Question

22. According to ghost hunters, large cities have, on average, 9 haunted houses each. As far as we know, it is unpredictable which houses will be haunted; and just because one house is haunted, that does not mean any of its neighbors will or will not be haunted.a) What is the standard deviationof the number of haunted houses in a large city? b) What is the probability that 5or fewerhaunted houses will be found in a city?

Answer 1

Answer:

a) The standard deviation is 3.

b) 11.56% probability that 5 or fewer haunted houses will be found in a city

Step-by-step explanation:

We only have the mean number of events, and the events are independent. So we use the Poisson distribution to solve this question.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval, which is the same as the variance.

According to ghost hunters, large cities have, on average, 9 haunted houses each.

This means that $\mu = 9$

a) What is the standard deviation of the number of haunted houses in a large city?

The variance is the same as the mean, which is 9. The standard deviation is the square root of the variance, so it is 3.

b) What is the probability that 5 or fewer haunted houses will be found in a city?

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-9}*(9)^{0}}{(0)!} = 0.0001$

$P(X = 1) = \frac{e^{-9}*(9)^{1}}{(1)!} = 0.0011$

$P(X = 2) = \frac{e^{-9}*(9)^{2}}{(2)!} = 0.0050$

$P(X = 3) = \frac{e^{-9}*(9)^{3}}{(3)!} = 0.0150$

$P(X = 4) = \frac{e^{-9}*(9)^{4}}{(4)!} = 0.0337$

$P(X = 5) = \frac{e^{-9}*(9)^{5}}{(5)!} = 0.0607$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0001 + 0.0011 + 0.0050 + 0.0150 + 0.0337 + 0.0607 = 0.1156$

11.56% probability that 5 or fewer haunted houses will be found in a city

Answer 2

The true statements are:

• The standard deviation is 3
• The probability that 5 or fewer haunted houses will be found in a city is 11.56

How to determine the standard deviation

The mean of the distribution is given as:

$\mu = 9$

For a Poisson distribution, we have:

$\sigma^2 = \mu = 9$

So, the standard deviation is:

$\sigma = \sqrt{\sigma^2$

This gives

$\sigma = \sqrt{9$

$\sigma = 3$

Hence, the standard deviation is 3

How to determine the probability

The probability that 5 or fewer haunted houses will be found in a city is represented as:

$P(x \le 5) = P(0) + P(1) + .... +P(5)$

Where:

$P(x) = \frac{e^{-\mu}\mu^x}{x!}$

This gives

$P(x) = \frac{e^{9} * 9^x}{x!}$

So, we have:

$P(0) = \frac{e^{9} * 0^1}{0!} = 0.0001$

$P(1) = \frac{e^{9} * 9^1}{1!} = 0.0011$

...

.

$P(5) = \frac{e^{9} * 9^5}{5!} = 0.0607$

The equation becomes

$P(x \le 5) = 0.0001 + 0.0011 + .... +0.0607$

$P(x \le 5) = 0.1156$

Hence, the probability that 5 or fewer haunted houses will be found in a city is 11.56

Read more about probability at:

https://brainly.com/question/25870256