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2 years ago

The missing number in the arithmetic sequence 60, 51, ___, 33 is:

Mathematics

1 answer:

andrew-mc [135]2 years ago

8 0

Answer:

Step-by-step explanation:

In this question, we have the find the rate of change in the sequence.

To find this, we need to find how much the sequence is changing by.

We know from 60 to 51 is -9, so we know that our sequence is changing by -9.

To find our unknown number, we would subtract 9 from 51.

51 - 9 = 42

Our missing number in the sequence is 42.

We can check our answer by subtracting 9 from 42. This gives us our next number, which is 33.

42 would be your answer.

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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]

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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

2 years ago