NYJC 2026 FM TP - FM Stats 1 (Solutions)

2026 J2 H2 FM FM Statistics 1 - Discrete and Continuous Random Variable – Solutions 1* The piano shop opens at 9 am every day. The waiting time (in hours) between successive arrivals of customers at the shop has an exponential distribution with mean 1  . Records show that on average, the shop sees no customers in the first 5 minutes on 20% of the days. (a) Show that the mean number of customers in one hour is 12ln 5 . [2] (b) Given that the first customer has not arrived by 9.05 am, find the probability that the shop sees its first customer by 9.15 am. [2] (c) Find the probability that the 15th customer arrives after 10.30 am. [2] Records show that, on average, 1 in 10 of its customers bought a piano. Let W be the random variable denoting the number of customers served before the customer who bought the second piano. (d) Find P( ).Ww= [2] (e) By considering the mean and variance of an appropriate geometric distribution, find E( ).W [3] 1 DHS_HCI_RI_TMJC Prelim 9649/2020/02/Q8 (a) Method 1: Use Exponential Distribution Let X be the waiting time (in hours) between successive customers. ( )~ E xpX  5P 0.2 60X = 12e 0.2 − = 12ln 5 = Mean number of customers in 1 hour = 12ln 5 Method 2: Use Poisson Distribution Let be A be number of customers arriving in 5 min. 5~ Po 60A   ( )P 0 0.2A== 12e 0.2 − = 12ln 5 = Mean number of customers in 1 hour = 12ln 5 M1: Distribution + Probability Statement AG1: Apply formula to show answer M1: Distribution + Probability Statement AG1: Apply formula to show answer (b) Method 1: Use Exponential Distribution 15 5 1 1P 1 P 60 60 4 12X X X X     = −         11P 6X= −  M1

1P 6X=  12ln5 61 e 0.96 − = − = Method 2: Use Poisson Distribution ( )P First customer arrives by 9.15 am 0A= ( )P1B= where B is the number of customers who arrive at the shop in 10 mins ( )~ Po 2ln 5B 12ln5 61 e 0.96 − = − = Method 3: Use Conditional Probability ( ) /4 /12 /12 11P15 5 12 4P 160 60 P 12 11PP 4 12 1P 12 1 e 1 e e 0.96 X XX X XX X   −− −     =       −       =   − − − = = A1 M1 A1 (iii) Let C be the number of customers arriving in 1.5 hours. ( )~ Po 18ln 5C ( )P 14 0.00162C= B1 B1 (b) Let W be the number of customers served before the customer who bought the second piano. For 1,2,3,w=  ( ) ( )( )( ) ( ) ( ) ( ) 1 1 2 P 0.1 0.9 0.1 0.9 0.11 wwwW w w −− = = = Method 2: Let Y be the number of customers served up to and including the one who bought a piano. ( )~ Geo 0.1Y M1 A1 Exactly 1 out of w customers bought a piano. Need to ensure that the (w+1)th customer bought a piano.

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 12 12 1 2 1 2 1 1 1 12 1 independent of 12 P P 1 P1 P P 1 since , are independent 0.9 0.1 0.9 0.1 0.9 0.1 0.9 0.1 w r w r w r r w w r r w W w Y Y w Y Y w Y r Y w r Y Y w = −− = − = − = = + − = = + = + = = = + − = = =     (c) ( ) ( )( ) ( )( ) 21 1 21 1 E 0.9 0.1 0.1 0.9 0.1 w y w y W w w w  − =  − = = =   Consider ( )~ Geo 0.1Y : ( ) ( )( ) 1 1 1E 0.9 0.1 10 0.1 y y Yy  − = = = = ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) 22 22 2 1 2 2 1 Var E E E Var E 1 0.10.9 0.1 10 190 0.1 y y Y Y Y Y Y Y y  − = =−  =+  −= + = Thus ( ) ( )E 0.1 190 19W == Alternatively, a better method (Note that question specified the use of mean and variance of geometric distribution, hence we should not do like this) Let 12 1W Y Y= + − , where ( )12, ~ Geo 0.1YY ( ) ( ) ( )12E E 1 2E 1 121 0.1 19 W Y Y Y= + − = − =− = M1 M1 A1

2* The continuous random variable X has probability density function f given by 2 25 ( 1), 1 4,f ( ) 0, otherwise. xxx  + −  =  The random variable Y is defined by (a) Show that 4 25 P( ) for 0 1.Y y y y =   [3] (b) Find the cumulative distribution function of Y . [4] (c) Find the probability density function of Y. [2] 2.YX= 2 EJC JC2 Prelim 9649/2019/02/Q7 (a) Given 2 ) For 0 1, P P() ( y Y yy X  =   2 ( 1) 0 1, 1 1 () 4 (shown) P( ) 2 d Note that for 25 2 2 25 25 2 2 5 2 2 y y y y y yy x y x y X x x x yyy − − = −   = =+   +   −   = +− −  =   B1 M1 A1 (b) Note, for (a) we considered which corresponds to Now, , For , P( 1)P (1 )() PYYy yY =  +   P( 1) P(1 )Y X y=  +   1 ( 1) 4)42 1 d ( 125 25 y xy x=+ +    1 242 25 25 2 42 25 25 2 1( 1)2 y x x y y = + +  + + + − =  ( ) 21 OR 2 125 2 2 25 1 yy yy   ++  +    +  = M1 M1 A1 2.YX= 0 1 y 1 1,x−   1 4 1 16xy     1 16y

3 The number of distinct uranium deposits in a given area is a Poisson random variable with parameter 10. Independently, the number of distinct plutonium deposits in this area is a Poisson random variable with parameter 5. In a fixed period of time, each dep osit is discovered independently with probability 1 50 . (a) Find the probability that no less than 2 deposits are discovered during this period.[2] (b) Given that no more than 3 deposits are found in this period of time, find the probability that there are at least 1 uranium deposit. [4] (a) Let U and V be the number of uranium deposits and plutonium deposits discovered during this period of time. ~ (0.2)U Po and ~ (0.1)V Po . Let T U V=+ . ~ (0.3)T Po . Prob. Req d P( 2) 1 P( 1) 0.0369 T T  = = −  = M1 A1 (b) P( 1| 3) P( 1 and 3) P( 3) P(1 3 and 0) P(1 2 and 1) P( 1 and 2) P( 3) 0.1810065 0.1810.9997342 UT UT T U V U V U V T  =    = +   = + = ==  == M1 M1 M1 A1 ( ) 1 2 0, 0 4 , 0 125G( ) 1 1 , 1 1625 , 16 y yy y yy y y    = + +     A1 (c) 2 , 0 1 25 1g( ) G ( ) 0 otherwi 1 1 , 1 1 e 625 s ' y y yy y y     = +   =       M1 - differentiate A1

4 A particle moves along a horizontal line, relative to a fixed origin, so that its position at a given time t is 0 sinx x t= , where 0x is a positive constant. Let X be the continuous random variable denoting the particle’s position. (a) Show that the cumulative density function of X is given by ( ) 0 1 00 0 0 1, 11F sin , 2 π 0, xx xx x x x x xx −   = + −      − and hence, find the probability density function of X. [5] (b) Find the exact probability that the magnitude of X is at most 1 02 x . [2] (c) Show that the expectation of X is 0, and find the exact variance of X using the substitution 0 sinxx = . [7] 4 ACJC JC2 Prelim 9649/2019/02/Q10 (a) 1 0 0 sin sin xx x t t x −=  = Consider one period. Thus ~ U(0, 2 )T  , 1f ( ) 2t = . If 0xx− , ( )P0Xx= 0If 0,xx−   ( ) ( ) ( ) ( ) ( ) 00 0 0 11 00 11 2 0 111 2 P P sin 2 sin 2sin by symmetry sin xx xx x x x x X x T     −−  −  −   = −   +  =+  =+  

0If 0 , xx ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 00 11 1 0 1 0 P P 0 P 0 P π 2π P 0 sin or sin 11 π 2 sin by symmetry2π 2π 11 sin2 π xx xx X x X X x T T T x x x x −− − −  =  +   =   +   −    = +     =+   Hence, ( ) 0 1 00 0 0 1, 11F sin , 2 π 0. xx xx x x x x xx −   = + −      − ( ) ( ) 22 0 0022 0 00 dfF d 1 π 1 , π 0 or xx x xx x x x xx x x x x = = −  −  = −  −  (b) ( ) ( ) ( ) ( ) ( ) ( )( ) 1 1 1 0 0 02 2 2 11 0022 11 11 22 1 3 PP FF 1 sin sinπ X x x X x xx −−  = −   = − − = − − = M1 A1

5 Three fair six -sided die are thro