2019 H2 Prelim (Normal Distribution) Qns with Sol

Birth weight can be used to predict short and long-term health complications for babies. Studies show that the birth weight of babies born to mothers who do not smoke in a certain hospital can be assumed to follow a normal distribution with mean 3.05 kg and variance  2 kg 2 . The hospital classifies babies based on their birth weight as shown in the table below. Birth weight Classification Less than 1.5 kg Very low birth weight 1.5 kg to 2.5 kg Low birth weight 2.5 kg to 4.0 kg Normal birth weight More than 4.0 kg High birth weight (i) A sample showed that 20.2% of the babies born to mothers who do not smoke have low birth weight. If this is true for the entire population, find two possible values of  , corrected to 2 decimal places. Explain clearly why one of the values of  found should be rejected. [4] Studies also show that babies born to mothers who smoke have a lower mean birth weight of 2.80 kg. For the remaining of the question, you may assume the birth weight of babies born to mothers who smoke is also normally distributed and that the standard deviations for the birth weight of babies born to mothers who do not smoke and mothers who smoke are both 0.86 kg. (ii) Three pregnant mothers-to-be who smoke are randomly chosen. Find the probability that all their babies will not be of normal birth weight. State an assumption that is needed for your working. [3] (iii) Find the probability that the average birth weight of two babies from mothers who do not smoke differs from twice the birth weight of a baby from a mother who smokes by less than 2 kg. [3] Babies whose birth weight not classified as normal will have to remain in hospital for further observation until their condition stabilises. Depending on the treatment received and length of stay, the mean hospitalisation cost per baby is $2800 and standard deviation is $500. (iv) Find the probability that the average hospitalisation cost from a random sample of 50 babies, whose birth weight not classified as normal, exceeds $3000. [2] ASRJC JC2 Prelim 9758/2019/02/Q6 (Solutions) 6 (i) Let X be random variable representing the birth weight of a baby in kg.  2 ~ N(3.05, ) X  P(1.5 2.5) 0.202 X    By plotting the graph of P(1.5 2.5) 0.202 y X     using GC, we have 0.70 (2 dp) or 1.56     (2 dp) If  = 0.70, P(2.5 < X < 4) = 0.70 If  = 1.56, P(2.5 < X < 4) = 0.37

1 2 3 1 2 3 4 Required probability = P( 50) 7 0.446 X X X Y Y Y Y         (iv) The random variables X and Y are independent of each other. (v) Let C be the score of a randomly chosen student in Group C. If normal distribution is assumed, P( 0) 0.0548 C   A significant proportion would fall under an inadmissible area, hence normal distribution is not suitable as a model. DHS JC2 Prelim 9758/2019/02/Q10 The speeds of an e-scooter ( km/h) X and a pedestrian ( km/h) Y measured on a particular stretch of footpath are normally distributed with mean and variance as follows: mean variance X 12.3 9.9 Y  2  It is known that P( 5.2) P( 7.0) 0.379. Y Y     (i) State the value of  and find the value of .  [2] (ii) Given that the speeds of half of the e-scooters measured are found to be within a km/h of the mean, find a . [2] (iii) A LTA officer stationed himself at the footpath and measured the speeds of 50 e-scooters at random. Find the probability that the 50th e- scooter is the 35th to exceed LTA’s legal speed limit of 10 km/h. [3] (iv) On another day, the LTA officer randomly measured the speeds of 6 e-scooters and 15 pedestrians. Find the probability that the mean speed of the e-scooters is more than twice the mean speed of the pedestrians captured. [3] (v) Find the probability that the mean speed of n randomly chosen e-scooters is more than 10 km/h, if n is large. [2] DHS JC2 Prelim 9758/2019/02/Q10 (Solutions) Qn Suggested Solution 10 (i) By symmetry, 1 5.2 7. 6. 0 2    

(ii) 3 4 ~ N(0,14) X X  3 4 3 4 P( ) 1 ) P( 0 2 X X X X      (by symmetry) (iii) 1 2 2 1 ) Var( Var( ) n n X n Y X X       1 2 2 1 Var( ) Var( ) Var( ) n X X X n       2 1 2 4 6 2 n n        2 1 1 n n n    (Sum of A.P.) 1 1 n   Since the n X ’s are independent Normal distributions with common mean, 1 1 ~ N , n Y n         (NB: The variance of n Y decreases as n increases.) Either 2 P 1 1) 3 ( n Y        1 1 1 1 1 2 3 1 1 P 1 n n n n Y                         1 1 1 1 2 3 1 1 P n n Z               1 0.9 1 1 6742 n   Solving this inequality, 14.6017 n  Hence, the smallest possible value of n is 15. Alternatively

