The probability distributions for 2 variables are defined as followsX ~ N$(120,{\sigma}^{2})$ and Y ~ N$(\mu ,2{\sigma}^{2})$ and $P(X<124)$ = $P(Y>124$. Calculate $\mu $.

Lisa Hardin
2022-07-23
Answered

The probability distributions for 2 variables are defined as followsX ~ N$(120,{\sigma}^{2})$ and Y ~ N$(\mu ,2{\sigma}^{2})$ and $P(X<124)$ = $P(Y>124$. Calculate $\mu $.

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yermarvg

Answered 2022-07-24
Author has **19** answers

Let $W:=(X-120)/\sigma ,\phantom{\rule{thinmathspace}{0ex}}Z:=(Y-\mu )/(\sigma \sqrt{2})$ so $Z,\phantom{\rule{thinmathspace}{0ex}}W\sim N(0,\phantom{\rule{thinmathspace}{0ex}}1)$ and

$P(W<4/\sigma )=P(Z>(124-\mu )/(\sigma \sqrt{2}))=P(Z<(\mu -124)/(\sigma \sqrt{2})).$

Hence

$\frac{4}{\sigma}=\frac{\mu -124}{\sigma \sqrt{2}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mu =124+4\sqrt{2}.$

$P(W<4/\sigma )=P(Z>(124-\mu )/(\sigma \sqrt{2}))=P(Z<(\mu -124)/(\sigma \sqrt{2})).$

Hence

$\frac{4}{\sigma}=\frac{\mu -124}{\sigma \sqrt{2}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mu =124+4\sqrt{2}.$

asked 2022-04-07

Let ${Y}_{1},{Y}_{2},...,{Y}_{n}$ be random samples from a normal distribution where the mean is 2 and the variance is 4. How large must n be in order that $P(1.9\le \overline{Y}\le 2.1)\ge 0.99$?

My attemp^

We are computing for the sample mean of our random sample that was given in the problem. By definition, $z=\frac{\overline{Y}-\mu}{\frac{\sigma}{\sqrt{n}}}$. So rewrite the equation so we can transform the data to make the mean 0 and the standard deviation 1. If I do that, I get,

$P(\frac{1.9-2}{\frac{2}{\sqrt{n}}}\le z\le \frac{2.1-2}{\frac{2}{\sqrt{n}}})\ge \mathrm{0.99.}$ This means that

$P(\frac{-0.1}{\frac{2}{\sqrt{n}}}\le z\le \frac{0.1}{\frac{2}{\sqrt{n}}})\ge \mathrm{0.99.}$ This also means that

$P(-0.05\sqrt{n}\le z\le 0.05\sqrt{n})\ge \mathrm{0.99.}$ I am not really sure what to do after this step. I am trying to use the definition of the normal distribution, however that was too difficult to do.

Do you guys know what to do after this step?

My attemp^

We are computing for the sample mean of our random sample that was given in the problem. By definition, $z=\frac{\overline{Y}-\mu}{\frac{\sigma}{\sqrt{n}}}$. So rewrite the equation so we can transform the data to make the mean 0 and the standard deviation 1. If I do that, I get,

$P(\frac{1.9-2}{\frac{2}{\sqrt{n}}}\le z\le \frac{2.1-2}{\frac{2}{\sqrt{n}}})\ge \mathrm{0.99.}$ This means that

$P(\frac{-0.1}{\frac{2}{\sqrt{n}}}\le z\le \frac{0.1}{\frac{2}{\sqrt{n}}})\ge \mathrm{0.99.}$ This also means that

$P(-0.05\sqrt{n}\le z\le 0.05\sqrt{n})\ge \mathrm{0.99.}$ I am not really sure what to do after this step. I am trying to use the definition of the normal distribution, however that was too difficult to do.

Do you guys know what to do after this step?

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What are the two types of hypotheses used in a hypothesis test? How are they related?

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Let X denote the lifetime (in hours) of a light bulb, and assume that the density function of X is given by

On average, what fraction of light bulbs last more than 15 minutes?

asked 2022-07-21

chromatic number of a graph versus its complement

What can be said about the rate of growth of f(n), defined by

$f(n)=\underset{|V(G)|=n}{min}[\chi (G)+\chi (\overline{G})],$

where the minimum is taken over all graphs G on n vertices.

Two observations.

(1) Either G or $\overline{G}$ contains a clique on roughly logn vertices by Ramsey theory, so $f(n)\ge {c}_{1}\mathrm{log}n$ for some constant ${c}_{1}>0$.

(2) If G=G(n,1/2) is a random graph, then $\chi (G)\approx \chi (\overline{G})\approx n/\mathrm{log}n$ n/ log n almost surely, so we also have $f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$ n/log n for some constant ${c}_{2}>0$.

These bounds seem hopelessly far apart.

Can we improve on the bounds

${c}_{1}\mathrm{log}n\le f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$

for all sufficiently large n?

What can be said about the rate of growth of f(n), defined by

$f(n)=\underset{|V(G)|=n}{min}[\chi (G)+\chi (\overline{G})],$

where the minimum is taken over all graphs G on n vertices.

Two observations.

(1) Either G or $\overline{G}$ contains a clique on roughly logn vertices by Ramsey theory, so $f(n)\ge {c}_{1}\mathrm{log}n$ for some constant ${c}_{1}>0$.

(2) If G=G(n,1/2) is a random graph, then $\chi (G)\approx \chi (\overline{G})\approx n/\mathrm{log}n$ n/ log n almost surely, so we also have $f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$ n/log n for some constant ${c}_{2}>0$.

These bounds seem hopelessly far apart.

Can we improve on the bounds

${c}_{1}\mathrm{log}n\le f(n)\le {c}_{2}\phantom{\rule{thinmathspace}{0ex}}n/\mathrm{log}n$

for all sufficiently large n?

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Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1. If $P(-e<z<e)=0.2128$, what is e?

asked 2022-06-24

Having some issue with the concept of Z score.

When exactly do I use $Z=\frac{\overline{X}-u}{\sigma}$

and when do I use $Z=\frac{\overline{X}-u}{\frac{\sigma}{\sqrt{n}}}$

I get very confused with in what situation should be using which Z calculation. Really appreciate it to have someone explain the concept

When exactly do I use $Z=\frac{\overline{X}-u}{\sigma}$

and when do I use $Z=\frac{\overline{X}-u}{\frac{\sigma}{\sqrt{n}}}$

I get very confused with in what situation should be using which Z calculation. Really appreciate it to have someone explain the concept

asked 2022-06-15

For $NB(r,p)$ where $r$ is the number of successes and the $p$ is the probability of success. Also we have:

$f(x)={\textstyle (}\genfrac{}{}{0ex}{}{r+x-1}{r-1}{\textstyle )}{p}^{r}{q}^{x}$

Shown that $f(x)=\frac{(r+x-1)q}{x}f(x-1)$. So:

$f(x)>f(x-1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{(r-1)q}{p}>x$

$f(x)<f(x-1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{(r-1)q}{p}<x$

And after a long calculation, I've found that the mode is the integer part of $\frac{(r-1)q}{p}$. Where I did wrong?

$f(x)={\textstyle (}\genfrac{}{}{0ex}{}{r+x-1}{r-1}{\textstyle )}{p}^{r}{q}^{x}$

Shown that $f(x)=\frac{(r+x-1)q}{x}f(x-1)$. So:

$f(x)>f(x-1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{(r-1)q}{p}>x$

$f(x)<f(x-1)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\frac{(r-1)q}{p}<x$

And after a long calculation, I've found that the mode is the integer part of $\frac{(r-1)q}{p}$. Where I did wrong?