ACJC 2024 H1 LCP MidYear Solutions

1 Anglo-Chinese Junior College 2024 JC 2 H1 Mid-Year Assessment Solution Qn Solutions 1 (a) ( ) 2 2 d2ln 5d5 xxxx += + 1(b) 2d7 d xx x x  −  13 22d 7d xxx =−   11 227 3 7 3 7 3 7 3 or or or2 2 2 2 22 x x x xx x x xxx − −−= − − Alternative method (quotient rule – not in H1 Syllabus) ( ) ( ) 2 2 d7 d 17 2 7 2 xx x x x x x x x x  −  − − − = ( ) ( ) ( ) ( ) 2 2 22 2 17 2 7 2 2 7 2 7 2 14 4 7 2 7 3 7 3 7 3 or 222 x x x x x x x x x x xx x x x x xx x x x x x x xx x x − − − = − − − = − − += − − −== 1(c) 2 2 2 3 1 d4 11 d16 2 1 1 1 16 2 3 xxx xxx x x cx  + = + + =− + + +   2(a) O x y y = 0 4e3+1

2 2(b) 324e e xxy −−=+ 32d 8e ed xxy x −−=− − When x = –3, ( ) ( )3 2 3 3 34e e 5ey −− −= + = ( ) ( )3 2 3 3 3d 8e e 9ed y x −− −=− − =− Equation of tangent is ( ) 33 3 3 3 33 5e 9e 3 9e 27e 5e 9e 32e yx yx yx −− − − − −− − =− − =− + + =− + 3(a) ( ) 2 1200 18750007 12005Tx x= + + 2 1875000240 8400Tx x= + + 3(b) 3 d 3750000240d T xx=− 3 3750000240 0 x−= 3 15625 25 x x = = ( ) 2 1875000240 25 8400 17400 25T = + + = 2 24 d 11250000 0d T xx = Therefore T is minimum. OR x 24.9 25 25.1 d d T x -2.903 0 2.857 Sketch \ – / Therefore T is minimum. 3(c) 3.93(3 sf)t =

3 3(d) 3(e) ( ) ( ) 55 00 d 1 13ln 7 2 d 185.547 186 nearest integer P t t t= + + = =  The total profit over the first 5 years is $186000. 4(a) 24 40 40.380x= + = Unbiased estimate of the population variance 2 2 1 24 83.2 0.96280 1 80s = − = −  4(b) An estimator is unbiased if the expected value of the estimator is equal to the population value. 5(a) Let X be the random variable “ Heights of girls, in cm, in TI Junior College” ( ) 2~,XN  ( )P 0.4L X L −  −  = 4.0 P =     −−    LXL P 0.4 LL Z −   = 5244005.0= L = 0.524 (to 3 s.f) (b) ( ) 2P 2 P XLXL   − + =   =        LZ 2 P = ( ))5244005.0(2 P Z = ( )048801.1 P Z = 0.14713488 = 0.147 (to 3 s.f) (c) ( ) ( )1 2 1 2P 2 P 2 0 0.5 X X X X X X+  = + −  = O x y ( )12 22 ~ 0, 6X X X N +−

4 6(a) 0 1 H : 3.4 H : 3.4   =  at 10% sig level, where  represents the population duration Under 0H , 0.303~ 3.4, 70XN   approx. by Central Limit Theorem p-value: 0.12846 or 0.128 (z-value -1.5202) Since p-value = 0.12846> 0.10 ( 1.5202 1.64485− − ), we do not reject H0. There is insufficient evidence at 10% level of significance to reject the claim. Manager’s claim should not be rejected. 6(b) There is no need to assume normal distribution as the distribution can be approximated to normal using Central Limit Theorem since the sample size (n = 70) is large. 6(c) 0 1 H : 3.4 H : 3.4   =  Under 0H , 0.8~ 3.4, 50XN   approx. by Central Limit Theorem Finding the expression 3.28 3.4-value= 0.8 50 z − For manager’s claim to not be supported, we do not reject 0H p value100 17.1 (3 sf)     p value100 0 17.1(3 sf)      

5 7(a) (i) Let X be the r.v “number of green or yellow light sticks, out of 8 light sticks.” ( )~ 8, 0.27XB Required probability = ( ) ( )P 3 1 P 2 0.371827 0.372XX = −  = = (3s.f) (shown) (ii) Let Y be the r.v “number of light sticks that are not green or yellow, out of 8 light sticks.” ( )~ 8, 0.73YB Expected number of light sticks that are not green or yellow = ( )EY = 8 (0.73) = 5.84 7b(i) Let W be the r.v “number of packets of glowing sticks with at least 3 green or yellow light sticks, out of 80 packets of glowing sticks.” ( )~ 80, 0.371827WB Required probability = ( ) ( )P 30 P 29 0.481269 0.481WW =  = = (3s.f) (bii) ( )~ 8, 0.73YB ( )E 5.84Y = ; ( )Var 1.5768Y = Since n is large, by Central Limit Theorem, 1 2 3 80 1.5768~ 5.84,80 80 Y Y Y Y N+ + + +   approx Required prob. = 1 2 3 80P 5.8 80 Y Y Y Y+ + + +  = 0.388 (3s.f) 8(a) (i) p (0.7) + (1 – p) (0.3) = 0.4p + 0.3 (ii) P (Benjamin dives in different direction as the ball is kicked) = 1 – (0.4p + 0.3) = 0.7 – 0.4p Required prob = (0.4p + 0.3) (0.65) + 0.7 – 0.4p = 0.895 – 0.14p (bi) (a) If Benjamin is the goalkeeper: 6 5 5 442 750CCC = If Benjamin is the forward: 6 5 5 4 4 1 375CCC = Number of ways required = 1125 (bi) (b) If no sibling selected: 6 5 2 4 4 2 75C C C = If 1 of the siblings included: 6 5 3 2 4 4 1 1 450C C C C = Number of ways required = 525 (bii) Number of ways required = 4 3 3! 5! 3! 6! 2!C = 24883200 (bii) Alternatively, Number of ways required = ( ) 246 333! 5! 3! 3! 2!CC = 24883200