SOLUTION: At a particular hospital, records shows that each day, on average, only 80% of the people keep their appointments at the outpatient clinic. Find the probability that when 200 appo

Question 1086776: At a particular hospital, records shows that each day, on average, only 80% of the people keep their appointments at the outpatient clinic. Find the probability that when 200 appointments have been booked: more than 170 patients keep their appointments
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
At a particular hospital, records shows that each day, on average, only 80% of the people keep their appointments at the outpatient clinic. Find the probability that when 200 appointments have been booked: more than 170 patients keep their appointments

Solution:
The situation calls for the binomial distribution. Before proceeding, we need to go through the checklist to qualify the use of the distribution.
1. Bernoulli trials, i.e. exactly two possible outcomes (success = keep or failure=not keep appointments)
2. Number of trials is known before and constant throughout the experiment,
i.e. independent of outcomes. (n=200)
3. All trials are independent of each other. (assumed from context)
4. Probability of success is known, and remain constant throughout trials. (historical information, p=0.8)

Since all criteria are satisfied, we can model with binomial distribution, where the probability of x successes out of N trials each with probability of success p is given by
P(x)=C(n,x)(p^x)(1-p)^(n-x)
and,
C(N,x) is number of combinations of selecting x objects out of N.

Substituting values,
n=200
p=0.8
x>170, or x=171, 172,173,...200.
P(x>170)
=1-P(x<=170)
=1-(sum P(x=i), i=0 to 170)
=1-0.97172 [obtained from software, such as R, pbinom(170,200,0.8)=0.9717234]
=0.02828

Alternatively, for large values of n, we can use normal approximation as long as the binomial distribution is not too skewed np(1-p)>5 (preferably 10), or 200(.8)(.2)=32 > 10.

Steps:
1. find mean=np=200*0.8=160
2. find standard deviation, σ²=np(1-p)=32, σ=√32=4√2=5.656
3. apply continuity correction, x=170+0.5=170.5
find Z=(170.5-160)/5.656=1.85616
4. find P(z 5. P(x>170)=1-0.96828=0.03172

Note that error of approximation is greater than 12%, compared to the exact value using binomial distribution. Values of p closer to 0.5 would give much better approximations.

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