who have this disease 5 35 5 1 21 1 14 4 35 5 6 2 21 2 14 3 35 5 3

# Who have this disease 5 35 5 1 21 1 14 4 35 5 6 2 21

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who have this disease Prob(X) 0 ( 21 0 )( 14 5 ) ( 35 5 ) 1 ( 21 1 )( 14 4 ) ( 35 5 ) 6
2 ( 21 2 )( 14 3 ) ( 35 5 ) 3 ( 21 3 )( 14 2 ) ( 35 5 ) 4 ( 21 4 )( 14 1 ) ( 35 5 ) 5 ( 21 5 )( 14 0 ) ( 35 5 ) The expected number of women who have the disease given that 5 persons have the disease is 3 10. The voltage in a certain circuit is a random variable with mean 120 and standard deviation 5. Sensitive equipment will be damaged if the voltage is not between 112 and 128. Use Chebyshev’s inequality to bound the probability of damage. X = voltage in a certain circuit. Probability of damage = 1 - P(112 < X < 128) P ( 112 < X < 128 ) = P ( 8 < X 120 < 8 ) = P ( 1.6 5 < X 120 < 1.6 5 ) 1 1 1.6 2 P ( 112 < X < 128 ) 0.61 1 P ( 112 < X < 128 ) 1 0.61 = 0.39 There is less than 39% probability that the equipment is damaged. 11. I am the only bank teller on duty at my local bank. I need to run out for 20 minutes, but I don’t want to miss any customers. Suppose the arrival of customers can be modeled by a Poisson distribution with mean 2 customers per hour. What are the chances that I can take a customer-free 20 minute break? X = number of customer arriving in 20 minutes P(2/3) P ( X = 0 ) = e 2 3 = 0.51 There is 51% chance that I can take a customer-free 20 min break. 7
12. On an average, the number of hits on a certain web page are 20 hits per day. What is the probability of observing more than 30 hits per day? (Hint: write the formula using Poisson distribution). Ans: E(X) =20 P(X>30) = x = 31 20 x e 20 x ! Probability of exactly 30 hits? 13. A family has 5 natural children and has adopted 2 girls. Each natural child has equal probability of being a girl or a boy, independently of the other children. Let X denote the number of girls out of the 7 children. Find the PMF of X , E(X) and V(X) . Ans: Given the information in question X=Y+2 where Y∼Binom(5, 1/2 ) . Consequently, PMF of X is p X (t)=p Y (t−2)= 5 C t-2 ( 1 2 ) 5 t=2,3,4,5,6,7 E(X)=E(Y+2)=E(Y)+2= 5/2 +2= 9/2 V(X)=V(Y+2)=V(Y)= 5/4 14. Mark works for an organization committed to raising money for Alzheimers research. From past experience, the organization knows that about 20% of all potential donors will agree to give something if contacted by phone. They also know that of all people donating, about 5% will give \$100 or more. On average, how many potential donors will he have to contact until she gets her first \$100 donor or more. X = number of donors to be contacted to get first big donation G(p) p = probability of getting big donation =0.2*0.05 = 0.01 E(X) = 1/p = 100 On average Mark has to contact 100 donors before he gets his first \$100 donor or more 15. A researcher studying crime is interested in how often people have gotten arrested. Let X be a Poisson random variable with parameter 3. X is the number of times that a random person got arrested in the last 10 years. However, data from police records are being used for the researcher’s study, and people who were never arrested in the last 10 years do not appear in the records. In other words, the police records have a selection bias: they only contain information on people who have been arrested in the last 10 years. So averaging the numbers of arrests for people in the police records

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• Summer '18
• Sagar Arora