Question

5 Peu DISUutions Graded (20%) Due: 15/11/2019 Question 1 State three (3) conditions for a situation to described using a Binomial model. Write down the pdf for the binomial distribution and give a brief description of each term A random variable X has the distribution Bin( 12,p). (a) Given that p = 0.25 find (0) P(X<5) (ii) PCX >7) (b) Given that P(X = 0) = 0.05, find the value of p to 3 decimal places. (c) Given that the variance of X is 1.92, find the possible values of p. Question 2 In a large restaurant an average of 3 out of every 5 customers ask for water with their meal. A random sample of 10 customers is selected. (a) Find the probability that () exactly 6 ask for water with their meal, (ii) less than 9 ask for water with their meal. (b) A second random sample of 50 customers is selected. Find the smallest value of n such that P(X<n) 0.9 where the random variable X represents the number of these customers who ask for water

Answer 1

The & Conditions for a situation that can be described using Binomial distributions are & i Events must be of the from Bernoulli events le. 1x le a random variable, which ban take value o & with probability pand up. Low Each of each observation other is indepeodest an The humber of observation - is fixed. 2 The por of Binomial distribution | Plant ( or qu-x. where - is the nunmber of observation, P- probability that eveot is occurring in a Bernoull ercot q = (1-P) ? 2 is the number, which definen a succes, Such'w P(x26) implies probabilite of obtaining & successe out of in trials : ② Griveo that Xo Bin (12, p) Whel p2o.25 UD PC**5)= 3 (13) 10.36%20 nzo T z 0.542356 e 6.8424

CD PCX73) - 2 13) (oud? (0. A PCX71) > 0.014252 20.0143 (round 6 - Gneo that P(x 20) = 0.05 > (13) Pº(10) 2005 2 - ( 1² 20.05 42 - (1P) & (0.05) = 0 3 790 R = -0.7790 zoa2o9. i @) V(x192 - up qz 1.92 to 12x pei-p = 1.92 . 121-12p² 21.92 p-p² = 1.92 = z b.16 12-7 F) p²p +0.1620 +1 + VI 4*0.16 1+ 6.6 Rz 2

either on Pe 0.4 - P-0.8 or p=0.2 .. the possible values are P2o.5 or pro a Here P2315 100 P(x=6) = ( 3 ) (9) (22 uzlo. Xou Bin (101315) P(x26) z 9 2 0.2508 (17 p(x(9) = z 1107 13157/21 No 2 (21 z 0.95364 Here a represents the events that a customes asly for the water (6 Given n250 pz 3/5 E p/Xen) 209. 120 Adding cumulatively, we have I p(xan) 20.8428 ang

2 Р(х ги) 2 Иго 0:4644 бinco P(x2 и) zorao49 Ксиаvе ие 30 . и 36/1

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