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### 45-733 PROBABILITY AND STATISTICS I Notes #6

February 2000

#### Central Limit Theorem

1. Central Limit Theorem:
If X1 + X2 + ... + Xn is a random sample from a distribution with mean m and finite variance, s2, then for any constant k:
```                 _
lim n ® +¥ P[(Xn - m)/s/n1/2 £ k] = F(k)
```
2. Note that this theorem holds for all distributions, continuous and discrete.
3. A useful alternative way of writing the formula is:
P{[(åi=1,n Xi) - nm]/ sn1/2 £ k} » F(k)
4. For finite sample size this is an approximation. If the distribution from which the random sample is drawn is unimodal and symmetric, then the sample size can be "small" and the approximation will be fairly good. If the distribution is skewed, then the sample size must be "larger" for the approximation to be fairly good. There are no hard and fast rules.
5. Example: Sampling from a Bernoulli distribution.
Suppose we flip a coin 900 times. What is the probability that the number of heads will be greater than 495?
This is taking a random sample of size 900 from a Bernoulli distribution
with parameter p = 1/2. Recall that:
```

æ  p  x = 1
ç
f(x) = ç 1-p x = 0
ç
è  0 otherwise
```

Where E(X) = p = m    and     VAR(X) = p(1 - p) = s2
In this case m = p = 1/2    and     s2 = p(1 - p) = 1/4
Hence: P[(åi=1,900 Xi) > 495] =
P{[(åi=1,900 Xi) - 900*1/2]/(1/2)*30 > (495 - 450)/15} = P(Z > 3) = 1 - F(3)
= .00135
6. Problem 7.27 p.308
Let Xi = "Life of Lamp i". We are given m = 50 hours, and s = 4 hours. We are to calculate:
P(System will work longer than 1300 Hours) = P[(åi=1,25 Xi) > 1300] =
P{[(åi=1,25 Xi) - 25*50]/(4*5) > (1300 - 1250)/20} =
P(Z > 2.5) = 1 - F(2.5)
= .0062
So, 62 out of every 10,000 times you came back from a trip to Paris a lamp would still be burning in the greenhouse!

7. Problem 7.32 p.308
Let Xi = "Time Person i Waits to go Through the Checkout Once he/she has reached it". We are given m = 2.5 minutes, and s = 2 minutes. We are to calculate:
P(It will take more than 4 hours to serve a 100 people) = P[(åi=1,100 Xi) > 240] =
P{[(åi=1,100 Xi) - 100*2.5]/(2*10) > (240 - 250)/20} = P(Z > -1/2) = 1 - F(-1/2)
=
1 - .3085 = .6915

Suppose we have 600 graduating seniors. Based upon records we know that, on average, 1/3 of seniors have 2 parents at their graduation, 1/3 of seniors have 1 parent at their graduation, and 1/3 of seniors have no parents at their graduation. What is the probability that less than 650 parents show up at the graduation?
Let Xi = "Number of Parents for the ith Senior". To get m and s we have to get the distribution of Xi. Clearly f(xi) is a discrete uniform distribution. That is:
```

æ 1/3  x = 0, 1, 2
f(x) = ç
è 0 otherwise
```

So that E(Xi) = åi=1,3 xif(xi) =
0*1/3 + 1*1/3 + 2*1/3 = 1
and E[(Xi)2] = åi=1,3 (xi)2f(xi) =
02*1/3 + 12*1/3 + 22*1/3 = 5/3
and VAR(Xi) = 5/3 - 1 = 2/3
Therefore, m = 1 parent and s2 = 2/3 parent.
P[(åi=1,600 Xi) < 650] = P{[(åi=1,600 Xi) - 600*1]/[(600)1/2(2/3)1/2] < (650 - 600)/20} =
P(Z < 2.5) = F(2.5) = 1 - F(-2.5)
= 1 - .0062 = .9938