The public transport system is aware that one person in twenty does not buy a ticket for their trip. The manager decides to challenge 200 travelling people as to charger they have a valid ticket or not. (a) calculate the probabilty that everyone had a ticket, (b) calculate the probabilty that at most 3 people had no ticket, and (c) calculate the probabilty that between 7 and 14 (inclusive) had no ticket. I thought binomial? But then 200 is too big for it? Thanks for any help!
Yes, you are right in considering the binomial distribution for this problem. However, you are also correct in noting that with a sample size of 200, it becomes impractical to use the binomial distribution directly due to computational limitations. Fortunately, there is a well-known approximation called the normal approximation to the binomial distribution that can be used in cases like this.
Before we proceed, let's define the variables:
- n: sample size (number of people), which is 200 in this case.
- p: probability that a person does not buy a ticket, which is 1/20 or 0.05 (since one in twenty does not buy a ticket).
- q: probability that a person buys a ticket, which is 1 - p = 19/20 or 0.95.
(a) To calculate the probability that everyone had a ticket, we need to find the probability that none of the 200 people did not buy a ticket. As per our defined variables, this probability corresponds to p^200.
Probability (everyone had a ticket) = p^200 = (0.95)^200, which is approximately 0.005 (rounded to three decimal places).
(b) To calculate the probability that at most 3 people had no ticket, we need to find the sum of the probabilities of having 0, 1, 2, and 3 people without tickets.
Probability (at most 3 people had no ticket) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
For each term, we can use the binomial probability formula, but as mentioned earlier, we will approximate it using the normal distribution due to the large sample size.
We use the continuity correction here, which means we consider the values between 0 and 0.5 as 0 and between 3 and 3.5 as 4.
P(X = 0) ≈ P(0 – 0.5 < X < 0 + 0.5) ≈ P(-0.5 < X < 0.5)
Using the normal approximation, we can find the approximate probability using the z-score formula:
P(-0.5 < X < 0.5) ≈ P((-0.5 - np) / sqrt(npq) < Z < (0.5 - np) / sqrt(npq))
Plugging in the values, we get:
P((-0.5 - 200*0.05) / sqrt(200*0.05*0.95) < Z < (0.5 - 200*0.05) / sqrt(200*0.05*0.95))
Simplifying further, we get:
P(-0.5 < Z < 0.5)
Now, we can look up the values in the standard normal distribution table or use statistical software to find the probability. In this case, it is approximately 0.382.
Similarly, you can calculate P(X = 1), P(X = 2), and P(X = 3), and sum them up to get the desired result.
(c) To calculate the probability that between 7 and 14 (inclusive) people had no ticket, we proceed similarly to part (b).
Probability (7 ≤ X ≤ 14) = P(X = 7) + P(X = 8) + ... + P(X = 14)
Again, we can use the normal approximation to estimate each term using the continuity correction. Then, sum up the probabilities to get the desired result.
Keep in mind that while the normal approximation provides reasonable estimates for large sample sizes, it may not be absolutely precise.