Example 1. Assume that X follows the normal distribution N(2, 2 2 ). Estimate the probabilities: (a) P (X 3); (b) P (X 1); (c) P (1 X 3).

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1 Example 1. Assume that X follows the normal distribution N(2, 2 2 ). Estimate the probabilities: (a) P (X 3); (b) P (X 1); (c) P (1 X 3). First of all, we note that µ = 2 and σ = 2. (a) Since X 3 is equivalent to Z = X = 0.5, we find from the table that P (X 3) = P (Z 0.5) = Φ(0.5) (b) Since X 1 is equivalent to Z = X = 0.5, we find from the table that P (X 1) = P (Z 0.5) = P (Z 0.5) = Φ(0.5) (c) It follows from that P (X 1) = 1 P (X 1) = P (1 X 3) = P (X 3) P (X 1) = Exercise 1. Assume that X follows the normal distribution N(1, 9). Estimate (a) P (X 1.4); (b) P (X 1.22); (c) P ( 1.22 X 1.4). 1

2 Example 2. Estimate k such that P (Z k) = 0.1, where Z follows the standard normal distribution. Note that for such k, we have From the table we find Φ(k) = P (Z k) = = 0.9. Φ(1.28) and Φ(1.29) , which means that k can be approximated by either 1.28 or Knowing that k (1.28, 1.29), we can have a better approximation of k using the following linear interpolation: = k = 0.18, which implies that k Exercise 2. Estimate k such that P ( k Z k) = 0.97 where Z is the standard normal random variable. 2

3 12.1 Some Special Probability For any Z following N(0, 1), we have P ( 1 Z 1) 0.68 and and P ( 2 Z 2) 0.95 P ( 3 Z 3) So, for any X following N(µ, σ 2 ), we have P (µ σ X µ + σ) 0.68 and and P (µ 2σ X µ + 2σ) 0.95 P (µ 3σ X µ + 3σ)

4 13 Applications of the Normal Distribution Example 3. Suppose the salary of a group of 1000 civil servants follows the normal distribution N(10000, ). (a) Estimate the number of civil servants having salary less than (b) Estimate the lowest salary of the top 200 civil servants. First of all, we note that µ = and σ = Let X be the salary of a civil servant, and let Z = (X 10000)/1000. (a) Then we have P (X 10500) = P ( Z So, the desired number of civil servants is ) = P (Z 0.5) }{{} from the N(0,1) table 4

5 (b) We are supposed to estimate k such that P (X k) = = or equivalently, ( P Z k ) ( ) k = Φ = Φ(K) = From the table, we find Φ(0.84) and Φ(0.85) , which means K can be approximated by either 0.84 or 0.85, and accordingly k can be approximated by either or To have a better approximation, we apply the linear interpolation to obtain K , which means K can be approximated by and accordingly k by

6 13.1 From Binomial or Poisson to Normal This subsection presents the relationship between the binomial or Poisson distributions and the normal distribution. Let us first recall the central limit theorem. Proposition 1. Let X 1, X 2,..., X n be a sequence of independent, identically distributed random variables with mean µ and variance σ 2. Then the cumulative distribution function of the following random variable tends to that of the standard normal random variable as n : Z n = X 1 + X X n nµ σ. n Proposition 2. For a large λ, P oisson(λ) can be approximated by N(λ, λ). Proof. It follows from the fact that P oisson(λ) is the sum of λ independent P oisson(1) and the central limit theorem. Proposition 3. For a large n, Binomial(n, p) can be approximated by N(np, np(1 p)). Proof. It follows from the fact that Binomial(n, p) is the sum of n independent Bernoulli(p) and the central limit theorem. 6

7 Example 4. A fair coin is tossed 100 times. We can find the probability that the number of heads obtained is between 48 and 52 by using the normal approximation: The number of heads obtained from the 100 tosses follows Binomial(100, 0.5), which, by Proposition 3, can be approximated by N(50, 25). Letting X N(50, 25) and Z = (X 50)/5, we then deduce that the area of the following five rectangles 1 are approximated by P (47.5 X 52.5), which can be computed as ( ) P Z = P ( 0.5 Z 0.5) 1 2( ) = Exercise 3. A fair coin is tossed 200 times. Find the probability that the number of heads obtained is between 98 and 102 by using the normal distribution approximation. 1 Since we are approximating a p.m.f. of a discrete random variable by a continuous one, some adjustments have been made so as to get a better approximation. 7

8 Example 5. Let Y 1, Y 2,..., Y n be n independent Poisson random variables having the same parameter 1, that is, each Y i follows P oisson(1). Then Z n = Y 1 + Y Y n, the sum of the n Poisson random variables with parameter 1 is a Poisson random variable with parameter n, namely, Z n follows P oisson(n). By the central limit theorem, for large n, (Z n n)/ n approximately follows the normal distribution N(0, 1). Then, we have P (Z n = n) = P (n 1 < Z n n) = P ( 1 < Z n n 0) n n 0 e x2 2 dx 2π 1 1 n 0 dx = 1, 2π 2πn 1 1 n 8

9 where we have used the fact that for large n, e x2 2 1, for x ( 1/ n, 0). Now, since Z n is a Poisson random variable with parameter n, we have P (Z n = n) = e n n n, n! which, together with the derived approximation P (Z n = n) 1/ 2πn, implies that n! n n+1 2e n 2π, which is Stirling s formula. So, using the connection between the Poisson distribution and normal distribution, we have given a heuristic argument for Stirling s formula. 9

10 A Summary 1. Bernoulli experiment 2. Bernoulli distribution Bernoulli(p) 3. Geometric distribution Geometric(p) 4. Binomial distribution Binomial(n, p) 5. Exponential distribution Expo(λ) 6. Poisson distribution P oisson(λ) 7. Normal distribution N(µ, σ 2 ) 8. The standard normal distribution N(0, 1) 9. Linear interpolation method 10. Normal approximates binomial 11. Normal approximates Poisson 10