# The Central Limit Theorem Part 1

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1 The Central Limit Theorem Part. Introduction: Let s pose the following question. Imagine you were to flip 400 coins. To each coin flip assign if the outcome is heads and 0 if the outcome is tails. Question: If we average all of those values (all 400 of those 0s and s) together, what s the probability that the result will be 0.57 or more? 2. Building an Understanding We could approach the question by starting simple and looking at one coin. We assign X to be the average number of heads for the one flip, but of course for one coin this is just the number of heads. The flip results are T and H, corresponding to 0 and respectively, so we have distribution as follows. Remember this is the distribution for the average number of heads. which corresponds to the histogram x 0 p(x) /2 /2 Now let s do two coins. We assign X 2 to be the average number of heads. The flip results are TT, TH, HT and HH, corresponding to 0, /2, /2 and respectively (remember, average the number of heads per pair of rolls), so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /2 p(x) /4 /2 /4

2 Now let s do three coins. We assign X 3 to be the average number of heads. The flip results are TTT, TTH, THT, THH, HTT, HTH, HHT and HHH, corresponding to 0, /3, /3, 2/3, /3, 2/3, 2/3 and respectively, so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /3 2/3 p(x) /8 3/4 3/8 /8 Almost last but not least, before we jump to our result, let s do four coins. We assign X 4 to be the average number of heads. The flip results are TTTT, TTTH, TTHT, TTHH, THTT, THTH, THHT, THHH, HTTT, HTTH, HTHT, HTHH, HHTT, HHTH, HHHT and HHHH, corresponding to 0, /4, /4, 2/4, /4, 2/4, 2/4, 3/4, /4, 2/4, 2/4, 3/4, 2/4, 3/4, 3/4 and respectively. so we have distribution as follows. Remember this is the distribution for the average number of heads. x 0 /4 2/4 3/4 4/4 p(x) /6 4/6 6/6 4/6 /6 What do you see? The probability distribution is becoming a normal distribution! The more coins we flip the closer it becomes! Just to convince you here s the eight coin histogram. We ve scaled it up vertically so you can see what s going on - the highest rectangle is actually only 70/256 = high.

3 3. The Central Limit Theorem: The CLT states that our observation is accurate. More rigorously it states that if we have an event with random variable X (think a single coin toss, counting heads) and if we repeat it n times and assign X n to be the average of the results (think n coin tosses with X n being the average number of heads) then as n gets larger the distribution approaches a normal distribution. What s fascinating about this is that it s true no matter what the probability distribution function for the single variable is. It could be coin tosses, dice rolls, exponential distributions, whatever. In the long term the average outcome approaches the normal distribution. This is useful for us because it says that if we are interested in the average outcome of a very large number of flips (or any random variable at all) we can get an approximation by looking at the corresponding normal distribution rather than by actually flipping lots of coins. For large n the probability distribution for the average outcome is approximately a normal distribution. Since a normal distribution is determined completely by the mean µ and standard deviation σ we just need to determine what those would be for the normal distribution corresponding to our very large n. The CLT gives us the result as: The corresponding normal distribution has µ = E(X). The corresponding normal distribution has σ = σ(x)/ n. Note: You shouldn t look at this and say It s obvious! because it s not. The CLT is not easy to prove - we are unable to prove it with the mathematics we currently know.

4 4. Application: Some examples: Example : We flip 400 coins, assigning for head and 0 for tail each time. What s the probability that the average outcome will be greater than or equal to 0.57? If X is the outcome for a single coin and X 400 is the average outcome of 400 coins then for X we have E(X) = /2() + /2(0) = /2 σ(x) = V ar(x) = /2( /2) 2 + /2(0 /2) 2 = /2 The CLT says that the corresponding approximating normal distribution has µ = /2 σ = (/2)/ 400 = In friendly terms 400-Coin Flip Average Outcome Distribution Normal Distribution with µ = /2 and σ = So our desired value is P(0.57 X 400 < ) but by the CLT we can this approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 0.57 = z = ( )/0.025 = 2.8 and so P(0.57 X 400 < ) P(2.8 Z < ) = = 0.26% This means that the probability of getting an average outcome of over 0.57 from flipping 400 coins is approximately 0.26%.

5 Example 2: Suppose the random variable X corresponds to rolling a die with winnings and losings given as follows: Rolling a wins \$2, rolling a 2 or 3 wins \$7, rolling a 4 wins nothing and rolling a 5 or 6 loses \$6. Suppose we roll 2000 dice and average our winnings from each roll. What s the probability that we ll win under \$0.75? Here X is the random variable that gives the average of one roll. Of course the average for just one roll is the same as the winnings for that one roll and so E(X) = 2(/6) + 7(2/6) + 0(/6) 6(2/6) = 4/6 = 2/3 σ(x) = (2 2/3) 2 (/6) + (7 2/3) 2 (2/6) + (0 2/3) 2 (/6) + ( 6 2/3) 2 (2/6) = 257/9 Let X 2000 be the random variable corresponding to the average earnings for 2000 rolls. The CLT says that the corresponding approximating normal distribution has µ = 2/3 σ = 257/9/ 2000 = 257/8000 In friendly terms 2000-Dice Roll Average Earnings Distribution Normal Distribution with µ = 2/3 and σ = 257/8000 Our desired value is P( < X ) but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 0.75 = z = (0.75 2/3)/ 257/ and so P( < X ) Z( < Z < 0.70) = 75.80% Note that this makes sense. If the average winnings per roll is \$2/3 \$0.66 then after a lot of rolls we ought to be pretty close to that. Having under \$0.75 on average is pretty likely!

