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adell [148]
3 years ago
7

Find the volume of the ellipsoid x^2+y^2+9z^2=64

Mathematics
1 answer:
yuradex [85]3 years ago
4 0
Parameterize the ellipsoid using the augmented spherical coordinates:

\begin{cases}x=\frac18\rho\cos\theta\sin\varphi\\\\y=\frac18\rho\sin\theta\sin\varphi\\\\z=\frac38\rho\cos\varphi\end{cases}

Then the Jacobian for the change of coordinates is

\mathbf J=\dfrac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}=\begin{bmatrix}\frac18\cos\theta\sin\varphi&-\frac18\rho\sin\theta\sin\varphi&\frac18\rho\cos\theta\cos\varphi\\\\\frac18\sin\theta\sin\varphi&\frac18\rho\cos\theta\sin\varphi&\frac18\rho\sin\theta\cos\varphi\\\frac38\cos\varphi&0&-\frac38\rho\sin\varphi\end{bmatrix}

which has determinant

\det\mathbf J=-\dfrac3{512}\rho^2\sin\varphi

Then the volume of the ellipsoid is given by

\displaystyle\iiint_E\mathrm dx\,\mathrm dy\,\mathrm dz=\iiint_E|\det\mathbf J|\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

where E denotes the spaced contained by the ellipsoid. In particular, we have the definite integral and volume

\displaystyle\frac3{512}\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\dfrac\pi{128}
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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

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