According to an airline flights on a certain route are on time 90% of the time s

Question

According to an airline flights on a certain route are on time 90% of the time suppose 8 flights are randomy selected and the number of on tre nons is recorded Use technology to find the probabilities. Use the Tech Help button for further assistance (a) Determine whether this is a binomial experiment (b) Find and interpret the probability that exactly 6 flights are on time (c) Find and interpret the probability that at least 6 flights are on time (d) Find and interpret the probability that fewer than 6 fights are on time (e) Find and interpret the probability that between 5 and 7 flights, inclusive, are on time (a) Does the probability experiment represent a binomial experiment? A. Yes. because the experiment satisfies all the criteria for a binomia eipenment B. No because the probability of success differs from trial to trial 0 C. No, because there are more than two mutualy exclusive outcomes for each tnal D. No. because the trials of the experiment are not independent

Explanation / Answer

1.

We have been given binomial distribution params as:

p( on time) = .90
n = 9

a. A is right. There is 2 outcomes, on time and late and all params of binomial distribution have been given

b. P(X=6) = 9C6(.90^6)(.1^3) = .4464. Probability to get exactly 6 flights on time is .4464

c. P(X>=6) = P(X=6,7,8,9) = 9C6(.90^6)(.1^3)+..+9C9(.90^9)(.1^0) = 0.9917. Probability to get atleast 6 flights on time is .9917

d. P(X<6) = 1-P(X>=6) = 1-.9917 = .0083. Probability to get fewer than 6 flights on time is .0083

e. P(X=5 , 6, 7) = 9C5(.90^5)(.1^4)+9C6(.90^6)(.1^3)+9C7(.90^7)(.1^2) = .2243. Probability to get between 5 and 7 (inclusive both) flights on time is 0.2243.

2.

We have been given binomial distribution params as:

p = .50
n= 30

Hence, P(X=20) = 30C20*(.5^20)*(.5^10) = 0.0280

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