A popular resort hotel has 300 rooms and is usually fully booked. About 4% of the time a reservation is cancelled before 6:00 PM deadline with no penalty. What is the probability that at least 280 rooms will be occupied? Use binomial distribution to find the exact value and the normal approximation to the binomial and compare your answers.
Using Binomial probability function, we are looking for:
P(280≤X≤300)
=Σ C(300,r)p^r*(1-p)^(300-r)
for r=280, 281, ...300, and
C(300,r)=300!/((300-r)!r!)
=0.00896+0.01530+0.02474+0.03777+0.05427+...+0.00004+0.0000
=0.98992
Using normal approximation with continuity correction, we have
n=300, p=0.96,μ=np=288,σ=sqrt(npq)=3.3941
P(280≤X≤300)
=1-Φ(279.5,288,3.3941)
=0.99386
Error = (0.99386-0.98992)/0.98992=0.4%.
If we look at the tail, the performance is much worse. The reason for the poor performance (if we look at the tail) is because
np=300*0.96=288 >> 10
npq=300*0.96*0.04= 11.52 ~ 10.
If we use Camp Paulson approximation, we get
1-P(279≤X≤300)
=0.98982 (error=-0.0096%, much better)
Ref:
http://www.johndcook.com/blog/camp_paulson/
To find the probability that at least 280 rooms will be occupied, we can use the binomial distribution and its normal approximation.
Let's start by using the binomial distribution to find the exact probability.
The binomial distribution formula is given by:
P(X=k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- P(X=k) is the probability of getting exactly k successes
- n is the number of trials
- k is the number of desired successes
- p is the probability of success in each trial
In this case, the number of trials is 300 (number of rooms), and the probability of success is 1 - 0.04 = 0.96 (probability that a reservation is not cancelled).
Now, we want to find P(X>=280), which is the probability of having at least 280 occupied rooms. This can be calculated by summing the probabilities for all values of k from 280 to 300.
P(X>=280) = P(X=280) + P(X=281) + ... + P(X=300)
Using the binomial distribution formula, we can calculate these probabilities step by step.
Let's calculate P(X=280) first:
P(X=280) = C(300, 280) * (0.96)^280 * (1-0.96)^(300-280)
Using the binomial coefficient formula C(n, k) = n! / (k! * (n-k)!), we can calculate C(300, 280) = 300! / (280! * (300-280)!) = 300! / (280! * 20!)
Now, let's calculate P(X=281), P(X=282), and so on until P(X=300) using the same formula.
After calculating all these probabilities, we can sum them up to get P(X>=280).
Now let's compare this result with the normal approximation to the binomial.
To use the normal approximation, we need to calculate the mean (μ) and the standard deviation (σ) of the binomial distribution.
The mean is given by μ = n * p = 300 * 0.96 = 288.
The standard deviation is given by σ = sqrt(n * p * (1-p)) = sqrt(300 * 0.96 * 0.04) = sqrt(11.52) ≈ 3.39.
Using these values, we can approximate P(X>=280) using the normal distribution.
P(X>=280) = 1 - P(X<280) ≈ 1 - P(Z<(279-288)/3.39)
In the equation above, Z is a standard normal variable (mean = 0, standard deviation = 1).
Now, we can use a standard normal distribution table or a calculator to find P(Z<(279-288)/3.39) and subtract that value from 1 to get the approximate probability.
Comparing the exact probability from the binomial distribution with the approximate probability from the normal distribution will allow us to see how well the normal approximation fits the binomial distribution in this case.
To solve this problem, we will use the binomial distribution to find the exact probability and then compare it with the normal approximation to the binomial.
The binomial distribution is used when we have a fixed number of independent trials, with each trial having the same probability of success. In this case, each reservation can be considered as a trial, and the probability of a reservation being cancelled is given as 4%.
The binomial distribution is defined by two parameters: the number of trials (n) and the probability of success (p). In our case, the number of trials is 300 (the number of rooms) and the probability of success is 0.04 (the probability of a cancellation).
To find the probability that at least 280 rooms will be occupied, we need to calculate the cumulative probability from 280 to 300 using the binomial distribution.
P(X ≥ 280) = P(X = 280) + P(X = 281) + ... + P(X = 300)
Where X is the random variable representing the number of occupied rooms.
To calculate each individual probability, we can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where (n choose k) represents the number of combinations of n items taken k at a time.
Using this formula, we can find the exact probability.
Now let's calculate the exact probability using the binomial distribution.
P(X = 280) = (300 choose 280) * 0.04^280 * (1-0.04)^(300-280)
P(X = 281) = (300 choose 281) * 0.04^281 * (1-0.04)^(300-281)
...
P(X = 300) = (300 choose 300) * 0.04^300 * (1-0.04)^(300-300)
Then, sum up all these probabilities to find P(X ≥ 280).
Now, let's calculate the normal approximation to the binomial.
The normal approximation to the binomial distribution uses the mean (μ) and standard deviation (σ) of a normal distribution, which are given by:
μ = n * p
σ = sqrt(n * p * (1-p))
In our case, μ = 300 * 0.04 = 12 and σ = sqrt(300 * 0.04 * 0.96) ≈ 3.096.
To approximate the probability P(X ≥ 280) using the normal distribution, we can use the cumulative distribution function (CDF) of the normal distribution.
P(X ≥ 280) ≈ 1 - P(Z < (280 - μ)/σ)
Where Z is a standard normal random variable.
Now, let's calculate both the exact probability using the binomial distribution and the approximate probability using the normal distribution, and compare the results.