Because all airline passengers do not show up for their reserved seat, an airline sells 125 tickets for a flight that holds only
1 answer:
Answer:
a) 0.9961
b) 0.9886
Step-by-step explanation:
If X is the variable define by people who do not show up to take their flight, and considering that the probability of success (the person shows up) is 0.90, and the probability of fail (the person doesn't show up) is 0.10, the variable X has a binomial distribution.
The binomial distribution formula would be:
××
a) What is the probability that every passenger who shows up can take the flight?
This happens when X≥5
P( X≥5) = 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)) = 0.9961
a) What is the probability that the flight departs with empty seats?
This happens when X>5
P(X>5) = 1 - P(X≤5) = 1 - (P(X=0) + P(X=2) + P(X=3) + P(X=4) + P(X=5)) = 0.9886
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Answer:
The score that separates the lower 5% of the class from the rest of the class is 55.6.
Step-by-step explanation:
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.
In this question:
Find the score that separates the lower 5% of the class from the rest of the class.
This score is the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.
The score that separates the lower 5% of the class from the rest of the class is 55.6.