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Svetlanka [38]
3 years ago
9

An airline has a policy of booking as many as 15 persons on an airplane that can seat only 14. ​(Past studies have revealed that

only 88.0​% of the booked passengers actually arrive for the​ flight.) Find the probability that if the airline books 15 ​persons, not enough seats will be available. Is it unlikely for such an overbooking to​ occur?
The probability that not enough seats will be available is___
Is it unlikely for such an overbooking to​ occur?
Mathematics
1 answer:
aksik [14]3 years ago
7 0

Answer: No, it is unlikely for such an overbooking to occur.

Our required probability is 0.146.

Step-by-step explanation:

Since we have given that

Number of persons on an airplane book = 15

Probability of booked passengers actually arrive for the flight = 88%

We need to find the probability that if the airline books 15 persons, not enough seats will be available.

We will use "Binomial distribution":

Probability would be

P(X=15)=(0.88)^{15}=0.146

No, it is unlikely for such an overbooking to occur.

Hence, our required probability is 0.146.

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kvv77 [185]

X²= 9

X POWER 3= 27

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1 year ago
What is the slope of the line that passes through the points (3,2) and (-1,-4)? Write your answer as a fraction in simplest form
Nataly_w [17]

Answer:

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7 0
2 years ago
When 258 college students are randomly selected and surveyed, it is found that 106 own a car. Find a 99% confidence interval for
juin [17]

Answer: 0.332 < p < 0.490

Step-by-step explanation:

We know that the confidence interval for population proportion is given by :-

\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

, where n= sample size

\hat{p} = sample proportion

z* = critical z-value.

As per given , we have

n= 258

Sample proportion of college students who own a car = \hat{p}=\dfrac{106}{258}\approx0.411

Critical z-value for 99% confidence interval is 2.576. (By z-table)

Therefore , the  99% confidence interval for the true proportion(p) of all college students who own a car will be :0.411\pm (2.576)\sqrt{\dfrac{0.411(1-0.411)}{258}}\\\\=0.411\pm (2.576)\sqrt{0.00093829}\\\\= 0.411\pm (2.576)(0.0306315197142)\\\\=0.411\pm 0.0789=(0.411-0.0789,\ 0.411+0.0789)\\\\=(0.3321,\ 0.4899)\approx(0.332,\ 0.490)

Hence, a 99% confidence interval for the true proportion of all college students who own a car : 0.332 < p < 0.490

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3 years ago
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zepelin [54]

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6 0
3 years ago
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