2023 JC2 H2 Normal Distribution BMQ Solutions (RVHS)

Statistics 5 Tutorial: Normal Distribution Basic Mastery Questions 1. Learning Experience Activity (20-30 mins) Explore the characteristics of a normal c urve: shape, centre, spread, the probability as area under the curve, and the empirical rule using a graphing tool. (1) Shape, centre and spread Carry out the following investigation: 1. Start with a fixed , say 7. 2. Study the behaviour of the normal curve as you change the values of , say from 1, 1.5, 2, or 2.5. 3. Repeat (1) and (2) with different values of , say 8, 9 and 10. 4. How is the shape of the normal curve affected by changing the mean? Changing the standard deviation? 5. Explain why this behaviour is to be expected. (2) Probability as area under the curve Discuss: Can you think of an analogy why probability that a value lies between two values x = a and x = b is given by the area under the curve from x = a and x = b? (3) The empirical rule (ref pt. 6 of section 5.2.1) Sketch a normal curve using any of the values of and given in Q(1), and find the percentage of the total area which is within: (a) 1 standard deviation of the mean (b) 2 standard deviations of the mean (c) 3 standard deviations of the mean     

(1) The mean µ controls the location of the peak of the curve. Varying it will move the entire distribution to the left or right. On the other hand, t he standard deviation σ controls whether the curve is broad and flat (larger standard deviation) or narrow and tall (smaller standard deviation). Varying it will not move the distribution left or right, but will affect the spread of the distribution. (2) Use the analogy of a dartboard where the probability that a dart falls inside a defined compound is given by the ratio of the area of the defined compound to the total area of the dartboard. (3) (a) 68.3% , (b) 95.4%, (c) 99.7% (The empirical rule or the 68-95-99.7 rule: Practically all of the population (99.7%) lies in the interval μ ± 3σ, about 95% of the population lies in the interval μ ± 2σ, and about 68% of the population lies in the interval μ ± σ.) 2. Given X and Y are independent with and . State the distribution of: (a) 2X, (b) , where are two independent observations of X, (c) X − 2Y, (d) , where are three independent observations of Y. (a) 2 ~ N(6,16)X (b) 12 ~ N(6,8)XX+ (c) 2 ~ N( 9,68)XY−− (d) 1 2 3 16~ N 6,33 Y Y Y++   ( )~ N 3, 4X ( )~ N 6,16Y 12XX+ 12 and XX 1 2 3 3 Y Y Y++ 1 2 3, and Y Y Y Why must X and Y be independent? If X and Y are not independent, ( ) ( ) ( )Var 2 Var Var 2X Y X Y−  +

3. Given , find a if . P( ) 0.1182 Using GC (invNorm) 152 Xa a = = 4. The independent random variables R and S each have normal distributions. The means of R and S are 10 and 12 respectively, and the variances are 9 and 16 respectively. Find the following probabilities (i) , (ii) , where is the mean of a sample of 4 independent observations of R, (iii) , (iv) , where S1 and S2 are two independent observations of S. (i) ( )P 12 0.748 R= (ii)      4 9,10 N~R ( )P 12 0.909 R= (iii) )25 ,2( N~ −− SR ( ) ( )P P 0 0.655 R S R S = −  = (iv) )68 ,4( N~2 21 −−− SSR ( ) ( )1 2 1 2P 2 P 2 0 0.314 R S S R S S + = − −  = ~ N(144,49)X P( ) 0.1182Xa= ( )12P R ( )12P R R ( )SRP ( )212P SSR +

5. Given that 2~ N(70, )X  , find the value of σ such that ( )P | | 102 0.2X = . Solution: ( )P | | 102 0.2 70 32P 0.2 32P 0.2 32 32P +P 0.2 32 32P 0.1 or P 0.1 32 1.28155 25.0 X X Z ZZ ZZ       =  −  =  =    −  =          − =  =         − =− =