6 Example 3: The lifespan of a bacterium is randomly distributed with mean 5 days and standard deviation 2 days. Suppose you have 60 bacteria in your lab. What s the probability that the average lifespan will be over 4.8 days? Let X be the random variable corresponding to the lifespan of an individual bacterium and let X 60 correspond to the average of the lifespans of the 60 bacteria. Since X is already given to be distributed with E(X) = 5 σ(x) = 2 the CLT says that the corresponding approximating normal distribution has µ = 5 σ = 2/ 60 In friendly terms 60-Bacteria Average Lifespan Distribution Normal Distribution with µ = 5 and σ = 2/ 60 Our desired value is P(4.8 X 2000 ) but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = 4.8 = z = (4.8 5)/(2/ 60).26 and so P(4.8 X 60 < ) P(.26 Z < ) = = 89.62% Note that this seems reasonable. If the expected lifespan of a single organism is 5 days and we have a bunch of them then very probably the average lifespan of all of them will be over 4.8 days.

7 Example 4: The lifespan of a bacterium is distributed with mean 5 days and standard deviation 2 days. Suppose you have 60 bacteria in your lab. Over what age can you expect 70% of the bacteria to live? We ve reversed the question here, giving the desired percentage and asking for the age. Initially nothing changes from the previous problem. Let X be the random variable corresponding to the lifespan of an individual bacterium and let X 60 correspond to the average of the lifespans of the 60 bacteria. Since X is already given to be distributed with E(X) = 5 σ(x) = 2 the CLT says that the corresponding approximating normal distribution has µ = 5 σ = 2/ 60 Now things change. We want to know M such that P(M X 000 < ) = 0.7 but by the CLT we can approximate with the normal distribution. We find the corresponding z-scores x = = z = x = M = z = (M 5)/(2/ 60) and so P(M X 000 < ) = P ( (M 5)/(2 ) 60) Z < We wish this probability to be as close to 0.7 as possible. Since this probability is the area to the right of (M 5)/(2 60) we want the area to the left to be as close to 0.3 as possible. We reverse-lookup on the table and get the closest value Thus (M 5)/(2/ 60) = 0.52 M 5 =.04/ 60 M = 5.04/ 60 M 4.92 and so 70% of the bacteria will live over 4.92 days. We should note that this is important if we re collecting bacteria to study and we want to collect a sufficient quantity so that enough will survive for our study.

8 Problems Approximate all final answers to four places beyond the decimal point. Any intermediate value which is awkward (annoying square roots for example) may be approximated to four places beyond the decimal point. Any value which needs to be looked up in the standard normal table should be approximated as necessary first. Use when appropriate. Any value which needs to be reverse-looked up in the standard normal table should be reverse-looked up to the closest value.. Suppose you flip 2500 coins, assigning to heads and 0 to tails. (a) Calculate E(X) and σ(x) for a single flip. (c) What is the probability of the average outcome being between 0.49 and 0.52? (d) What is the probability of the average outcome being over 0.525? 2. Suppose you flip 000 coins. Each head wins \$2 and each tail loses \$.50. (a) Calculate E(X) and σ(x) for a single flip. (c) What is the probability of your average earnings being \$0.27 or more? What would your total earnings be if this were the case? (d) What is the probability of your average earnings being a dime or less? 3. Suppose you roll 3600 dice. Rolling a,2 or 3 wins \$3. Rolling a 4 or 5 loses \$4. Rolling a 6 loses \$2. (a) Calculate E(X) and σ(x) for a single roll. (c) What is the probability of your average earnings being positive? (d) Your average earnings have 80% probability of being below what value? 4. A honeybee drone has an expected lifespan which is exponentially distributed with mean 45 days. You collect one thousand honeybee drones. (a) What is the standard deviation? (b) If you pick a random bee, what is the probability it will live over 47 days? Note: This has nothing to do with the CLT! (c) Calculate µ and σ for the corresponding approximating normal distribution. (d) What is the probability that the average lifespan of all the bees in your collection will be over 47 days? (e) Under what average age will 0% of your honeybees live?

9 5. A certain drug is to be rated either effective or ineffective. Suppose lab results indicate the drug is effective 75% of the time and ineffective 25% of the time. An effective drug increases the lifespan of a patient by 5 years while an ineffective drug causes a complication which decreases the lifespan of a patient by year. As part of a study you administer the drug to 0000 patients. (a) Calculate E(X) and σ(x) for a single patient. (c) What is the probablity that the average lifespan increase will be 3.55 or more years? (d) What gain in lifespan will 25% of your patients experience? 6. A certain event is a random variable with PDF given by f(x) = x for x e. (a) Find E(X) and σ(x) for this event. (c) If this event occurs 900 times what is the probability of the average outcome being below.72? (d) There is a 90% probability that the average outcome is above which value?

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