According to an airline, flights on a certain route are on time 80% of the time.
ID: 2907373 • Letter: A
Question
According to an airline, flights on a certain route are on time 80% of the time. Suppose 10 flights are randomly selected and the number of on-time flights is recorded. (a) Explain why this is a binomial experiment. (b) Determine the values of n and p (c) Find and interpret the probability that exactly 7 flights are on time (d) Find and interpret the probability that fewer than 7 flights are on time. (e) Find and interpret the probability that at least 7 flights are on time (f) Find and interpret the probability that between 5 and 7 flights, inclusive, are on time. (b) n = | | (Type an integer or a decimal. Do not round.) p-D | (Type an integer or a decimal. Do not round.) (c) The probability that exactly 7 flights are on time is (Round to four decimal places as needed.) Interpret the probability In 100 trials of this experiment, it is expected that aboutwill result in exactly 7 flights being on time (Round to the nearest whole number as needed.) (d) The probability that fewer than 7 flights are on time is (Round to four decimal places as needed.) Interpret the probability In 100 trials of this experiment, it is expected that about will result in fewer than 7 flights being on time Click to select your answer(s)Explanation / Answer
a) It is bonomial distribution, because it has n = 10 independent trials and the probability value 0.08 for flights on a certain route are on time remains the same for each trial.
b) n = 10
p = 0.8
c) P(X = 7) = 10C7 * (0.8)^7 * (0.2)^3 = 0.2013
Expected value = 100 * 0.2013 = 20.13 = 20
In 100 trials of experiment, it is expected that about 20 will result in exactly 7 flights being on time.
f) P(5 < X < 7)
d) P(X < 7) = 1 - P(X > 7)
= 1 - (P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10))
= 1 - (10C7 * (0.8)^7 * (0.2)^3 + 10C8 * (0.8)^8 * (0.2)^2 + 10C9 * (0.8)^9 * (0.2)^1 + 10C10 * (0.8)^10 * (0.2)^0)
= 1 - 0.8791 = 0.1209
Expected value = 100 * 0.1209 = 12.09 = 12
In 100 trials of experiment, it is expected that about 12 will result in fewer than 7 flights being on time.
f) P(5 < X < 7)
e) P(X > 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
= 10C7 * (0.8)^7 * (0.2)^3 + 10C8 * (0.8)^8 * (0.2)^2 + 10C9 * (0.8)^9 * (0.2)^1 + 10C10 * (0.8)^10 * (0.2)^0
= 0.8791
Expected value = 100 * 0.8791 = 87.91 = 88
In 100 trials of experiment, it is expected that about 88 will result in at least 7 flights being on time.
f) P(5 < X < 7)
= P(X = 5) + P(X = 6) + P(X = 7)
= 10C5 * (0.8)^5 * (0.2)^5 + 10C6 * (0.8)^6 * (0.2)^4 + 10C7 * (0.8)^7 * (0.2)^3
= 0.3158
Expected value = 100 * 0.3158 = 31.58 = 32
In 100 trials of experiment, it is expected that about 32 will result in between 5 and 7 flights, inclusive being on time.
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