      P 12 0.158655 0.159 3sf P 6 X Y       a “6 - inch” loaf more likely to be less than 6 inches. (ii)         1 2 2 2 1 2 E 2 6.1 12.2 0 Var 2 0.1 2 0.2 0.08 Y Y X Y Y X             1 2 ~ N 0,0.08 Y Y X       1 2 1 2 1 P P 0 2 Y Y X Y Y X        (iii) Let A denote the number of “6 - inch” sandwiches less than 6 -inches in length in a week.   ~ B 3,0.239750 A     P 1 0.855122 0.855 3sf A    Alternative             2 3 2 3 required probability P 6 P 6 3 P 6 0.23975 1 0.23975 3 1 0.23975 0.855 3sf Y Y Y                     (iv) Let B and C denote the number of “6 - inch” sandwiches less than 6 -inches in length in a 4-week and 3-week period respectively.   ~ B 12,0.239750 B and   ~ B 9,0.239750 C                 P 1 P 3 P 1 4 P 4 P 1 1 P 3 1 P 4 0.41571 0.14710 0.13721 0.445674 0.446 3sf A C A B B A C B                     JPJC JC2 Prelim 9758/2019/02/Q9 In this question you should state the parameters of any distributions that you use. The masses of Grade A and Grade B strawberries are normally distributed with mean 18 grams and 12 grams respectively and standard deviation 3 grams and 2 grams respectively. (i) Find the probability that a randomly chosen Grade A strawberry has mass between 17 and 20 grams. [1]

( Other possible answer: The event that a mass of a box of 4 dark truffles has mass more than 70g is independent of other boxes.) 11(iii) Let W be the mass, in g, of a randomly chosen salted caramel ganache.   2 ~ N , W   Given     P 12 P 15 W W    and   P 15 0.97 W   Method 1 By symmetry, 12 15 13.5 2       P 15 0.97 W   2 2 15 13.5 P 0.97 1.5 1.88079 0.797537 0.636065 0.636 (3 sf) Z                   Method 2   P 12 0.03 12 P 0.03 12 1.88079 1.88079 12 ---(1) W Z                         P 15 0.97 15 P 0.03 15 1.88079 1.88079 15 ---(2) W Z                      2 2 Solving equation (1) and (2), 13.5, 0.797537 0.636065 0.636 (3 sf)         11(iv)   ~ N 13.5, 0.636065 W Find     1 2 6 P 0.28 ... 0.34 W W W T     . Let   1 2 6 0.28 ... 0.34 S W W W T     

Required probability = P( 70000 S  ) = 0.0297 (correct to 3 s.f.). 9(iv) [3]   1 2 3 3 R R R S    ~ N( 12000  , 2 12(8000 ) ) Required probability = P     1 2 3 3 R R R S    = P     1 2 3 3 0 R R R S     = 0.667 (to 3 s.f.) 9(v) [3] Consider P( 50000 S  ) > P( 50000 R  ) P 50000 68000 Z          > 0.9599408865 P 18000 Z          > 0.9599408865 18000 1.750000503      0 <  < 10286 (to nearest whole number) [ or 0 <   10285 ] 9(vi) [1] Assume that the working lifespan of all the tyres are independent. RVHS JC2 Prelim 9758/2019/02/Q12 An orchard produces apples and pears. (i) The masses of apples produced by the orchard have mean 70 grams and standard deviation 40 grams. Explain why the masses of apples in the orchard are unlikely to be normally distributed. [1] (ii) The masses of pears produced by the orchard are normally distributed. It is known that one-third of the pears produced weigh less than 148 grams, and one-third weigh more than 230 grams. Find the mean and standard deviation of the masses of pears produced in the orchard. [4] During each production period, the orchard produces 300 apples and 400 pears and the owner is able to sell all the apples at $0.005 per gram and all the pears at $0.008 per gram. It is further known that the owner needs to spend a fixed amount of $ c as the operating cost for each production period and that the orchard makes a profit 90% of the time. (iii) Explain why the total mass of the 300 apples is approximately normally distributed. [1]

(i) Find the probability that two standard packets weigh more than a large packet. [3] (ii) Find the probability that the mean mass of two standard packets and one large packet of sugar is between 680g and 700g. [3] SAJC JC2 Prelim 9758/2019/02/Q5 (Solutions) 5 Let X be the mass of a standard packet of sugar in grams. Let Y be the mass of a large packet of sugar in grams.     2 2 N 520,8 N 1030,11 X Y   1 2 N 10,249 X X Y       1 2 1 2 P P 0 0.73687 0.737 (3 s.f.) X X Y X X Y         1 2 83 N 690, 3 3 X X Y         1 2 P 680 700 0.94272 3 0.943 (3 s.f.) X X Y             TJC JC2 Prelim 9758/2019/02/Q9 A Wheel Set refers to a set of wheel rim and tyre. The three types of wheel sets are the Clincher Bike Wheel Set, Tubular Bike Wheel Set and Mountain Bike Wheel Set. The weight of a rim of a Clincher Bike Wheel Set follows a normal distribution with mean 1.5 kg and standard deviation 0.01 kg. The weight of its tyre follows a normal distribution with mean 110 g and standard deviation 5 g. (i) Let C be the total weight in grams of a randomly chosen Clincher Bike Wheel Set in grams. Find   P 1620 C  . [3] (ii) State, in the context of the question, an assumption required in your calculation in (i) . [1] Let T be the total weight in grams of a Tubular Bike Wheel Set, where   2 N , 15 T  . (iii) The probability that the weight of a randomly chosen Clincher Bike Wheel Set exceeds a randomly chosen Tubular Bike Wheel Set by more than 150 g is smaller than 0.70351 correct to 5 decimal places. Find the range of values that  can take. [5]

Let M be the total weight in grams of a randomly chosen Mountain Bike Wheel Set with mean 1800 g and standard deviation 20 g. (iv) Find the probability that the mean weight of 50 randomly chosen Mountain Bike Wheel Sets is more than 1795 g. [3] TJC JC2 Prelim 9758/2019/02/Q9 (Solutions) 9(i) C ~ N(1500 + 110, 10 2 + 5 2 )   P 1620 0.185546 0.186 A    (3sf) (ii) Assume that the weight of a randomly chosen Clincher Bike rim and tyre are independent of each other. (iii)   2 1610 , 125 15 C T N        1610 , 350 C T N      P 150 0.70351 C T    150 1610 P 0.70351 350 Z            1460 P 0.70351 350 Z            1460 0.534523 350      1449.9999   1450   (iv) Since 50 n  is large, by Central Limit Theorem, 400 ~ 1800, approxiately 50 M N         ~ 1800, 8 approximately M N   1795 0.96146 0.961 P M    (3sf) TMJC JC2 Prelim 9758/2019/02/Q9 The masses in grams of Envy apples have the distribution   2 N ,   . (i) For a random sample of 8 Envy apples, it is given that the probability that the sample mean mass is less than 370 grams is 0.25, and the probability that the total mass of these 8 Envy apples exceeds 3000 grams is 0.5. Find the values of  and  . [4] For the rest of the question, use 380   and 20   .

(ii) Find the probability that the total mass of 8 randomly chosen Envy apples is between 2900 grams and 3100 grams. [2] The masses in grams of Bravo apples have the distribution   2 N 250,18 . To make a fruit platter, a machine is used to slice the apples and remove the cores. After slicing and removing the cores, the mass of an Envy apple and the mass of a Bravo apple will be reduced by 30% and 20% respectively. A fruit platter consists of 8 randomly chosen Envy apples and 12 randomly chosen Bravo apples. (iii) Find the probability that the total mass of fruits, after slicing and removing the cores, in a fruit platter exceeds 4.5 kg. [4] (iv) State an assumption needed for your calculations in parts (ii) and (iii) . [1] To beautify the fruit platter, fruit carving is done on the apples after slicing and removing their cores. The carving reduced the masses of each apple (after slicing and removing its core) by a further 10%. Let p be the probability that the total mass of fruits in a fruit platter, with carving done, exceeds 4.1 kg. Without calculating p , explain whether p is higher, lower or the same as the answer in part (iii) . [1] TMJC JC2 Prelim 9758/2019/02/Q9 (Solutions) Qn Solution Mark Scheme 9 Normal and Sampling Distributions [12] (i) Let A be the mass of a randomly chosen Alpha apple in grams. Given   2 N , A   2 N , 8 A         and   2 1 2 3 8 ... N 8 ,8 A A A A         1 2 8 P ... 3000 0.5 8 3000 375 A A A μ μ         

  2 2 P 370 0.25 370 P 0.25 8 370 375 0.67449 8 21.0 (3s.f) A μ Z σ σ σ                      (ii)   2 ~ N 380, 20 A Let     2 1 2 8 ... ~ N 8 380, 8 20 N 3040, 3200 S A A A          P 2900 3100 0.849 (3s.f.) S    (iii) Let B be the mass of a randomly chosen Beta apple in grams.   2 ~ N 250,18 B Let     2 1 2 12 ... ~ N 12 250,12 18 N 3000, 3888 T B B B        . Let 0.7 0.8 C S T   .   2 2 E( ) 0.7 3040 0.8 3000 4528 Var( ) 0.7 3200 0.8 3888 4056.32 N 4528, 4056.32 C C C              P 4500 0.670 (3 s.f) C   (iv) Assume that the distributions of the masses of all apples are independent of one another.       P 0.9 4100 P 4555.6 P 4500 p C C C       Thus p is lower than the answer in part (iii) . VJC JC2 Prelim 9758/2019/02/Q10 In this question, you should state clearly the values of the parameters of any distribution you use. A bus service plies from a point A in the city, through a point B , and then to its terminal station at point C . Journey times in minutes from A to B have the distribution . (i) Find the probability that a randomly selected bus journey from A to B is completed within 35 minutes. [1]   2 N 28,4

