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3 years ago

Someone please help me!

Mathematics

1 answer:

Vaselesa [24]3 years ago

7 0

Look at 3 on Variable B's axis, then go to the right until you reach the line.

After you reach the line, go all the way down and see what number you get to on Variable A's axis. You get 8, so that is your answer.

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An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insu

ankoles [38]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

An insurance company examines its pool of auto insurance customers and gathers the following information: (i) All customers insure at least one car. (ii) 70% of the customers insure more than one car. (iii) 20% of the customers insure a sports car. (iv) Of those customers who insure more than one car, 15% insure a sports car. Calculate the probability that a randomly selected customer insures exactly one car, and that car is not a sports car?

Answer:

P( X' ∩ Y' ) = 0.205

Step-by-step explanation:

Let X is the event that the customer insures more than one car.

Let X' is the event that the customer insures exactly one car.

Let Y is the event that customer insures a sport car.

Let Y' is the event that customer insures not a sport car.

From the given information we have

70% of customers insure more than one car.

P(X) = 0.70

20% of customers insure a sports car.

P(Y) = 0.20

Of those customers who insure more than one car, 15% insure a sports car.

P(Y | X) = 0.15

We want to find out the probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

P( X' ∩ Y' ) = ?

Which can be found by

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

From the rules of probability we know that,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y ) (Additive Law)

First, we have to find out P( X ∩ Y )

From the rules of probability we know that,

P( X ∩ Y ) = P(Y | X) × P(X) (Multiplicative law)

P( X ∩ Y ) = 0.15 × 0.70

P( X ∩ Y ) = 0.105

So,

P( X ∪ Y ) = P(X) + P(Y) - P( X ∩ Y )

P( X ∪ Y ) = 0.70 + 0.20 - 0.105

P( X ∪ Y ) = 0.795

Finally,

P( X' ∩ Y' ) = 1 - P( X ∪ Y )

P( X' ∩ Y' ) = 1 - 0.795

P( X' ∩ Y' ) = 0.205

Therefore, there is 0.205 probability that a randomly selected customer insures exactly one car, and that car is not a sports car.

6 0

2 years ago

Which if the following rules best describes the matrix below?

crimeas [40]

Answer:

Step-by-step explanation:

i just know i guess

7 0

1 year ago

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a norma

nata0808 [166]

Answer:

a) $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b) Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c) $P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d) $P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e) $P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

Step-by-step explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

$X \sim N(110000,40000)$

Where $\mu=110000$ and $\sigma=40000$

If we select a sample size of n =35 the standard error is given by:

$\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}= \frac{40000}{\sqrt{50}}= 5656.85$

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

$P(X > 112000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >112000) = P(Z>\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>0.354)$

And we can use the complement rule and we got:

$P(Z>0.354) = 1-P(Z$

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

$P(X > 100000)$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P( \bar X >100000) = P(Z>\frac{100000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z>-1.768)$

And we can use the complement rule and we got:

$P(Z>-1.768) = 1-P(Z$

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

$P(100000$

And we can use the z score given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$P(100000< \bar X$

And we can use the complement rule and we got:

$P(-1.768$

8 0

3 years ago

Name one real-world objeslct that suggests points

JulijaS [17]

Answer:

magnets

Step-by-step explanation:

counter

bottle cap

pushpin

microwave button

6 0

3 years ago

Solve it please with explanation Marty simplified the expression 2(x - 2) - 2(3) as 2x - 8. Is he correct? Explain why or why no

Kamila [148]

Answer: Marty is not correct

Step-by-step explanation:

2(x-2)-2(3) use distributive property

2x-4-6

2x-10

Marty is not correct

3 0

2 years ago

Read 2 more answers