Mathematics, 05.04.2020 00:51 Jmarie08

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a normal distribution with a standard deviation of $40,000.

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Answer from: sanchez7489

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) $P( \bar X 112000) = P(Z\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z0.354)$

And we can use the complement rule and we got:

P(Z0.354) = 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:

$P( \bar X 112000) = P(Z\frac{112000-110000}{\frac{40000}{\sqrt{50}}}) = P(Z0.354)$

And we can use the complement rule and we got:

P(Z0.354) = 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

Answer from: Quest

ian gone lie this got me confused

step-by-step explanation:

Answer from: Quest

the workers who do no drive has no reasonable reason to buy gasoline due to they don't have anything to use it for

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