Question

In a pediatrician's office, the probability of a "no show" for any checkup appointment on any given day is 0.10. Suppose there are 10 appointment for one day. What is the random variable in this situation? What is defined as a "success"? What is the probability of success? What is defined as a "failure"? Calculate the probability of failure. Write the probability mass (distribution) function.

Answer 1

2) What is the random variable in this situation ? In the given situation, the random Variable is Binomial rondom Varialde because 1 # of appoiutusut Ce triels) = 10 is fixed 2) All the appoints of are independent & identical in the aspect of 'no show Considering an appoiutuneit, it results in two outcomes either no show (success) or not (failure 7 Probability of "no show for any

appoint appointment (-e probability of success) = 0.10 is same for every appointment ( what is success ? As explained above, "no show" is a success (c) What is probability of success? As explained above, Probability of success = Probability of "no show = 0.10 (d) What is defined as a failure" ? In the context of our problem, failure is that "doctor sees patient (e) What is probability of failure? Probability of failure = 1-0-10 = 0.9 tre (4 Probability Mass function: n=10 patients, x takes 0, 1, 2, 3, 4 -- Formula: P(x=x) = ng kr, a Hese p=o.l, q=0.9 P(x=0) = loç..(0:1?(0.9) P(x=1) = 100.Co.D' Co.glby P (X=2) = 106. C.)? 6.9102 p (X=3) = 103. 6.0574 JO. И-Х x 1o-o = 0.3487 0.3874 0.1937 (.. (0.9) 103

10.4.CO.D". (0.90-4 & Co.gl8 the < 0.00000) P(X=4) = 0.0112 P(X=5) = 100,(6.")*5 = 0.0015 P(X=6) 109. (6.1) (0.4) 06 = 0.0001 P(x=7 a roç. (6.1? (04) 67 0.00000g P(x=8) P(X=9) = 109. (..)". Co.go.9 PX=0) loc.Co.)" (0.9) NOTE: We can find the above probabilities using Binomial online calculator Probability Mass function: X 4 P(x=) 0.3487 0.3874 0.1937 0.0574 0-01120-0015 0-000) 71 < 0.00000) 10-10 (0.9) < 0.000001 0 2 3 10 7 8. 9 <0-0000D 1 <0.000001 0.000009 <0.000001 P(X<4 : P(X<4) = P(x =o) + P(x=1) + P(x=2) +P(X=3) = 0.3487 +0.3874 +0.1937 +0.0574 0.9872 Ans P(x2): P(X72) = |- [PCX=0) + P(X=D]

|- [0.3487 +0.3874 = (-0.7361 = 0.2639 Ars P(X=6): From the above probability distribution, P(x=6) = 0.0001 Ans DUX P C4 < X <8) PC4< X<B) = P(X=5)+P(x=6)+ P(x =7) = 0.0015+ 0.0001 + 0.000009 = 0.0016 (nearly Probability that doctor sees every patient scheduled le PCX - 0): P(X=0) =0.3487 Expected number of ho shows for a given day : For a Binomial distribution Expected number of successes = Mean nop loxool It means there is there is l "no show & out of to patients, on average

Vanlance for the number of no shows" For a Binonial chistribution, Variane = n. p. q = 10 x0.180.q=0.9 Ans standard deviation for the number of "no shows" We have, Standard deviation =W Variance No.9 0.9487 Ars