A distribution in which the frequency is either constantly increasing or constantly decreasing is called a(n):_______

Naval intelligence reports that 4 enemy vessels in a fleet of 17 are carrying nuclear weapons. If 9 vessels are randomly targete

icang [17]

Answer:

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

Step-by-step explanation:

The vessels are destroyed without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

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

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

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

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

In this question:

Fleet of 17 means that $N = 17$

4 are carrying nucleas weapons, which means that $k = 4$

9 are destroyed, which means that $n = 9$

What is the probability that more than 1 vessel transporting nuclear weapons was destroyed?

This is:

$P(X > 1) = 1 - P(X \leq 1)$

In which

$P(X \leq 1) = P(X = 0) + P(X = 1)$

So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,17,9,4) = \frac{C_{4,0}*C_{13,9}}{C_{17,9}} = 0.0294$

$P(X = 1) = h(1,17,9,4) = \frac{C_{4,1}*C_{13,8}}{C_{17,9}} = 0.2118$

Then

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0294 + 0.2118 = 0.2412$

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.2412 = 0.7588$

0.7588 = 75.88% probability that more than 1 vessel transporting nuclear weapons was destroyed

8 0

3 years ago

If the probability density of a random variable is given by find the probabilities that a random variable having this probabilit

kolezko [41]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

3 0

4 years ago

If a graph of a system of two equations shows two lines that coincide on a coordinate plane, how many solutions does the system

olga55 [171]

If a graph of a system of two equations shows two lines that coincide on a coordinate plan there would be C.)Infinitely many solutions

I hope this helps, have a great day.

3 0

3 years ago

Read 2 more answers

A veterinarian polled her clients own dogs as to whether or not they used dry food only for their dogs which of the following de

vaieri [72.5K]

Answer:

Can you describe this question

Step-by-step explanation:

3 0

3 years ago

Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample

scoundrel [369]

Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

8 0

3 years ago

A distribution in which the frequency is either constantly increasing or constantly decreasing is called a(n):________