Naval intelligence reports that 5 enemy vessels in a fleet of 18 are carrying nuclear weapons. If 9 vessels are randomly targeted and destroyed, what is the probability that no more than 1 vessel transporting nuclear weapons was destroyed

Answer:

0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

Step-by-step explanation:

Vessels are chosen without replacement, which means that we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

Naval intelligence reports that 5 enemy vessels in a fleet of 18 are carrying nuclear weapons.

This means, respectively, that [tex]k = 5, N = 18[/tex].

9 vessels are randomly targeted and destroyed

This means that [tex]n = 9[/tex]

What is the probability that no more than 1 vessel transporting nuclear weapons was destroyed?

This is:

[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]

In which

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,18,9,5) = \frac{C_{5,0}*C_{13,9}}{C_{18,9}} = 0.0147[/tex]

[tex]P(X = 1) = h(1,18,9,5) = \frac{C_{5,1}*C_{13,8}}{C_{18,9}} = 0.1324[/tex]

[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0147 + 0.1324 = 0.1471[/tex]

0.1471 = 14.71% probability that no more than 1 vessel transporting nuclear weapons was destroyed.

A business school is conducting a study to investigate whether a students’ scores on a placement test can be used to predict students’ starting salaries. Based on a random sample of 200 graduates of the business school, a 98 percent confidence interval for the slope of the linear regression line relating placement scores and starting salary is calculated to be (315,336). Assume the conditions for inference are met.

Which of the following is a correct interpretation of the confidence interval?

a. There is a 98 percent probability that the slope of the population regression line is between $315 per point and $336 per point.
b. Ninety-eight percent of the time, a 10-point increase in placement score will result in an average increase between $3,150 and $3,360 in starting salary.
c. We are 98 percent confident that a 10-point increase in placement score will result in a predicted increase in starting salary of between $3,150 and $3,360.
d. We are 98 percent confident that predicted starting salaries will be between $315,000 and $336,000.
e. We are 98 percent confident that the regression equation can be used to make accurate predictions for placement scores between 315 and 336.

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H
«[?]
4748
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