Phoenix is a hub for a large airline. Suppose that on a particular day, 8000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1400 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 430 connecting passengers missed their connecting flight and were delayed in Phoenix. Of the 430 who were delayed, 75 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8000 passengers. (Round your answers to three decimal places.)(a)Compute the probability that the selected passenger had Phoenix as a final destination.(b)Compute the probability that the selected passenger did not have Phoenix as a final destination.(c)Compute the probability that the selected passenger was connecting and missed the connecting flight.(d)Compute the probability that the selected passenger was a connecting passenger and did not miss the connecting flight.(e) Compute the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight in Phoenix.(f)An independent customer satisfaction survey is planned. The company carrying out the survey plans to contact 50 passengers selected at random from the 8000 passengers who arrived in Phoenix on the day described above. The airline knows that the survey results will not be favorable if too many people who were delayed overnight are included in the survey. Should the airline be worried? Explain whether or not you think the airline should be worried, using relevant probabilities to support your answer.

Step-by-step explanation:

Hello!

In a day a total of n= 8000 passangers arrived to Phoenix.

1400 of these passangers final destination was Phoenix. (A)

The remaining 6600 passangers were all taking connection flights to other cities. (B)

Due to several flights being late, 430 connecting passengers missed their connecting flight and were delayed in Phoenix. (C)

Of these 430 delayed passengers, 75 were delayed overnight. (D)

If there was one passenger chosend at random:

a) You have to calculate the probability of this passegners final destination to be Phoenix. Symbolically: P(A)

To calculate this probability you have to divide the number of  passengers whose final destination was Phoenix by the total number of pasengers:

b) The passangers whose final destination was not phoenix are those taking connection flights, the probability of this event, called B, is calculated dividing the total of connecting passengers by the total of passengers:

c) You need to calculate the probability of "the passenger was connecting and missed the connecting flight"

The amount of passengers that fit this situation are 430 of 8000, so the probability is calculated as:

d) Of all 6600 connecting passengers, only 430 missend the flight wich means that 6170 took the connecting flight, the probability of this event "E" is

e) The passenger either had Phoenix as a final destination or was delayed overnight, symbolically:

P(A∪D) = P(A) + P(D) - P(A∩D)

The probability of the passenger being delayed overnight is:

The events "A" and "D" are mutually exclusive, this means that they cannot occur both at the same time, so their intersection is void, P(A∩D)= ∅

Then

P(A∪D) = P(A) + P(D) = 0.175 + 0.009= 0.184

f) If using the same data we turn this event into a binomial variable were our success will be "the passenger missed his flight and was delayed overnight" with probability p=0.009 and the sample n= 50

Let's say that the airline should be worried if more than half of the surveyed passengers were delayed overnight then:

P(X>25)= 1 - P(X ≤ 25) = 1 - 0.999999 = 0.000001

The probability of most of the surveyed passenger were delayed overnight is too low so the company should not be worried.

I hope it helps!

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get