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melomori [17]
3 years ago
7

Put 45x+32=45-32 in slope interslept formthankyou.

Mathematics
1 answer:
Lilit [14]3 years ago
7 0
Easy peasy pumpkineasy the answer issssssssssssssss 455555
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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
(10-1)(10-2)(10-3)(10-10)
Goryan [66]

Answer:

0

Step-by-step explanation:

(10-1=9)(10-2=8)(10-3=7)(10-10=0)

9*8*7*0=0

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Need geometry post course test done ASAP
hram777 [196]
<h2>Answer:</h2>

All the answers have been attached in the file below.

Download pdf
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