The journey times in minutes from A to C , have the distribution . (ii) The journey times in minutes from B to C have the distribution . Given that the journey times from A to B are independent of the journey times from B to C , find the value of and show that = 7.04. [3] (iii) Find the set of values of k such that at least 90% of all journey times from A to C can be completed within k minutes. [2] The performance of the bus operation is deemed as “unreliable” if a random sample of 70 journeys from A to C yields a mean journey time exceeding 47 minutes. (iv) Two independent random samples of 70 journeys from A to C are taken. Find the probability that both samples will result in the performance of the bus operation to be deemed as “unreliable”. [3] To improve the reliability performance of the bus operation, more bus lanes are introduced and some bus stops along the bus route are removed. The journey times from A to B are now reduced by 10%, and the journey times from B to C now have the distribution . (v) Find the probability that two journeys from A to C are completed within a total of 90 minutes. [4] VJC JC2 Prelim 9758/2019/02/Q10 (Solutions) 10i Let X min be the journey times from A to B. X ~N(28,4 2 )   P 35 0.959945 0.960 X    10ii Let Y min and W min be the journey times from B to C and from A to C respectively. Y ~ Since X and Y are independent, W = X + Y       E E E 46.2 28 18.2 W X Y              2 2 2 2 Var Var Var 4.8 4 23.04 16 7.04 W X Y          10iii W ~ N(46.2,4.8 2 )   P 0.9 W k   By GC,   P 52.351 0.9 W     2 N 46.2,4.8   2 N ,    2    N 1,8     2 N ,  

  : 52.4 k k   10iv Let W min be the mean bus journey time from A to C in 70 such journeys. 2 4.8 ~ 46.2, 70 W N           2 2 P 47 0.081593 0.0066574 0.00666 W         10v Let T min be the new journey times from A to C. E( T ) = 0.9 28 17.2 42.4    Var( T ) = 2 2 0.9 4 8 20.96        1 2 ~ N 42.4,20.96 ~ N 84.8,41.92 T T T    1 2 P 90 0.78905 0.789 T T     YIJC JC2 Prelim 9758/2019/02/Q9 A company produces car batteries. The life, in months, of a car battery of the regular type has the distribution   2 N ,   . The mean life of 4 randomly selected car batteries of the regular type is denoted by X . It is given that     P 36.1 P 49.1 0.03355 X X     . (i) State the value of  and show that 7.10   , correct to 2 decimal places. [4] (ii) Find the smallest integer value of k such that more than 90% of the car batteries of the regular type have a life less than k months. [2] (iii) Past experience shows that 25% of the car batteries of the regular type with lives less than 36 months are due to bad driving habits. A random sample of 100 car batteries of the regular type is selected. Find the expected number of these car batteries which will have lives each less than 36 months due to bad driving habits. [2] (iv) After research and experimentation, the company produces a premium type of car battery using an improved manufacturing process which is able to increase the life of each car battery by 10%. Find the probability that the total life of 5 randomly chosen car batteries 52.351 0.9 46.2

of the premium type is more than the total life of 6 randomly chosen car batteries of the regular type. [4] YIJC JC2 Prelim 9758/2019/02/Q9 (Solutions) 9(i) 2 ~ N , 4 X         36.1 49.1 42.6 2       P 36.1 0.03355 X   36.1 42.6 P 0.03355 2 Z             13 1.8310     7.0999 7.10 (2 d.p.) (shown)    9(ii) Let X be the life, in months, of a car battery of the regular type.   2 ~ N 42.6, 7.10 X P( ) 0.9 51.699 Smallest integer value of 52 X k k k      Alternative P( ) 0.9 51: P( ) 0.882 0.9 52: P( ) 0.907 0.9 Smallest integer value of 52 X k k X k k X k k             9(iii)   P 36 0.17629 X   Expected number of car batteries which are due to bad driving habits     0.25 100 P 36 4.41 (3 s.f.) X    9(iv) Let Y be the life, in months, of a car battery of the premium type. 1.1 Y X      2 2 ~ N 1.1(42.6), 1.1 (7.10 ) ~ N 46.86, 60.9961 Y Y

      1 2 5 1 2 6 2 Let ~N 5(46.86) 6(42.6), 5(60.9961) 6(7.1) ~N 21.3, 607.4405 A Y Y Y X X X A A              P 0 0.194 A  