All categories

4 years ago

What does a probability distribution indicate? choose the correct answer below.

1 answer:

6 0

The correct answer is C) both a and b.

A probability distribution shows all of the possible values of the experiment and the probabilities that go with them.

You might be interested in

Like charges repel and I like charges attract Coulomb’s law states that the force F of attraction or repulsion between two charg

olga_2 [115]

Step-by-step explanation:

For the charges that have same sign of charges will repel each other while for the charges that have different charges will attract each other. So, we can say that like charges repel and unlike charges attract each other.

The Coulomb's law of attraction of repulsion states that force between charges is directly proportion to the product of charges and inversely proportional to the square of distance between them. Mathematically, it is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Hence, all the given statements are true.

8 0

3 years ago

Whats the reciprocal of <br> 4 3/7

Angelina_Jolie [31]

Answer

0.22580645161

Step-by-step explanation:

you just have to do 4 3/7 in a calculator

6 0

3 years ago

HELP ASAP <br> What is the slope of the line on the graph?

ollegr [7]

The slope of this line is -1/6.

2 Point: (6,2) (-6,6)
x₁ y₁ x₂ y₂

Remember:

rise change in y y₂ - y₁ 6 - 2 4
---------- = ------------------- = ------------- = ----------- = ----------
run change in x x₂ - x₁ -6 - 6 -12

Simplify 4/-12

4 ÷ 4 1
------- = ------------- = -1/3
-12 ÷ 4 -3

The slope of the line is -1/3.

3 0

4 years ago

When solving a system of two linear equations algebraically, how can you tell if the system has exactly one solution?

Artyom0805 [142]

Answer:

ther will be exactly one value. the answer is b

4 0

3 years 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

4 years ago

What does a probability distribution​ indicate? choose the correct answer below.

What does a probability distribution indicate? choose the correct answer below.