … … … … … 15 … Other cases: drawing helps! Case 1: 3<x<4 f(x) Case 3: x<4 P=0.2486 f(x) 4 3 x Case 4: x>2 f(x) 4 4 P=0.5+0.2486=0.7486 Case 2: 2<x<3 f(x) f(x) x x 3 Case 5: x>4 P=0.5-0.2486=0.2514 Case 6: x<2 f(x) P=0.2486 x 2 3 x x 2 2 3 P=0.5+0.2486=0.7486 3 P=0.5-0.2486=0.2514 BUS304 – Chapter 5 Normal Probability Theory 16 Advanced Cases f(x) f(x) f(x) x x 3 4 5 3 4 5 x 3 4 5 f(x) ? x 1 2 3 BUS304 – Chapter 5 Normal Probability Theory 17 More advanced case f(x) x 2 5 Use the normal table, you should be able to figure out any probability! BUS304 – Chapter 5 Normal Probability Theory 18 Exercise. Page 210 Problem 5.45 Problem 5.52 BUS304 – Chapter 5 Normal Probability Theory 19