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sineoko [7]
3 years ago
14

When someone buys a ticket for an airline​ flight, there is a 0.098 probability that the person will not show up for the flight.

A certain jet can seat 18 passengers. Is it wise to book 20 passengers for a flight on the​ jet?
Determine whether or not booking 24 passengers for 22 seats on the jet is a wise decision. Select the correct choice below and fill in the answer box in that choice with the probability that there are not enough seats on the jet.

​(Round to four decimal places as​ needed.)

A.It is a wise decision because the probability that there are not enough seats on the jet is __. So, overbooking is an unlikely event.

B. It is not a wise decision because the probability that there are not enough seats on the jet is __. So, overbooking is not an unlikely event.

C. It is not a wise decision because the probability that there are not enough seats on the jet is __. So, overbooking is an unlikely event.

D. It is a wise decision because the probability that there are not enough seats on the jet is __. ​So, overbooking is not an unlikely event.

Mathematics
1 answer:
DedPeter [7]3 years ago
5 0

Answer:

Step-by-step explanation:

Probability that a person will not show up is,

P(person not show up)=0.098.

Probability that they will show up is

1-0.098=0.902

P(person will show up)=0.902

Jet passenger seat=18.

a. But is it wise to book 20 passenger seat

If the flight attendant book 20.

We want to know the probability that the two person will not show up.

The probability that the 2 person  will not show up out of the 20 is given as

Using binomial distribution

Check attachment for formulas.

P(2 not showing up) =

20C2 • 0.098² • 0.902^20=0.2319

Which is 23.19% that the 2 person  will not show up.

b. But is it wise to book 24 passengers for 22 jet seat.

The probability that the 2 will not show up out of the 24 is given as

Using binomial distribution

P(2 not showing up) =

24C2 • 0.098²• 0.902²²=0.2741

Which is 27.41% that the four person will not show up.

c. Probability of no enough seat

For 20 passengers and 18seats on jet

1-0.2319=0.7631.

For 24 passengers and 20 seats on jet

1-27.41=0.7259.

Then,  It is not a wise decision because the probability that there are not enough seats on the jet is 0.7259 (72.59%). So, overbooking is an unlikely event.

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For each of the following binomial random variables, specify n and p. (a) A fair die is rolled 50 times. X = number of times a 5
Keith_Richards [23]

Answer:

a) n = 50, p = \frac{1}{6}

b) n = 16, p = \frac{1}{100}

c) n = 26, p = 0.25, \mu = 6.5

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

(a) A fair die is rolled 50 times. X = number of times a 5 is rolled

The die is rolled 50 times, so n = 50.

Each roll can have 6 outcomes. So the probability that 5 is rolled is p = \frac{1}{6}

(b) A company puts a game card in each box of cereal and 1/100 of them are winners. You buy sixteen boxes of cereal, and X = number of times you win.

You buy 16 boxes of cereal, so n = 16.

1 of 100 are winners. So p = \frac{1}{100}.

(c) Jack likes to play computer solitaire and wins about 25% of the time. X = number of games he wins out of his next 26 games.

He plays 26 games, so n = 26.

He wins 25% of the time, so p = 0.25

We have that \mu = np. So \mu = 26*0.25 = 6.5

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An integer is a whole number (not a fractional number) that can be positive, negative, or zero.

Since it's a decimal it can not be an integer.

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Answer:

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