For reference:

Consider the hospital patients in Example 2.6. Two patients are selected randomly, with replacement, from the total patients at Hospital 1. What is the probability mass function of the number of patients in the sample who are admitted?

Determine the cumulative distribution function for the random variable in Exercise 3.1.12; also determine the following probabilities:

a. P(X 1.5)

b. P(X 3)

c. P(X > 2)

d. P(1 X 2)

Problem I need help with:

How do you do this in matlab?

Determine the mean and variance of the random variable in Exercise 3.1.12.

I tried;

p=0.2413 x=0:2 p_x=geopdf(x,p) P334M=sum(x.*p_x) var=sum(x.^2.*p_x)*(P334M)^2

But I'm not getting what I got on paper calculations

Determine the mean and variance of the random variable in Exercise 3.1.12.

calculated as the number of visits in A plus the number of visits in B minus the number of visits An B (that would otherwise be counted twice) = 5292 +953 - 195 = 6050. Practical Interpretation: Hospitals track visits that result in LWBS to understand resource needs and to improve patient services. EXAMPLE 2.6 | Hospital Emergency Visits The following table summarizes visits to emergency depart- ments at four hospitals in Arizona. People may leave with- out being seen by a physician, and those visits are denoted as LWBS. The remaining visits are serviced at the emergency department, and the visitor may or may not be admitted for a stay in the hospital. Let A denote the event that a visit is to hospital 1, and let B denote the event that the result of the visit is LWBS. Calculate the number of outcomes in An B.A', and A UB. The event A B consists of the 195 visits to hospital 1 that result in LWBS. The event A' consists of the visits to hospitals 2, 3, and 4 and contains 6991 +5640 + 4329 = 16,960 visits. The event A UB consists of the visits to hospital 1 or the visits that result in LWBS, or both, and contains 5292 + 270+ 246 +242 = 6050 visits. Notice that the last result can also be Total LWBS Admitted Not admitted Hospital 2 3 4 5292 6991 56404329 195 270 246 242 1277 1558 666 984 3820 5163 4728 3103 Total 22,252 953 4485 16,814 We can see that 1277 of the 5292 total patients at hospital 1 are admitted. Using this, we can say that the probability that a randomly selected patient at hospital 1 is admitted is Number of patients admitted 1277 p = = 0.2413 Total number of patients 5292 Two patients are selected randomly, with replacement from the total patients at Hospital 1. Let X be the number of patients out of 2, who are admitted. Since we are selecting the 2 patients with replacement, the probability that a randomly selected patient is admitted remains the same for each of the 2 choice we make. We can say that X has a Binomial distribution with parameters, number of trials (number of patients randomly selected) n=2 and success probability (the probability that a randomly selected patient is admitted) p=0.2413 The probability that X=x patients in the sample are admitted is (given by the Binomial distribution) 0.2413"(1 -0.2413)- T =0,1,2 That is P(X=0) = 0.2413(1 -0.2413)2-0 2! = 0112 - 070.2413(1 -0.2413) 2-0 = 0.5756 P(X = 1) = (10.2413'(1 -0.2413)2-1 2! 1!(2-1)! 1,0.2413'(1 -0.2413) 2-1 = 0.3661 P(X =2) = 0.2413(1 -0.2413)2-2 2!(2 - 270.2413(1 -0.2413) 2-2 = 0.0582 P(X =2 the table cumulative mass probabilities of the number of patients in the sample who are admitted is (running sum of P(x)) Cumulative probability PIX2 the table cumulative mass probabilities of the number of patients in the sample who are admitted is (running sum of P(x)) Cumulative probability PIX