THE ROYAL STATISTICAL SOCIETY HIGHER CERTIFICATE EXAMINATION MODULE 2

Transcription

1 THE ROYAL STATISTICAL SOCIETY HIGHER CERTIFICATE EXAMINATION NEW MODULAR SCHEME introduced from the eaminations in 7 MODULE SPECIMEN PAPER B AND SOLUTIONS The time for the eamination is ½ hours The paper contains four questions, of which candidates are to attempt three Each question carries marks An indicative mark scheme is shown within the questions, by giving an outline of the marks available for each part-question The pass mark for the paper as a whole is 5% The solutions should not be seen as "model answers" Rather, they have been written out in considerable detail and are intended as learning aids For this reason, they do not carry mark schemes Please note that in many cases there are valid alternative methods and that, in cases where discussion is called for, there may be other valid points that could be made While every care has been taken with the preparation of the questions and solutions, the Society will not be responsible for any errors or omissions The Society will not enter into any correspondence in respect of the questions or solutions RSS 6

2 (i) The random variable X follows the Poisson distribution with probability mass function f( ) λ λ = e, =,,,! (a) Show that the mean and variance of X are both equal to λ (b) State the Poisson approimation to the binomial distribution, indicating the circumstances in which it is appropriate (5) (ii) A civil servant calculates weekly social security payments for unemployed adults These payments vary according to claimants' circumstances, and errors may occur Over a long period of time, the probability of a wrong calculation has been found to be 75 Find to decimal places the eact probability that a sample of contains (a) wrong calculation, (b) wrong calculations (7) (iii) Repeat part (ii) using the Poisson approimation to the binomial distribution In each case find to two significant figures the percentage error in the Poisson calculation Comment briefly on your results (8)

3 Among the inhabitants of Altamania, height, H say, is distributed Normally with mean 6 cm and standard deviation cm, ie N(6, 6) (i) (ii) Find the proportion of the population whose heights are within one standard deviation of the mean Find also the proportion of the population who are more than 68 cm tall (5) The Altamanian Police Force (APF) is restricted to persons who are more than 68 cm tall, and may be assumed to consist of a random sample of Altamanians satisfying this condition Find (a) the median height of members of the APF, (b) the proportion of members of the APF who are more than 7 cm tall () (iii) Assume that the mean and standard deviation of height among members of the APF are 695 cm and 35 cm respectively Find an approimate value for the probability that the mean height of a random sample of 5 members of the APF is more than 7 cm (5) 3 The events A, B and C have respective probabilities, and, and A, B 3 and C are, respectively, the complements of A, B and C (i) Given that A, B and C are mutually independent, find (a) P( A B C), (b) P( A C A B) (8) (ii) Now let A and B be independent, A and C be independent and B and C be independent, so that A, B and C are pairwise independent, and let P( A B C) = Find in terms of (a) P( A B C), (b) P( A C A B), (c) P A B C Also find the maimum and minimum possible values of ()

4 The random variable X has the geometric probability mass function (pmf) given by f (), where f = p p, =,,,, < p< (i) Sketch the graph of f () for the case p = /3, for 5 (ii) Obtain the mean and variance of X (5) () (iii) For any non-negative integer, show that P( X ) = ( p), and deduce that for any non-negative integers l and m ( ) P X l+ m X l = P X m Interpret this result (5) (iv) The random variable Y has pmf g(y), where g y y = θ θ, y =,,,, < θ < X and Y are independent, and the random variable Z is defined as the minimum of X and Y, ie Z = min(x, Y) By noting that P(Z z) = P(X z and Y z), find an epression for P(Z z), where z is any nonnegative integer By considering P(Z z) P(Z z + ), or otherwise, show that ( θ) z ( θ θ) P Z = z = p p+ p, z =,,, Identify the form of this distribution and hence write down E(Z) and Var(Z) (6)

5 SOLUTIONS Question (i) (a) λ λ λ e λ λe λ e λ EX = = = λ = λ = λ! ( )!! = = = We have Var(X) = E(X ) {E(X)} E(X ) can be found similarly, but it is easier to use E(X ) = E[X(X )] + E(X) and thus Var(X) = E[X(X )] + E(X) {E(X)} λ λ λ e λ λ e λ e λ EX [ ( X )] = ( ) = = λ = λ = λ! ( )!! = = = Thus Var( X ) = λ + λ λ = λ [The probability generating function or moment generating function could also be used this work is in Module 5] (i) (b) The binomial distribution with parameters n and p may be approimated by the Poisson distribution with parameter np if n is large and p is small As a "rule of thumb", ½ np gives an indication of how large n should be and how small p should be (If np >, a Normal approimation to the binomial may be better) (ii) Let X = number of wrong calculations We have X ~ B(, 75) P X = = = = 3353() P X = = ( ) ( 75) ( 995) () = = 3 Solution continued on net page

