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MATHEMATICS II (SHF 1124) CHAPTER 6 THE NORMAL DISTRIBUTION 6.1 : PROPERTIES OF THE NORMAL DISTRIBUTION ο§ In this chapter, the distribution of a continuous random variable will be discussed. The most widely used are the normal and standard normal distributions. ο§ Probability of a continuous random variable assumes within a certain interval is given by the area under curve within the interval itself. NORMAL DISTRIBUTION ο§ A large number of phenomena in the world can be described or approximately described by normal distribution. For example, the examination scores of students, the weight of newborn babies, the length of a type of leaves and the amount of soda in drinks. ο§ A normal probability distribution is a bell-shaped curve with the following characteristics: a) The total area under the curve is 1.0. b) The curve is symmetric about the mean π. c) The 2 tails of the curve extend indefinitely. d) The curve is described below ο§ Normal distribution is identified by 2 parameters, mean π and variance π 2 . Given the values of π and π 2 , the area under the curve can now be obtained. ο§ Each different values of π and π 2 give different normal distribution ο§ The value of mean π determines the center of the distribution whereas different values of π 2 determines how much the observation is dispersed. ο§ A normal random variable is a continuous random variable X that has a normal distribution π~π(π, π 2 ). DEPARTMENT OF MATHEMATICS Page 1 MATHEMATICS II (SHF 1124) 6.2 : STANDARD NORMAL DISTRIBUTION ο§ The normal distribution with mean π = 0 and standard deviation π = 1 is called the standard normal distribution denoted by π(0,1) described below: ο§ The Z value of Z scores are the units marked on the horizontal axis of standard normal curve. The continous random variable Z is usually used to represent the standard normal random variable, π~π(0,1) To standardized a normal random variable X to a standard normal random variable Z, the following formula is used: ο§ ο§ π= πΏβπ π where π = ππππ π = π π‘ππππππ πππ£πππ‘πππ ο§ ο§ The standard normal table can be used to find the area under the standard normal curve (probability) The table that will be used in this chapter is π·(π β€ π β€ π) DEPARTMENT OF MATHEMATICS Page 2 MATHEMATICS II (SHF 1124) EXAMPLE: If π~π(4,9) find using standard normal table: a) b) c) d) e) f) g) h) π(π β₯ 8) π(π β€ 5) π(4 β€ π β€ 6) π(π β₯ 3) π(π β€ 2) π(1 β€ π β€ 3) π(2 β€ π β€ 5) π(3.5 < π < 8.3) ο Note that for continous random π·(π. π β€ πΏ β€ π. π) i) the value of m if π(π β₯ π) = 0.1151 variable, 0.0918 0.6293 0.2486 0.6293 0.2514 0.2120 0.3779 0.4911 π·(π. π < π < 8.3) = π = π. π EXERCISE: 1. Compute the value of π so that i. π(π < π) = 0.3974 ii. π(βπ < π < π) = 0.8324 iii. π(π > π) = 0.3 iv. π(π < π) = 0.25 v. π(β1 < π < π) = 0.6 vi. π(0.3 < π < π) = 0.1 vii. π(π < π < 1.75) = 0.88 π = βπ. ππ π = π. ππ π = π. ππ π = βπ. ππ π = π. π π = π. ππ π = βπ. ππ 2. Let π be a normal random variable with mean π and standard deviation π. a) If π(π β₯ 50) = 0.2514 , π = 10 find π π = ππ. π b) If π(25 < π < 30) = 0.3944 , π = 25, find π π=π 3. Find the area of the curve below by using standard normal table a) Normal curve b) Normal curve 0.0684 0.3574 4. π~π(π , π 2 ). If π(π > 12) = 0.3 and π(π < 6) = 0.4, find the value of π and π. π = π. ππ , π = π. ππ 6.3 : APPLICATIONS OF THE NORMAL DISTRIBUTION Refer hondout given to the students DEPARTMENT OF MATHEMATICS Page 3 MATHEMATICS II (SHF 1124) 6.4 : THE CENTRAL LIMIT THEOREM ο§ ο§ ο§ The central limit theorem is used to solve problems involving sample means for large samples A sampling distribution of sample means Distribution using the means computed from all possible random samples of a specific size taken from a population Sampling error The difference between the sample measure and the corresponding population measure due to the fact that the sample is not a perfect presentation of the population Properties: i. mean of sample mean, πΜ will be the same as population mean π ππΏΜ = π ii. π = population standard deviation ππΜ = sample mean standard deviation ππΏΜ = ο§ π βπ The central limit theorem As sample size n increasing, the shape of distribution of ππΜ taken with replacement from π and π will approach normal distribution where ππΏΜ = π ππΏΜ = π βπ The Z-values : Μ βπ πΏ π= π β π β where Μ = π¬ππ¦π©π₯π π¦πππ§ πΏ π = π¦πππ§ π¨π π¬ππ¦π©π₯π π¦πππ§ DEPARTMENT OF MATHEMATICS Page 4 MATHEMATICS II (SHF 1124) 6.5 : THE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION ο§ ο§ ο§ A binomial distribution with parameters π and π can be approximated by a normal distribution with mean π = ππ and standard deviation π = βπππ. The normal approximation to the binomial distribution is good if π is large and π close to 0.5. The condition to use normal approximation is when : ππ β₯ π and π. π β€ π β€ π. π ο§ ο§ Note that π~π΅(π, π) is a discrete random variable whereas π~π(π , π 2 ) is continuous. Before using the normal approximation, a correction for continuity must be carried out because a continuous distribution is used as an approximation for a discrete distribution A correction for continuity a) b) c) d) e) f) g) π·(πΏ β₯ π) = π·(πΏ > π β π. π) π·(πΏ > π) = π·(πΏ > π + π. π) π·(πΏ β€ π) = π·(πΏ < π + π. π) π·(πΏ < π) = π·(πΏ < π β π. π) π·(π β€ πΏ β€ π ) = π·(π β π. π < πΏ < π + π. π) π·(π < πΏ < π ) = π·(π + π. π < πΏ < π β π. π) π·(πΏ = π) = π·(π β π. π < πΏ < π + π. π) Guideline to use normal approximation to binomial distribution 1. 2. 3. 4. 5. Check for ππ β₯ π Find π and π where π = ππ and π = βπππ Write the problem in probability notation Use continuity correction (show area under the curve) Find z-values from standard normal table EXAMPLE: 1. 80% of the books in a cupboard are academic books. If there are 50 books in the cupboard, find the probability that 40 books taken at random from the cupboard are academic books, by using a) binomial distribution, b) the normal approximation to the binomial distribution. DEPARTMENT OF MATHEMATICS 0.1398 0.1428 Page 5