Cetral Limit Theorem the Meaig ad the Usage Covetio about otatio. N, We are usig otatio X is variable with mea ad stadard deviatio. i lieu of sayig that X is a ormal radom Assume a sample of measuremets (or couts) that are distributed with the same probability distributio that has mea ad stadard deviatio. *** Example: Take a sample of 30 GPAs of high school seiors. Now look at the average x of your sample. It has a particular value. 3.3 3.11.35 1.39 3.80 3.1.35.76 3.43 1.91.7.3.10 3.4 3.55.34 3.78 1.66 3.56 3.45.87 3.10 3.07 3.8 1.64 3.01 3.45 3.10.4 1.98 Average (sample mea) x.79. Take aother sample of equal size ad average of that oe has aother value. 3.9.1.78.19 3.50 3.30.35.75 3.3.8..3.10 3.4 3.45.34.35.45.56 3.45.87 3.40 3.43.8 1.89 3.11.55 3.3.55 1.67 Average x.73. Ad so o. The sample average x itself is a radom variable with a probability distributio. We shall hoor that fact by writig X. What kid of probability distributio? Fact A. If the origial measuremet X is (approximately) ormally distributed the the sample mea X is (approximately) ormally distributed with the mea ad the stadard deviatio ( i.e., variace ), if the sample size is.
But eve if the origial measuremet is ot ormally distributed we still have: Fact B. Cetral Limit Theorem: As the sample size () icreases the closer to the ormal radom variable the sample mea X is, regardless of the ature of the iitial radom variable. This ormal is with mea ad the stadard deviatio ( i.e., variace ). The rule of thumb we use for the safe applicatio of CLT is whe the sample size is at least 30 or more. This allows us to use ormal distributio to assess probabilities of sample mea beig i some particular rage. Note: The umber Example 1. is ofte called stadard error of the mea. The height (X) of 18-year-old me is approximately ormally distributed with the mea X N 70,3. 70 iches ad stadard deviatio 3 iches, i.e., (a) What is the chace that a 18-year old ma selected at radom is betwee 67 ad 73 iches tall? Aswer: P X 67 73 0.683 by usig empirical rule. (b) If we select the sample of ie 18-year-olds what is the chace that the mea height X of the sample is betwee 67 ad 73 iches? 3 9 Aswer: Usig the Fact A above we have that X is N 70, N70,1 P 67 X 73 0.997 by empirical rule agai., therefore Example. The white blood cell cout per cubic millimeter of blood is approximately ormal with mea 7500 ad stadard deviatio 1750. Doctor uses two-test average X. What is the chace that X 4500?
Aswer: Usig Fact A above we have that X is approximately ormal 1750 N 7500, thus 4500 7500 P X 4500 PZ PZ.4 0.0078. 1750 This is very ulikely, i less tha 1% of the tests i geeral populatio. Example 3. Average Math SAT i the US for college boud seiors i 004 was 518 with stadard deviatio of 118. We assume approximately ormal distributio of the scores. Professor Witt advises 4 freshme. Let X be their average Math SAT. What is the chace that this is (a) better tha 600? (b) less tha 400? (c) better tha 500 but less tha 600? 118 4 (a) 600 518 600 518 P X 600 PZ P Z PZ 1.39 0.083 118. 59 4 Aswers: First ote that X is approximately ormal, N 518, N518,59 (b) 400 518 P X 400 P Z P Z 0.08. 59 (c) 500 518 600 518 P 500 X 600 P Z P 0.31 Z 1.39 0.5394. 59 59.
