HCI 8865 2022 Prelim Solutions vetted

Suggested Solutions for 2022 C2 H1 Preliminary Examinations No. Suggested Solutions 1 2 30x kx k− + + = has real roots if Discriminant 0 Discriminant ( ) ( )( ) 2 2 43 4 12 26 kk kk kk = − + = − − = + − For discriminant 0 , we have { : 2 or 6 }k k k −  2a ( ) 33 15 266 61dd 52 xx x x x x x x C x −− = − = − + 2(b) (i) ( ) ( )( ) 3 2 2 2 2 d 6ln 3 4d 6d ln 3 43d 62 34 12 34 xx xx x x x x + =+ =  + = + Method 2 ( ) ( ) 3 2 2 2 3 3 2 2 d 6ln 3 4d 1 3 4 636 34 12 34 xx xx x x x − +  + =  + = + (ii) ( ) ( ) 22 3 2 3 22 1 12 d d3 4 12 3 4 1 6ln 3 412 11 = ln 3 4 or ln 3 426 xx xxxx xc x c x c =++ = + + + + + +  3(i) 22 2AD x x x= + = ( )1 2 2 2 4 102PerimeterAEBCGFD x AD DF x= + + + = 2 2 2 4 10x x DF x + + + = -2 6

( )10 2 2 4 2 =5 2 22 x DF x   − + + = − + + (ii) Let area of window be A ( ) ( ) 2 22 2 2 2 2 2 2 2 2 11 2 4 422 1 3 5 2 2 422 1 3 20 2 4 2 82 3 5 4 2 202 A x x x x DF x x x x x x x x x x x x x x x     = + + +  = + + − + +   = + + − − − =− − − + ( ) 2 2 d 3 10 8 2 20d d 20For max , 0 d 3 10 8 2 d 3 10 8 2 0d A x x xx AAx x A x    =− − − + =  = ++ =− + +  2   is maximum whe 20 3 0 8 n 1 A x  = ++ 4(i) (ii) 33e xy −=− At x-axis, 3 10 e 3 ln 3 3 xyx −=  =  =− 3d 3ed xy x −= x y

(iii) When 1 ln 33x=− , 13 ln3 ln33d 3e 3e 9d y x −−= = = Equation of tangent, 10 9 ln 3 3 9 3ln 3 yx yx  − = − −   =+ Area ( ) ( ) ( ) ( ) ( ) 0 3 ln 3 3 03 2 ln 3 3 ln 3 2 2 2 2 113ln 3 ln 3 3 e d23 1e ln 3 323 1 1 eln 3 ln 32 3 3 1 1 3ln 3 ln 32 3 3 12ln 3 ln 3 units23 x x x x − − − − = − −  = − +   = − − − +  = − + − +  = − +  5(i) Let x be the selling price of a toy boat. Let y be the selling price of a toy car. Let z be the selling price of a toy plane. x y

2 9 5 3 335 3 4 38 2 xy x y z x y x z = + + = + = + + Rearranging the equations, we get 2 0 0 9 5 3 335 2 4 2 38 x y z x y z x y z − + = + + = + − = From GC, 22.4, 11.2, 25.8x y z= = = Hence, the selling price of a toy plane is $25.80. (ii) Substitute D = 20 and p = 0 into 0.083e pDa −=+ , 020 3 e 17 a a =+ = (iii) (iv) 0.083 17e pD −=+ 0.08 0.08 d 17( 0.08)ed 1.36e p p D p − − =− =− (v) ( ) 2 0.5 9 312pt= − + p (dollars) D (thousands) D = 3

( ) ( ) ( ) ( ) ( ) 1 2 2 1 2 2 2 d1 0.5 9 312 (0.5)(2) 9d2 1 9 0.5 9 3122 9 2 0.5 9 312 p ttt tt t t − − = − + − = − − +  −= −+ ( ) 0.08 2 d d d d d d 91.36e 2 0.5 9 312 p D D p t p t t t − = −=−  −+ When 20t= , ( ) 2 0.5 20 9 312 19.30025907 p= − + = ( ) ( ) 0.08 19.30025907 2 d 20 9 1.36ed 2 0.5 20 9 312 0.290388 0.2849703 0.0828 (3sf) D t − −=−  −+ =−  =− (vi) The monthly demand for toy trains at 20 months from now is decreasing at a rate of 82.8 toys per month. 6(i) No. of ways = 17 8 24310C = (ii) Method 1 Case 1: Husband at either end No. of ways = 7! 2 8! = 406425600 Case 2: Husband not at either end No. of ways = 14 8! 7! = 2844979200 Total number of ways = 406425600 + 2844979200 = 3251404800 Method 2

