The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
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Answer:
The probability is 0.31
Step-by-step explanation:
To find the probability, we will consider the following approach. Given a particular outcome, and considering that each outcome is equally likely, we can calculate the probability by simply counting the number of ways we get the desired outcome and divide it by the total number of outcomes.
In this case, the event of interest is choosing 3 laser printers and 3 inkjets. At first, we have a total of 25 printers and we will be choosing 6 printers at random. The total number of ways in which we can choose 6 elements out of 25 is , where
. We have that
Now, we will calculate the number of ways to which we obtain the desired event. We will be choosing 3 laser printers and 3 inkjets. So the total number of ways this can happen is the multiplication of the number of ways we can choose 3 printers out of 10 (for the laser printers) times the number of ways of choosing 3 printers out of 15 (for the inkjets). So, in this case, the event can be obtained in
So the probability of having 3 laser printers and 3 inkjets is given by
Answer:
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears
Step-by-step explanation:
For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. 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 ,
.
In this problem, we have that:
So
What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears
This probability is the pvalue of Z when X = 82. So
has a pvalue of 0.0885.
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears