* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download The Central Limit Theorem (CLT)
Survey
Document related concepts
Transcript
The Central Limit Theorem Consider the following probability distribution . The random variable X is the length of stay in the hospital for a randomly selected patient undergoing a certain medical procedure. X: p(x): 2 .1 4 .2 5 .4 6 .3 The mean and standard deviation of this distribution are = 4.8 and = 1.166. Knowing the appropriate model for a population is like knowing the entire population itself. In a real life situation we would not know which model would be the appropriate one for the population of interest, and we certainly would not know the actual value of . The problem then is to ESTIMATE in some way. One way to do this is to take a simple random sample from the population and then use it to calculate a value of the sample mean X . If we proceeded in this way, we would become aware of the fact that each time we took a new sample from the population and calculated X we would get a different value for our estimate of . Thus it is important to try to understand a bit more about the behaviour of X. Mean, Variance and Standard Deviation of the Sample Mean X Given: 1. A population with mean and standard deviation . 2. A random sample of size n from the population: X1 , X2 , … Xn with sample mean X = (X1 +X2 +… Xn)/n Then: = [ the mean of the distribution of X is equal to the population mean ] X = / n [ the sd of the distribution of X is equal to ( the population sd)/n] X 2 X = 2 /n [ the variance of the distribution of X is equal to (the population variance)/n ] (83) The Central Limit Theorem (CLT) is stated as follows: Given a large random sample of size n from a population with mean and standard deviation , then The sample mean X is approximately normally distributed with mean and standard deviation given by X = = / n X Note: 1. n >30 is usually large enough for the CLT to apply. 2. If the population from which we sample is normal thenX is exactly normally distributed with mean and standard deviation as above for any sample size. Example: The population of white males aged 35-44 in Canada has a mean systolic blood pressure of 130 with a standard deviation of 8. Find the probability that a random sample of 50 white males has a mean systolic blood pressure (a) below 133, (b) above 128, (c) within 2 of the population mean Solution: Here we are asked to calculate probabilities about the sample mean X. This suggests we use the CLT. Population: Systolic blood pressures of white Canadian males aged 35-44. Population Mean = 130; Population Stdev = 8. Since n =50 > 30, the CLT tells us that X is approximately normally distributed with = = 130 X (a) P ( X < 133 ) = (84) and = / n X (b) P( X > 128) = (c) Notice that “ X within 2 of the population mean “ means that X lies between 128 and 132. Thus, P ( 128 < X < 132 ) = (85) Example: The scores on a Standardized Reading Test are known to be normally distributed with a mean of = 10 and standard deviation of = 1.5. (a) Find the probability that a single individual selected at random scores between 9.4 and 10.6 on this test. (b) Find the probability that a random sample of 25 students has a mean score between 9.4 and 10.6 on this test. (c) Find the probability that a random sample of 64 students has a mean score between 9.4 and 10.6 on this test. Solution: Let X be the score of the randomly selected individual. Then X is normal = 10, = 1.5. (a) P( 9.4 < X < 10.6 ) = (86) (b) Since the population the sample came from is normal, then for any n ( hence for n =25), X is exactly normal with = 10 and = 1 .5/5 =.3 X X P( 9.4 < X < 10.6 ) = (c) For the same reason as in (b) above, for n =64, X is exactly normal , = = 10 X and = / n = 1.5/8 = 0.1875. X Therefore, P( 9.4 < X < 10.6 ) = (87)