Solutions for 02402 exam 15. December 20111

And since this is 0.3 we can findλ 15min = − log(0.3) = 1.2Since µ = λ in a poisson (page 106) this number also expresses the expected(average)number of arrivals during 15 minutes. The second part we find by finding the intensity(average) for a 60 minutes period:And hence:λ 60min = 4 · λ 15min = 4.8P (X ≥ 8) = 1 − P (X ≤ 7) = 1 − 0.887 = 0.113found in Table 4. Also to be found in R as 1-ppois(7,4.8). Henc the correct answeris 2: µ = 1.2 arrivals per 15 min and P (X 60min ≥ 8) ≈ 11%Exercise VOn a shelf 9 apparently identical ring binders are postioned. It is known that 2 of thering binders contain statistics exercises, 3 of the ring binders contain math problems and4 of ring binders contain reports. Three ring binders are sampled without replacement.Question V.1 (11) The random variable X describes the number of ring binders withstatistics exercises among the 3 chosen ones. The mean and variance for the randomvariable X is? This is a Hypergeometric distribution with a = 2, N = 9, n = 3. Themean and variance of this distribution is given on page 96 and page 98:µ = n · aN = 3 · 29 = 2 3σ 2 = n · aN (1 − a N )(N − nN − 1 ) = 3 · 29 (7 9 )(9 − 39 − 1 ) = 7 18Hence the correct answer is 1: µ ≈ 0.67 and σ 2 ≈ 0.39Question V.2 (12) The probability (P 1 ) that all the three chosen ring binders containreports and the probability (P 2 ) to chose exactly one of each kind of ring binder are?Since there are a = 4 that contain reports (and 5 that don’t) we use this hypergeometricdistribution (Y ) to find P 1 : (page 81)(43) (50)P 1 = P (Y = 3) =) = 1 21(93For P 2 we use the same principle as for the hypergeometric - the basic probabilityapproach - the denominator counts the total number of possibilities - the numerator6