SAJC 2021 JC1 H1 Math Normal Distribution (Teacher)

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 1 of 35 Chapter 4 (Statistics) Normal Distribution (Teacher’s copy) Objectives At the end of the chapter, students should be able to: (a) Understand the nature of the normal distribution which include  a normal distribution is used to model a continuous random variable  use normal curves to illustrate normal distributions with different  and   the shape and location of a normal curve are determined by the values of  and   the area under a normal curve between ax and bx is the probability  bxa P  the total area under a normal curve is 1 (b) Understand X and Y have independent normal distributions, then bYaX  has a normal distribution with mean yx ba   and variance 2222 yx ba   Contents 4.1 Introduction to Normal Distribution 4.2 Continuous Random Variable 4.3 Normal Distribution Curve 4.3.1 Properties of the Normal Distribution Curve 4.4 Finding Probabilities 4.5 Use of Inverse Normal 4.6 Standard Normal Distribution 4.7 Linear Combinations of Independent Normal Distributions 4.7.1 Properties of Expectation & Variance 4.7.2 Sum and Difference of Independent Normal Distributions 4.7.3 Sum of n independent observations from the same Normal Distribution 4.8 Summary and Checklist 4.9 Learning Experience

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 2 of 35 4.1 Introduction to Normal Distribution If you sketch a histogram corresponding to the heights of 17 years old male students in a SAJC, what would you observe? You will realized that majority of the students have about the same height with a few students who are taller than the average height and another few who are shorter than the rest. This is logical since there are fewer very short or very tall boys, and even fewer very very short or very very tall boys. The outline of the histogram generally shapes like a bell-shaped curve. Sample size = 400 Height Frequency Relative Frequency 1.650  1.675 112 112/400 = 0.28 1.675  1.700 128 128/400 = 0.32 1.700  1.725 116 116/400 = 0.29 In fact, such a ‘pattern’ of distribution can be observed for many other physical phenomena as well, such as weight of watermelons in a particular market , I.Q. scores of a population, time taken by students to run 100m, examination results of students in SAJC, household income of Singaporeans etc. Since it is so common, we shall name the distribution a normal distribution (and the bell-shape curve a normal curve) and study it with some depth. The normal variable is an example of a continuous random variable. Note: FrequencyRelative Frequency Total number in the sample height Relative frequency

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 3 of 35 4.2 Continuous Random Variable Recall from Chapter 3, a continuous random variable (CRV) is a random variable th at takes values in an interval. This differentiates from discrete random variable which can only takes a finite or countable infinite set of exact values. In theory, CRVs can be measured to any desired degree of accuracy. Examples of CRVs are:  The height of a randomly chosen 18-year-old student.  The actual mass of a packet of 1 kg rice.  The lifetime of a new battery.  The temperature of a student in SAJC on a particular day  The waiting time of bus number “142” at the bus -stop outside SAJC on a Monday morning. 4.3 Normal Distribution Curve Normal distribution curve of 17 years old students in SAJC 4.3.1 Properties of the Normal Distribution Curve 1. It is bell-shaped and symmetrical about the mean  . 2. It extends from  to  . 3. The total area under the curve is 1. 4. Mean = Median = Mode =  . 5. The shape of any Normal curve is determined completely by its mean  and standard deviation  . The mean gives the position of the axis of symmetry and standard deviation gives the spread. The smaller the standard deviation, the greater the clustering of values near the mean. Hence the curve will have a higher peak at the mean. 6. A continuous random variable X which follows a normal distribution with mean  and variance 2 can be denoted as   2~ N ,X  µ Most 17 years old have average height Heights Taller Shorter

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 4 of 35 7. For all normal curves a. Approx. 68% of the distribution lies within 1 standard deviations of the mean. b. Approx. 95% of the distribution lies within 2 standard deviations of the mean. c. Approx. 99.7% (nearly all) of the distribution lies within 3 standard deviations of the mean. Note: The above properties help us to determine whether the distribution of a random variable follows the normal distribution. Another method of checking whether a distribution follows a normal distribution is to use a graphical test, a common way to do so would be to plot the frequency distribution of the random variable. A normal curve would then be overlaid on the frequency distribution, so that we can see whether the normal curve is a good fit. Example 1 View the video https://www.youtube.com/watch?v=Wqw9cLRMPL0 (up to 07:50). Fill in the blanks. On the Normal curves, put in the markings on the horizontal axis and shade a region corresponding to (a) 68% of the distribution X, (b) 99.7% of the distribution Y, (c) 95% of the distribution V. Random Variable Normal Distribution Curve (a) Let X be the r.v. time taken in hour s to complete a marathon by a group of males. 223.8, 0.5   2~ N 3.8,0.5X (b) Let X be the r.v. height of a class of 18 year old students, measured in cm. 22180, 10   2~ N 180 , 10 Y (c) Let V be the r.v. volume in ml of a particular can of drink. 2330, 4   2~ N 330 , 2 V 180 Y V 330 + − −2 −3 +2 +3  X 3.8

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 5 of 35 4.4 Finding Probabilities Given that   2~ N ,X  , the probability that X lies between a and b is  P a X b .  P a X b = Area under the normal curve between xa and xb Note: 1. Probability that X takes a specific value is zero, i.e.  P0Xc As a result,    PPX c X c   ,    PPX c X c   . 2.        PPPPa X b a X b a X b a X b           3.    PPXX     0.5 Example 2 (GC: Normalcdf) Given X follows a normal distribution with mean 5 and standard deviation 2, i.e.  22,5N~X , calculate (i)  64P  X ; (ii)  8.7P X ; (iii)  3P X . Solution: (i)  P 4 6 3.83X   Steps Screenshot Remarks Press y½2 You should see this screen. Note the heading is ‘normalcdf’ If the lower limit,  and  are not specified, the default values used by the GC are -1E99, 0 and 1 respectively. Enter the values of : Lower bound (lower), Upper bound (upper), Mean (  ) and Standard deviation (  ) Press ENTER, ENTER.  a b X X

SAJC 2021 JC 1 H1 Mathematics Normal Distribution Page 6 of 35 (ii)   919.08.7P X Steps Screenshot Remarks Press y½2 Yo