No. of ways = ( ) 16 18! 7! 1 = 3251404800C   (iii) No. of ways = 89 44 8820CC= (iv) Case 1: includes wife (but not husband) 78 34 2450CC= Case 2: includes husband (but not wife) 78 43 1960CC= Total number of ways = 2450 + 1960 = 4410 Required Probability = 4410 1 8820 2= 7(i) ( )65 = 26 21 n ' 'x x x M S A− + + − +   ( )n ' ' 18M S A x  = + ( )40 26 14 n ' 'x x x S M A= − + + − +   ( )n ' 'S M A x  = ( )45 21 14 n ' 'x x x A S M= − + + − +   ( )n ' ' 10A S M x  = + (ii) Since M and S are independent, P( ) P( ) P( )M S M S =  18 + x x 10 + x A S M 26 - x 21 - x 14 - x x 3

26 65 40 92 92 92 26(92 ) 65 40 2600 2392 26 2600 26 208 8 x x x x x x x =+ + + + =  = += = = (iii) P( )MA is the probability that a customer likes mangoes, given that he or she likes apples. P( )P( ) P( ) 21 100 45 100 7 15 MAMA A = =  = (iv) Number of customers who like exactly two different fruits = 18 + 13 + 6 = 37 Number of customers who like only one fruit = 26 + 18 + 8 = 52 Required probability = P(3 like exactly two different fruits and 1 like only one fruit) 37 36 35 52 4 0.103100 99 98 97=     = (3 sf) or 37 52 31 100 4 0.103CC C == (3 sf)

8(i) ( ) 3540 6120442.5, 88 Using 1.834352 40.8 , 6120 1.834352 442.5 40.88 700.00608 = 700 (to nearest integer) kxy yx k k += = = =+ + =+ = (ii) 0.996r= . It indicates a strong positive linear correlation between the number of Nutella brownies sold and the profit. (iii) (iv) Sub x = 480 into 1.834352 40.8yx=+ , we obtain 921.28896y= $921.29 (to 2 d.p. for money)y The product moment correlation is close to 1 and 480x= is within the data range of [190, 620]. Thus, the estimate will be a reliable one. (v) As r measures the degree of scatter of the data points, an increase of 80 for all the values of the monthly profit (values of y) will not change the scatter of the data. Hence, there will be no change in the value of r. 9(i) Let X be the duration of a particular viral infection in a child. Let  be the mean population duration of a particular viral infection among children and 2 be the population variance of a particular viral infection among children. ( ) 22 120 1.215 1.48863119s == , 4.07x = H0 :  = 4.3 H1 :   4.3 at 5% level of significance (190,390) (620,1200) y x

Under H0, since 120n= is large, by Central Limit Theorem, 2 ~ N ,X n   approximately. XZ s n −= ~ N(0,1) approximately Using a one-tailed test, 4.07x = gives p-value = 0.038920609 < 0.05 We reject H0 at 5% level of significance and conclude that there is sufficient evidence there the mean duration of viral infection of the particular viral among children is different from the researchers’ claim. (ii) Not necessary. Since n = 120 is large, by Central Limit Theorem, the sample mean duration of a particular viral infection in children follows a normal distribution approximately. (iii) H0 :  = 4.3 H1 :  < 4.3 Since H0 is rejected, p-value 100 k 0.038920609 2 100 1.95 (3 s.f.) k k   (iv) Unbiased estimate of population variance, ( ) 2 2 77.88 1.3259 59 yys −= = = (v) H0:  = 4.3 H1:  < 4.3 There is a probability of 100  to conclude that the population mean duration of infection being treated with medicine M is less than 4.3 days when it is actually 4.3 days. 10 (i) Let X be the mass of a randomly selected Honeycrisp apple. ( ) 2 N 78,13X . ( )P 70 90 0.5528661 0.553X  =  (ii) Let Y be the mass of a randomly selected empty box. ( ) 2 N 10, 0.8Y

Let ( )1 2 3 24 ... ~ 1882,4056.64T X X X X Y N= + + + + + ( ) 0.9680354 0.968P 2000T = (iii) The mass of an apples is independent of the mass of another apple. The mass of an apple is independent of the mass of an empty box. (iv) k is the largest mass. ( )P 0.55Xk From GC, When ( ) ( ) ( ) 75, P 0.59125 76, P 0.56113 77, P 0.53066 k X k k X k k X k =  = =  = =  = 76k = Therefore largest 76k = Alternative method 78P 0.55 13 78 0.1256613 76.366 largest 76 kZ k k k −  − −  = (v) Let A be the cost of a randomly selected Honeycrisp apple. ( )N 5.46, 0.8281A Let F be the total cost of an empty box and 24 Honeycrisp apples. ( )N 131.24,19.8744F ( )1 2 3 N 393.72, 59.6232F F F++ ( )P 370 420 0.998604 0.999T  = 