ACJC 2026 Normal Distribution Summary

Anglo-Chinese Junior College 2026 H2 Mathematics 9758: Normal Distribution / Summary / Page 1 of 3 SUMMARY: Normal Distribution A normal distribution is a good model for continuous random variables such as weight, height, cost, duration, temperature, etc. If X has a normal distribution with mean  and variance 2 , we write it as ( ) 2~ N ,X  . 1 Properties of the Probability Density Function (i) It is symmetrical about its mean  . (This property is used quite often.) (ii) The mean, mode and median coincide at x = . (iii) The curve lies completely above the x-axis (since f ( ) 0x  for all real values x) and tapers off rapidly at each “tail” i.e. f( ) 0x +→ as x→ . (iv) Total area under the curve f ( ) d 1xx  − == . (v) The following probabilities are useful for determining if a random variable can be modelled by a normal distribution. • Approximately 68.3% of the data falls within one standard deviation about the mean. • Approximately 95.4% of the data falls within two standard deviations about the mean. • Approximately 99.7% of the data falls within three standard deviations about the mean. 2 Calculating Probabilities = P( area under the pdf c ) P( ) P( ) P( ) f ( ) d urve between and b a xa a X b a X b a X b a b Xb x xx   =   =   =  = =  = Probabilities are calculated using G C command normalcdf by specifying the lower and upper bounds, (i.e. a, b) and the mean and standard deviation. x   −   +  2 2 () 21f ( ) e 2 x x    −− =  (no need to know)

Anglo-Chinese Junior College 2026 H2 Mathematics 9758: Normal Distribution / Summary / Page 2 of 3 3 Finding Unknown x Values Given Probability When the probability is given, we use the calculator command invNorm to find the unknown x value. Examples Given that ~ N(10,4)X , find the value of the unknowns given that (i) P( ) 0.2Xa= (ii) P( ) 0.34Xb= (iii) ( )P 10 0.7Xc−  = 4 Standardisation Given that 2~ N( , )X  , then ~ N(0,1)XZ   −= . We use standardis ation only if either  and/or 2 are unknown or contains unknowns. Examples Given that 2~ N( , )X  , P( 5.4) 0.3X = and P( 2.1) 0.2X = , find  and 2 . 5 Linear Combination of Independent Normal Random Variables Linear combinations of ind ependent normal variables yield variables that are also normally distributed. Specifically, if X and Y are any two independent variables, and a and b are constants, then aX bY is also normally distributed with the means and variances calculated as follows: E( ) E( ) E( )aX bY a X b Y =  22Var( ) Var( ) Var( )aX bY a X b Y = + If 2 11~ N( , )X  and 2 22~ N( , )Y  , and a and b are constants, then • ( ) 2 2 2 2 1 2 1 2~ N ,aX bY a b a b    + + + • ( ) 2 2 2 2 1 2 1 2~ N ,aX bY a - b a b   −+ This result can be easily extended to linear combination of any number of independent normal variables. If 12, ,..., nX X X are independent observations of X where 2~ N( , )X  , then 12, ,..., nX X X have the same distribution as X. Therefore, the sum of n independent observations of X 2 12 ~ N( , )nX X X n n + + +  . Read and interpret the context carefully to see whether you are taking repeated observations of a variable, or a scalar multiple of a variable.

Anglo-Chinese Junior College 2026 H2 Mathematics 9758: Normal Distribution / Summary / Page 3 of 3 Examples Given that the height of boys in cm, 2~ N(168,5 )X and height of girls in cm, 2~ N(162,3.4 )Y , find the probabilities such that (i) the total height of two randomly selected boys exceeds 340 cm, (ii) the height of a randomly selected boy and girl differs by more than 5 cm, (iii) three times the height of a girl exceeds the total height of three randomly selected boys, (iv) the average height of 2 boys and 1 girl exceeds 167 cm. Solution (focus on formulating and calculating the combined normal variables) (i) 12P( 340)XX+ 22 12 ~ N(168 168,5 5 ) N(336, 50)XX+ + + = (ii) ( )P5 XY− 22~ N(168 162,5 3.4 ) N(6, 36.56)XY− − + = (iii) ( )1 2 3P3 Y X X X + + ( ) 22 1 2 33 ( ) ~ N 3(162) 3(168),9(3.4 ) 3(5 ) N( 18,179.04)Y X X X− + + − + = − (iv) ( )12 12P 167 or P 5013 X X Y X X Y++  + +  ( ) 2 2 2 12 2 168 168 162 5 5 3.4~ N , N 166, 6.843 3 3 X X Y ++ + + + + = or ( ) ( ) 2 2 2 12 ~ N 168 168 162, 5 5 3.4 N 498, 61.56X X Y+ + + + + + =