Question

A survey taken in a farming village in Nepal asked, "How many bullocks does your household have?" and "How many cows does your household have?" The probability distributions of number of bullocks B and number of cows C shown here are consistent with the data published by the Inter-University Consortium for Political and Social Research (ICPSR). 0 1 2 3 Probability 0.66 0.04 0.28 0.02 0 1 2 3 4 Probability 0.78 0.11 0.06 0.03 0.02 (a) If we draw separate probability histograms for B and C, then both shapes would be which of the following? skewed right skewed left symmetric (b) Which distribution could be called bimodal? the one for bullocks the one for cows both of the above neither of the above (c) which are more likely to be kept in pairs? cows bullocks (d) Find the probability of no more than two bullocks, PB < 2). (Round your answer to two decimal places.)

Answer 1

Probability distribution of bullocks is given by B 012 3 Probability (0.66 1004 | 0.28 0.02 Probability distribution of cows is given by Probability 078 0.11 0.06 0.03 0.02 2) If we draw seperate histograms for Band C, then both the shapes will be skewed night ds probabilities are decreasing in nahire C (except at B=2) Neither cow nor bullock distribution can be called bimodal. As both of them have single mode . Mode is the value at which probability of r.v. is maximum and it is called bimodal if the maximum probability is at two points). Hence correct option is neither of above. c) Bullocks are more likely to be kept in pairs. e) Plno more than two cows) je. PCC52) · PCCS2) - PCC=0) + PCC=1) + P(C=2) = 0.78 +0.11 +0.06 1. PCCC2) = 0.95 6) Plouring owning no cows and no bullocks) je PC Bro and C=0) PCB-0 and C=o) = PCB=0). P(CO) distribution of bullocks and = 0.66X076 cows is independent - 0.51 Hence correct option is res. d) PC no more than two bullocks) = PCBS2) PCBS2) - PCB=0) + PCB-1) +PCB-2) = 066 +0.04 +0.28 P(BS2) = 0.98

g) using lu=X, P(as) + 22 P(22) + ... Tu Plak) MB = 11.P (W ) + 12. PCX2=17) + 13. PLX3=13) +14 PCX214) = 0.(066)+1(0.04)+2(0.28) +310-02) = 0+0.04 +0.56+0.06 UB = 0.66 Mean no. of bullocks owned by a randomly chosen boushol are UB=0.66 h) Mic=x. P(x1=1) +2. P[x = x₂ 7+tz - P[X3=13] + 24. P[X4 X4]+ X5 P[xs-xs] - 0(0-78) + 1 (0-1) + 210:06) + 310-03) +460-02) - oto 11 +0.12 +0.09+0.08 llc = 0.4 Mean no. of bu cows owned by Tandomely chosen household Uc=0.4 1) By given information; PCB=0)- no. of households having o bullocks Total no. of households = 66 = 0.66 100 - 066 PCB=0) = 0.66 simillary PCB=1) = 4 = 0.04 PCB= 2) = 28 = 0-28 PCB=3) = Too = 0-02 Average = (0.66) + 110.04 ) + 2(0-28+ 3(0.02 ) = 0.66 Hence average is equal to the mean of bullocks 1) PCC-0-4) As the probability will be zero as the random variable under consideration is no of cows owned by a household, hence no of cows is a disvete riv. with take values 0,1,2... it can't be an inte 1. PCC= 0.4) - 0.0 fraction