Central Limit Theorem

In this class, we will be dealing with samples from a population. If the sample items are independent of one another and conditions remain the same while the sampling is being conducted then the distribution of the sample sum and sample average will be mound shaped as in our simple spinner example above. We state this next, with a little bit of notation:

Central Limit Theorem: Let X₁, X₂,..., X_n be a random sample from a population with mean $\mu$ and standard deviation $\sigma$ . Let $\bar{X}$ be the sample average of X₁, X₂,..., X_n. Then the distribution of $\bar{X}$ is approximately normal with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$ .

**Figure 6.7:** N(70, 16)
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Based on the empirical rule, 95% of adult males will have heights in the interval (62,78).

Next suppose we take a sample of 16 adult males. By the Central Limit Theorem $\bar{X}_16$ will be approximately normal with mean 70 and standard deviation $4/\sqrt{16}=1$ . A picture of this distribution is found in Figure 6.8.

**Figure 6.8:** N(70,1)
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Notice that this distribution is less variable than the original population. Based on the empirical rule, 95% of the time the average height of 16 adult males will fall in the interval (68,72).

Next suppose we take a sample of 64 adult males. By the Central Limit Theorem $\bar{X}_{64}$ will be approximately normal with mean 70 and standard deviation $4/\sqrt{64}=.5$ . A picture of this distribution is found in Figure 6.9.

**Figure 6.9:** N(70,.25)
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Notice that this distribution is less variable than the original population and the distribution of the average height of a sample of 16 adult males. Based on the empirical rule, 95% of the time the average height of 64 adult males will fall in the interval (69,71).

As the Central Limit Theorem dictates, as the sample gets large the distribution of the sample average becomes less and less variable (the noise is being cut by the $\sqrt{n}$ , i.e. the standard deviation of the sample average is $\sigma/\sqrt{n}$ .) Hence the sample average is getting closer to the population mean $\mu$ .

We will use the sample average to estimate $\mu$ . What's that you say? Speak louder. It's not the estimate but how much it misses by! Hey, you are right again. How much did it miss by? Hey, that's the topic of the next chapter. For now, lets solve a few interesting problems with the CLT.

Elevator Problem . Sixteen adult males approach an elevator on the 100th floor of Everett Tower. The elevator has the sign:

Looks hard but we can do it with the help of the CLT and Doctor Population. From Doctor Population we need the average weight and standard deviation of an adult male. There is plenty of information on this. So Doctor Population consults his blue book and tells us that the mean weight of an adult male is 170 pounds and the standard deviation is 15 pounds. This is all we need.

Since we have expressed the CLT for sample averages, first express the problem in terms of the sample average; i.e., find the probability that the sample average of 16 adult males exceeds 2900/16 = 181.25. By the CLT, the distribution of the sample average is approximately normal with mean 170 and standard deviation $15/\sqrt{16} = 3.75$ . The probability that we want is the shaded area in Figure 6.10.

**Figure:** $P(\bar{X} > 181.25)$
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The z-score is (181.25 - 170)/3.75 = 3. Using the probability module, the probability that the the sample average of 16 adult males exceeds 2900/16 = 181.25 is 1 - .9986 = .0014. Hence only once out of a 1000 times will the cable snap if 16 males enter the elevator at one time.

Note, we made certain assumptions to solve this problem. The 16 males should be independent of one another, no kin, friends, etc. They must be a random sample form the general population, no football team, etc.

As a final note. It is not odd for the weight of one adult male to exceed 181.25 pounds, ( z-score is $z = \frac{181.25 - 170}{15} = .75$ ); hence (assuming an approximate normal distribution), the probability that the weight of one adult male exceeds 181.25 pounds is 1 - .7733 = .2367. This happens 24% of the time. But it is odd that the average (based on a random sample) weight of 16 adult males exceeds 181.25 pounds.

Exercise 7.2.1

1.: The scores on a general test have mean 450 and standard deviation 50. It is highly desirable to score over 480 on this exam. A person can get into Smith's College prestigious MBA program if he/she scores over 480. In one location 25 people sign up to take the exam. The average score of these 25 people exceeds 490. Is this odd? Should the test center investigate? Answer on the basis of the CLT.
2.: A machine fills cereal boxes at a factory. Due to an accumulation of small errors (different flakes sizes, etc.) it is thought that the amount of cereal in a box is normally distributed with mean 22 oz. for a supposedly 20 oz. box. Suppose the standard deviation of the amount filled is 1.3 oz. A federal regulatory selects four of these boxes at random and finds that the average content of these boxes is less than 18 oz. This official knows that the company claims the mean content to be 22 oz. He promptly fines the company. Who is right? Use the CLT in your answer.
3.: Sixteen adult males are in a pit which is 98 feet deep. They decide to stand on one another (feet to head), hoping that the person on top can grip the top of the pit and get out, and, hence go for help. What's the probability that their plan succeeds? (Ans: .0003).