A local motel has 100 rooms. The occupancy rate for the winter months is 60%. Find the probability that in a given winter month more than 65 rooms will be rented. Use the normal distribution to approximate the binomial distribution.
To solve this problem, we need to use the normal distribution to approximate the binomial distribution.
First, let's find the mean (μ) and standard deviation (σ) for the binomial distribution.
The mean (μ) is given by μ = n * p, where n is the number of trials and p is the probability of success on each trial. In this case, n = 100 and p = 0.6, so μ = 100 * 0.6 = 60.
The standard deviation (σ) is given by σ = sqrt(n * p * (1 - p)). In this case, σ = sqrt(100 * 0.6 * 0.4) = sqrt(24) ≈ 4.9.
Next, we need to calculate the z-score for the given value of more than 65 rooms rented. The z-score is calculated using the formula: z = (x - μ) / σ, where x is the observed value, μ is the mean, and σ is the standard deviation. In this case, x = 65, μ = 60, and σ = 4.9.
So, the z-score is z = (65 - 60) / 4.9 ≈ 1.02.
Now that we have the z-score, we can use the standard normal distribution table or a statistical calculator to find the probability corresponding to the z-score. The probability of more than 65 rooms being rented is equal to 1 minus the cumulative probability up to the z-score.
P(x > 65) = 1 - P(z ≤ 1.02)
Using a standard normal distribution table or a statistical calculator, we can find P(z ≤ 1.02) to be approximately 0.8461.
Therefore, P(x > 65) = 1 - 0.8461 ≈ 0.1539.
So, the probability that in a given winter month more than 65 rooms will be rented is approximately 0.1539, or 15.39%.