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Mathematics

1 answer:

Damm [24]3 years ago

4 0

Answer:

-1

Step-by-step explanation:

The axes on your graph are not labeled, so we have to assume they follow the usual convention. That is, the vertical axis is the y-axis, and the horizontal axis is the x-axis.

The x-coordinate tells how far to the left or right of the y-axis the point is. Here, point L is 1 grid square to the left of the vertical line that is the y-axis. If you follow the vertical line through L down to where it crosses the x-axis, you will see an unlabeled open circle there. (We don't know the purpose of that circle, but we call it to your attention so you know you're looking in the right place.)

Looking 4 more grid squares to the left of that point, you see the marking "-5". This tells you each grid square corresponds to one unit. Then the first one to the left of the y-axis (where the open circle is) has a value of -1. That is the value of the x-coordinate of point L.

The x-coordinate of point L is -1.

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Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

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If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :