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gavmur [86]
3 years ago
10

According to the Department of Transportation, 27% of domestic flights were delayed in the last year at JFK airport. Five flight

s are randomly selected at JFK. What is the probability that all five flights are delayed?
Mathematics
1 answer:
Mars2501 [29]3 years ago
4 0

Answer:

The probability that all the five flights are delayed is 0.2073.

Step-by-step explanation:

Let <em>X</em> = number of domestic flights delayed at JFK airport.

The probability of a domestic flight being delayed at the JFK airport is, P (X) = <em>p</em> = 0.27.

A random sample of <em>n</em> = 5 flights are selected at JFK airport.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

P(X=x)={5\choose x}0.27^{x}(1-0.27)^{5-x};\ x=0,1,2...

Compute the probability that all the five flights are delayed as follows:

P(X=5)={5\choose 5}0.27^{5}(1-0.27)^{5-5}=1\times 1\times 0.207307=0.2073

Thus, the probability that all the five flights are delayed is 0.2073.

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Answer and Explanation:

Given : The random variable x has the following probability distribution.

To find :

a. Is this probability distribution valid? Explain and list the requirements for a valid probability distribution.

b. Calculate the expected value of x.

c. Calculate the variance of x.

d. Calculate the standard deviation of x.

Solution :

First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.25           0               0                0

1     0.20        0.20             1              0.20

2    0.15          0.3               4             0.6

3    0.30         0.9               9             2.7

4    0.10          0.4               16             1.6

   ∑P(x)=1     ∑xP(x)=1.8               ∑x²P(x)=5.1

a) To determine that table shows a probability distribution we add up all five probabilities if the sum is 1 then it is a valid distribution.

\sum P(X)=0.25+0.20+0.15+0.30+0.10

\sum P(X)=1

Yes it is a probability distribution.

b) The expected value of x is defined as

E(x)=\sum xP(x)=1.8

c) The variance of x is defined as

V=\sum x^2P(x)-(\sum xP(x))^2\\V=5.1-(1.8)^2\\V=5.1-3.24\\V=1.86

d) The standard deviation of x is  defined as

\sigma=\sqrt{V}

\sigma=\sqrt{1.86}

\sigma=1.136

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