on monday two hundred thirty-four students went on a trip to the zoo. All five buses were filled and four students had to travel
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Answer:
8.69% probability that at least 285 rooms will be occupied.
Step-by-step explanation:
For each booked hotel room, there are only two possible outcomes. Either there is a cancelation, or there is not. So we use concepts of the binomial probability distribution to solve this question.
However, we are working with a big sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
A popular resort hotel has 300 rooms and is usually fully booked. This means that
About 7% of the time a reservation is canceled before the 6:00 p.m. deadline with no pen-alty. What is the probability that at least 285 rooms will be occupied?
Here a success is a reservation not being canceled. There is a 7% probability that a reservation is canceled, and a 100 - 7 = 93% probability that a reservation is not canceled, that is, a room is occupied. So we use
Approximating the binomial to the normal.
The probability that at least 285 rooms will be occupied is 1 subtracted by the pvalue of Z when X = 285. So
has a pvalue of 0.9131.
So there is a 1-0.9131 = 0.0869 = 8.69% probability that at least 285 rooms will be occupied.