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At certain airport, 75%of the flights arrive on time. A sample of 10 flights is

ID: 3174917 • Letter: A

Question

At certain airport, 75%of the flights arrive on time. A sample of 10 flights is studied given that the number of flights that land is around 500 per day. Find the probability that all 10 of the flights were on time Find the probability that exactly four of the flights were on time Find the probability that more than eight flights were on time A traffic light at a certain intersection is green 40% of the time, yellow 25% of the time, and red 35% of the time. A car approaches this intersection once each day. Let X represent the number of days that pass up to and including the first time the car encounters a red light. Assume that each day represent an independent trial Find P(X = 3) Find P(X =

Explanation / Answer

1.

a.There are 100 flights studied, that is n=10 independent trials, with probability of success, that is arriving on time is denoted as p=0.75. The probability of success is constant throughout the trials. The specific number of success in n=100 trials is denoted by r. Use binomial distribution, P(X,r)=nCr(p)^r(1-p)^n-r to compute the required probability.

P(X=10)=10C10(0.75)^10(1-0.75)^0=0.0563 (ans)

b. P(X=4)=10C4(0.75)^4(1-0.75)^6=0.0162 (ans)

c. P(X>8)=1-P(X<=8)=1-[P(X=0)+P(X=1)+...+P(X=8)]=1-[10C0(0.75)^0(1-0.75)^10+10C1(0.75)^1(1-0.75)^9+...+10C8(0.75)^8(1-0.75)^2]

=1-0.7560

=0.2440 (ans)

2.

a. This accounts for geometric distribution with probability of success, that is encountering a red light is p=0.35, and X denote number of days until the first time the car encounters a red light. Use the formula, P(X=x0=q^(x-1)p, where, q is probability of success.

P(X=3)=0.65^(3-1)*0.35=0.1479

b. P(X<=3)=P(X=1)+P(X=2)+P(X=3)=0.65^(1-1)*0.35+0.65^(2-1)*0.35+0.65^(3-1)*0.35=0.7254 (ans)

c. E(X)=mu=1/p=1/0.35=2.86 (ans)

d. Variance, sigma=q/p^2=0.65/0.35^2=5.32 (ans)

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