### What is the inflection asymptote of the image of the probability density function of a normal distribution?

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As you can see from the graph above, the curves both intersect y=0 between (-5,0) and (5,10), respectively, so there are two solutions.

*Define the function and solve it

functiony=f(x)

y=1./6250.*exp(-1/50.*(x-3). ^2). *2.^(1/2). *(-16+x.^2-6.*x)/pi.^(1/2);

r=fzero(‘f’,-5)

r=-2

>>r=fzero(‘f’,5)

r=8.0000

This yields the two points of inflexion, x=8 and x=-2, which also i.e. mu+-sigma.

5) Verify that the curve is solved asymptotically with the x-axis as an asymptote:

A Vertical asymptote x=a is an asymptote of y=f(x)<==>limf(x)=∞ or limf(x)=∞

x->a+0x->a-0

While a is found in the Interrupts find – – ∞ type II interrupts B horizontal asymptotes

x+∞ (-∞) when y=b is an asymptote of y=f(x) <==>limf(x)=b (or limf(x)=b)

x->+∞x->-∞< /p>

Find the limit value b of its first-order inverse as x tends to infinity, if it exists, i.e., there is a horizontal asymptote y=b.

*First, define the function:

functiony=f(x)

symsx;%Define the symbolic variables.

y=exp(-1/2*((x-3)/5)^2)/(sqrt(2*pi)*5);

*Find the limit of the first-order derivative as x tends to infinity

limit(f,inf)

ans=0

6) Verify the 3sigma rule

Ideas: Solve the integral of the probability density function over the interval [mu-3*sigma,mu+3*sigma].

y=’exp(-1/2*((x-3)/5)^2)/(sqrt(2*pi)*5)’;

double(int(y,-12,18))

ans=0.9973