Example 4. Suppose the populatio of adult males had a mea weight of 18 pouds with a stadard deviatio of 3 pouds. The umber of males treated for heart related problems i local healthcare facility was 53 i 011. Their average weight was 191. Would this lead to coclusio that males with heart related problems have more weight? Aswer: 3 By the Fact B from CLT we have that X is approximately ormal N 18, 53 If we assume that the mea weight of 18 applies to this group of idividuals the the chace that the weight would be above 191 is 19118 P X 191 PZ PZ.8487 0.00. 3 53 This idicates that it is very ulikely that mea weight of 18 ad stadard deviatio of 53 is applicable o this group. It is far more likely that the me with heart related problems are heavier tha 18 pouds o average. Example 5. The mea lifetime of a certai type of icadescet light bulb is 900 hours with a stadard deviatio of 900 hours. We have a batch of 100 of these light bulbs. What is the chace these light bulbs are goig last more tha 100,000 hours? Aswer: Let s deote the lifetime of a sigle light bulb with X. The questio is posed as where P X1 X X100 100000 X i meas the lifetime of i-th light bulb. We ca write this questio as X1 X X100 P 1000 P X 1000 100. Now we ca use CLT because X is approximately with mea 900 ad stadard deviatio 90. 900 N 900,. That meas ormal 100
1000 900 P X 1000 P Z PZ 1.11 1 0.8665 0.1335 90 The batch of bulbs has about 13% chace to last more tha 100,000 hours. Example 6. A shippig compay charges extra if the average shippig load durig oe week period exceeds the compay limit of 0 tos per cotaier. Meat packig compay ships o average 18.5 tos per cotaier with a stadard deviatio of 7.4 tos. If they shipped 68 cotaiers this week, what is the chace the average load was above 0 tos? Aswer: The shippig load per cotaier is a radom variable X we do ot kow the ature of, apart from mea ad stadard deviatio. Therefore to aswer the questio, what is the probability that X, the average of 68 shippig loads exceeds 0 tos we use CLT. 7.4 Therefore X is approximately ormal, N 18.5,. 68 0 18.5 P X 0 PZ PZ 1.67 0.0475 7.4 68 The aswer is approximately 5%. Homework: Check olie.
Biomial Distributio by Normal Approximatio (Optioal) Remider: X = B(, p) the cout of successes for biomial distributio with the umber of idepedet idetical trials ad probability of success o oe trial p. Every trial ca ed with oly two possible outcome success ad failure. Probability of failure is q = 1 p. The probability mass formula is P X r p q r r r Accordig to Cetral Limit Theorem, biomial distributio i case of a large ca be approximated with a ormal distributio with the same mea (p) ad the same stadard deviatio ( pq ). The implicatio is that respective probabilities for a biomial radom variable (cout of successes) ca be approximated by usig z values of the variable i the charts for stadard ormal distributio. There are two requiremets the approximatio eeds to meet: p 5, q 5. Example 1. 1 out of 4 wome that smoke will die from smokig related diseases. What is the probability that out of 1,000 wome smokers more tha 50 will die from smokig related diseases? What is the probability that the death from smokig related disease will be betwee 40 ad 60 cases ( 40 X 60 )? Aswer: X, the cout of deaths is 1000,0.5 sice we have p = 50, q = 750. B. This ca be approximated by N 50,13.7 40 50 60 50 P X P Z P Z 13.7 13.7 40 60 0.73 0.73 0.5346. Example. O average % of batteries produced by a car battery maufacturer are defective. What is the chace that i a stock of 10,000 batteries there are o more tha 180 defective?
Aswer: X, the cout of defective batteries i a stock of 10,000 is B 10000,0.0 approximated by N 00,196. Obviously p = 00 > 5, q = 9800 > 5. 180 00 P X 180 PZ PZ 1.43 0.0764. 14. This ca be Example 3 (Small rages or specific value of the variable). O average oe out of 6 smokers is goig to die from the lug cacer. If there are 10 smokers i my eighborhood, what is the chace there will be 0 of them that will die of lug cacer? Aswer: X, the cout of rouds wo is B 10,1/ 6. This ca be approximated by 0, 16.67 19.5 0 0.5 0 P X P X P Z P Z 4.08 4.08 N. 0 19.5 0.5 0.1 0.1 0.0956 Oe ca calculate this probability directly from formula for the mass of B 100,1/ 38, 0 100 10 1 5 P X 3 0.0973 0. 6 6 Obviously the error of ormal approximatio to biomial was less tha 0.%. Homework: Check olie.