6 (iii) We approimate using X ~ Poisson( 75 = 5) With this, 5 ( = ) = 5e = 337( ) P X giving a percentage error of ( ) = 8%, and e P( X = ) = = 7 7! 5 5 giving a percentage error of ( ) = 58% [Note These percentage errors might come out slightly differently if more accuracy is kept in the binomial and Poisson probabilities] Both approimations are remarkably accurate, with percentage errors well below % The approimation for X = (one wrong calculation) is the more accurate of the two That approimation is an underestimate; the other is an overestimate

7 Question H N6, (i) P(56 < H 6) = P < Z ( [where Z N,) ] = P( < Z ) =Φ Φ( ) =Φ Φ( ) by symmetry, ie Φ() { } = 686 [Φ denotes the cdf of the standard Normal distribution as usual] P( Z ) P H > 68 = > = Φ = 8 (ii) (a) The relevant portion of the Normal distribution of H is that beginning at 68, which corresponds to the value Z = The median value m within this portion has ( 8) probability above it, ie, so Φ( m) = = 9886, corresponding to Z = 77 The corresponding value of H is μ + σ Z which is 6 + ( 77) = 69 cm (b) H = 7 corresponds to Z = = 5 ; we have Φ ( 5) = 9938, so P( H > 7) Φ ( 5) = 6 Conditional on H > 68, the probability is P H > 68 = 6 8 = 7 (iii) Mean height of 5 members 35 N 695, 5 P(mean > 7) = P Z > = Φ = Φ 89 ( 35 / 5) 35 = 9677 = 33

8 Question 3 P A P B P C 3 = = = (i) (a) By the given independence, P( A B C) = P( A) P( B) P( C) = = 3 (ii) (b) P( A C A B) ( ) ( ) ( ) P A C A B P A C B = = P A B P A B 3 = = 3 A B C is P(A)P(B), etc, and hence the values 3, 8 and 6 are found The others follow using P(A), P(B) and P(C) Using pairwise independence, the value of P( A B) P A B C = + from the diagram 6 (a) (b) P( A C A B) ( ) ( ) ( ) P A C A B P A C B = = P A B P A B 6 + = = 3 + ( 6 + ) + ( 6 ) 9 P A B C = + (c) Since all probabilities must lie in [,], we have and 8 8, ie

9 (i) Question Probability mass function: f = p p, =,,,, < p< /3 P(X = ) /9 /7 3 5 (ii) E( X) = ( p) p This can be found in several ways; the method shown = here (using geometric series) is perhaps not the most efficient, but should be easy to understand We have = 3 { } ( p) p = p + ( p) + ( p) + 3( p) + 3 = p{ + ( p) + ( p) + ( p) ( p) + ( p) ( p) + + } 3 p ( p) ( p) p p p = p = = ( p) ( p) ( p) p ( p) p Similarly, ( ) ( p) + ( p) = p EX = p p= Thus Var(X) = E(X ) {E(X)} = p + p p p = = p p p [The probability generating function or moment generating function could also be used this work is in Module 5] Solution continued on net page

10 r ( p) p (iii) P( X ) = ( p) p [geometric series] = = ( p r= ( p) ) (for = P( A B),,, ) We now use P( A B) = and take the event A as "X l + m" P( B) and the event B as "X l", so that A B = A Thus + ( p) l ( p) l m m ( ) ( P X l+ m X l = = p = P X m This is the "lack of memory" property of a geometric distribution ) z (iv) By independence, ( P Z z = P X z P Y z = p θ ( θ) z z+ { } {( θ) } P Z = z = P Z z P Z z+ = p p {( )( )} z ( ) ( )( ) z { } ( = p θ + p+ θ pθ = p θ p+ θ pθ, for z =,, This is a geometric distribution as given at the start of the question with p replaced by p + θ pθ Hence, from part (ii), p θ + pθ EZ = p + θ pθ p θ + pθ p+ θ pθ, Var(Z) = ) ) z