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Answer:
98.75% probability that every passenger who shows up can take the flight
Step-by-step explanation:
For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.
However, we are working with a large 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 ,
.
The probability that a passenger does not show up is 0.10:
This means that the probability of showing up is 1-0.1 = 0.9. So
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets
This means that
Using the approximation:
(a) What is the probability that every passenger who shows up can take the flight
This is , so this is the pvalue of Z when X = 120.
has a pvalue of 0.9875
98.75% probability that every passenger who shows up can take the flight
Answer:
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Step-by-step explanation:
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Let's think of the problem as follows.
Write all the 4-digit numbers that can be formed using the digits from 1 to 9, without repetition, in pieces of paper, and put them in a bag. What is the probability of picking the 4-digit number 1234, among these numbers.
The connection of the 2 problems is as follows:
The 4-digit number, for example 5489, represents drawing first 5, then 4, then 8, then 9 , in the original question.
we did not allow repetition, because for example the number 8918 would represent drawing 8, then 9, then 1 then 8 (again!!), which is not possible, so we lose the connection between the problems.
So there are in total 9*8*7*6= 3024 4-digit numbers, with non-repeating digits.
One of these numbers is 1234 (representing drawing 1, then 2, then 3, then 4)
among these 3024 numbers, the probability of picking 1234 is
We could have solved this problem also as :
P(drawing 1, 2, 3, 4 in order)=
Answer:
Write all the 4-digit numbers that can be formed using the digits from 1 to 9, without repetition, in pieces of paper, and put them in a bag. What is the probability of picking the 4-digit number 1234, among these numbers.
The connection of the 2 problems is as follows:
The 4-digit number, for example 5489, represents drawing first 5, then 4, then 8, then 9 , in the original question.
we did not allow repetition, because for example the number 8918 would represent drawing 8, then 9, then 1 then 8 (again!!), which is not possible, so we lose the connection between the problems.
So there are in total 9*8*7*6= 3024 4-digit numbers, with non-repeating digits.
One of these numbers is 1234 (representing drawing 1, then 2, then 3, then 4)
among these 3024 numbers, the probability of picking 1234 is
We could have solved this problem also as :
P(drawing 1, 2, 3, 4 in order)=
Answer: