Question 1088433: Hi there, this is the first question I asked (which was answered)
(1). Jack and Jill work in the school office. They are recording and analysing the number and length of phone calls received. The number of telephone calls that are answered by the office has a mean of 12 calls per hour.
a) Find the probability that there are no calls in any given 15 minute interval.
b) Find the probability there are no calls in half an hour.
Poisson distribution with calls proportional to the time. Expect 3 calls in 15 minutes, so lambda is 3.
Probability x=0 when lambda is 3:
formula is e^(-x)*lambda^x/x!
0!=1
e^(-3)*3^0/0!
=e^(-3)=0.0498.
I understand this but now I am struggling with another question (I'm not sure if you need the probablility from Q1 above to help solve)
Question:
(2). Jack and Jill have found that they are able to deal with 30% of calls. (70% are transferred to other people in the school).
a)Find the probability that, from the next 10 calls, Jack and Jill will be able to deal with fewer than 4 calls.
b) If Jack and Jill were able to deal with fewer than four of the calls then find the probability they dealt with 2 calls.
Thank-you for your help! I've tried butI'm not sure what to do! I really really appreciate it
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website! Question:
Jack and Jill have found that they are able to deal with 30% of calls. (70% are transferred to other people in the school).
a)Find the probability that, from the next 10 calls, Jack and Jill will be able to deal with fewer than 4 calls.
b) If Jack and Jill were able to deal with fewer than four of the calls then find the probability they dealt with 2 calls.
Thank-you for your help! I've tried butI'm not sure what to do! I really really appreciate it
Solution:
It looks like that you are working with different probability distributions at the same time. The first step in solving these problems is to identify the probability distribution that fits the situation. After that it would be just substitution of parameters and calculator work.
Jack and Jill's situation would fit the conditions for a binomial distribution because it must satisfy the following conditions (and it does):
1. Bernoulli trials, i.e. exactly two possible outcomes (can answer or not)
2. Number of trials is known before and constant throughout the experiment,
i.e. independent of outcomes (10 calls)
3. All trials are independent of each other (assumed from context)
4. Probability of success is known, and remain constant throughout trials (30%)
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.
Now N=10, p=0.30
(a) x=0,1,2,3
P(x<4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)
=∑C(N,x)p^x(1-p)^(N-p)
=C(10,0)0.3^0(0.7)^(10)+C(10,1)0.3^1(0.7)^(9)+C(10,2)0.3^2(0.7)^(8)+C(10,3)0.3^3(0.7)^(7)
=0.0282475249+0.121060821+0.2334744405+0.266827932
=0.6496107184
=0.650 approx.
(b) x=2
From part a,
P(x=2)=0.23347=0.233 approx.
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