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2 years ago

Given the mean of a random variable, X, is 10 and P(X < 11) = 0.67. Find the standard deviation.

Mathematics

1 answer:

Taya2010 [7]2 years ago

4 0

Answer:

Step-by-step explanation:

This is the problem we need to solve:

$z=\frac{x_i-\bar{x}}{\sigma}$ and we have everything but the z-score (which we find from a table) with our main unknown being the standard deviation.

If the probability that a random variable that is less than 11 is .67, we first have to find the z-score from the table that is closest to .67, and there are 2:

P(z < .43) = .66640 and P(z < .44) = .67003

We'll use z = .44

$.44=\frac{11-10}{\sigma}$ and

$.44=\frac{1}{\sigma}$ and

$\sigma=\frac{1}{.44}$ so

σ = 2.27 (check it; it works!)

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givi [52]

i) The given function is

$f(x)=\frac{x^2+4x-4}{x^2-2x-8}$

We can rewrite in factored form to obtain;

$f(x)=\frac{x^2+4x-4}{x^2-2x-8}$

$f(x)=\frac{(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)}{(x-4)(x+2)}$

The domain is

$(x-4)(x+2)\ne0$

$(x-4)\ne0,(x+2)\ne0$

$x\ne4,x\ne-2$

ii) To find the vertical asymptotes equate the denominator to zero.

$(x-4)(x+2)=0$

$(x-4)=0,(x+2)=0$

$x=4,x=-2$

iii) To find the roots, equate the numerator to zero.

$(x+2\sqrt{2}+2)(x-2\sqrt{2}+2)=0}$

$(x+2\sqrt{2}+2)=0,(x-2\sqrt{2}+2)=0}$

$(x=-2\sqrt{2}-2,x=2\sqrt{2}-2)}$

iv) To find the y-intercept, substitute $x=0$ into the equation.

$f(0)=\frac{0^2+4(0)-4}{0^2-2(0)-8}$

We simplify to obtain;

$f(0)=\frac{-4}{-8}$

$f(0)=\frac{1}{2}$

v) The horizontal asymptote is

$lim_{\to \infty}\frac{x^2+4x-4}{x^2-2x-8}=1$

The equation of the horizontal asymptote is y=1

vi) The function does not have a variable factor that is common to both the numerator and the denominator.

The function has no holes in it.

vii) The given function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

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Hope this helps :D

and what do you mean by BOTH of there?

there is only one question

<h2>From:</h2><h2 /><h2>kenny</h2>

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