Unit 13 Normal Distribution Reference: (Material source and pages) 2018/6/8 Unit 13 Normal Distribution IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP 13.1 Normal distribution A random variable X is said to have a Normal distribution, with parameters  and 2 if it can take any real value and has p.d.f. In this case, we write X ~ N(, 2). It can be shown that  is the expected value of X and 2 is the variance. IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP P(a < X < b) = = This integral cannot be done algebraically and its value has to be found by numerical methods of integration. The value of P(a < X < b) can also be viewed as the area under the curve of f(x) from x = a to x = b. IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

13.2 Characteristics of Normal Distribution 1. Bell shape 2. Symmetric 3. Mean = Mode = Median 4. The 2 tails extend indefinitely IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

13.3 Standard Normal Distribution A standard normal distribution is the normal distribution with  = 0 and 2 = 1, i.e. the N(0,1) distribution. A standard normal random variable is often denoted by Z. If Z ~ N(0,1), its c.d.f. is usually written as (z) = P(Z  z) = IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Note: (1) (z) may be interpreted as the area to the left of z under the standard normal curve. (2) P(Z < -z) = P(Z > z), since the standard normal curve is symmetrical about the line Z = 0. (3) The area between Z= -1 and +1 is 68% Z= -2 and +2 is 95% Z= -3 and +3 is 99% IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.3-1 (1/2) Given Z ~ N(0,1), find the following probabilities using the standard normal table. (a) p( Z  1.25) (b) P( Z > 2.33) (c) P(0.5 < Z < 1.5) (d) P( Z < -1.25) (e) p (-1.5< Z <-0.5) IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.3-1 (2/2) (a) P(Z  1.25) = 0.8944 P(Z > 2.33) = 1 - (2.33) = 1 – 0.9901 = 0.0099 (c) P(0.5 < Z < 1.5) = (1.5) - (0.5) = 0.9332 – 0.6915=0.2417 (d) P( Z < -1.25) = P(Z > 1.25) = 1 - (1.25) = 1 – 0.8944 = 0.1056 (e) P(-1.5 < Z < -0.5) = 0.2417 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.3-2 (1/4) Given Z ~ N(0,1), find the value of c if (a) P(Z  c) = 0.8888 (b) P( Z > c) = 0.37 (c) P(Z < c) = 0.025 (d) P(0  Z  c) = 0.4924 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.3-2 (2/4) (a) (c) = 0.8888 c = 1.22 (b) 1 - (c) = 0.37 (c) = 0.63 c = 0.332 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

Example 13.3-2 (3/4) (c) Obviously c is negative and the standard normal table cannot be used directly. Recall that P(Z < -z) = P(Z > z) 0.025 = P(Z < c) = P(Z > -c) = 1 –(-c) (-c) = 0.975 -c = 1.96 c = -1.96 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.3-2 (4/4) (d) P(0  Z  c) = (c) – 0.5 = 0.4924 (c) = 0.9924 c = 2.43 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

13.4 Standardization of normal random variables If X ~ N(, 2), then it can be shown that Z = ~ N(0,1) . In other words is the standard normal random variable. The process of transforming X to Z is called standardization of the random variable X. IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-1 (1/4) Suppose X ~ N(10, 6.25). Find the following probabilities (a) P(X < 13) (b) P(X > 5) (c) P(8 < X < 15) IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-1 (2/4) Let Z = . (a) P(X < 13) = P(Z < ) = P(Z < 1.2) = 0.8849 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-1 (3/4) (b) P(X > 5) = P(Z > ) = P(Z > -2) = P(Z < 2) = 0.9773 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-1 (4/4) (c) P(8 < X < 15) = P( ) = P(-0.8 < Z < 2) = (2) - (-0.8) = (2) – (1 - (0.8)) = 0.9772 – (1 – 0.7881) = 0.7653 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-2 (1/2) The lifetime of light bulbs produced in a certain factory is normally distributed with mean 3000 hours and standard deviation 50 hours. Find the probability that a randomly chosen light bulb has a lifetime less than 2900 hours. IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-2 (2/2) Let X be the lifetime of a randomly chosen light bulb. Then X ~ N(3000, 502). Let Z = . P(X < 2900) = P(Z < -2) = 1 – 0.9772 = 0.0228 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-3 (1/3) If a random variable has the normal distribution with  = 40 and  = 2.4, find the probabilities that it will take a value (a) less than 43.6 (b) greater than 38.2 (c) between 40.6 and 43.0 (d) between 35.8 and 44.2. IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-3 (2/3) Let X be the random variable. (a) P(X < 43.6) = P(Z < ) = P(Z < 1.5) = 0.9332 (b) P(X > 38.2) = P(Z > ) = P(Z > -0.75) = P(Z < 0.75) = 0.7734 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP

ITD1111 Discrete Mathematics & Statistics STDTLP Example 13.4-3 (3/3) (c) P(40.6 < X < 43.0) = P( ) = P(0.25 < Z < 1.25) = 0.2957 (d) P(35.8 < X < 44.2) = P( ) = P(-1.75 < Z < 1.75) = 0.9198 IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP