How do u work out graphs of normal distribution ie X~N(0,4).How do u know what values it reaches on x & y axis

Question

How do u work out values on x and y axis for a normal distribution ie X~N(0,4). I understand that the peak will be at 0 on this particular graph but at what level on y, and how do you know how long it needs to stretch to on the x axis.

Thanks

Answer 1

If the rv is normally distributed, its values go from -inf to +inf. I usually use the fact that the mean indicates where the peak of the graph is, and the standard deviation indicates where the inflexion point. So, your graph peaks at x = 0, and has inflexion points at x = +2 and -2. As for the value at the peak, I usually use the formula:

(1/sqrt(2*pi)*sigma)*exp(((x - mu)^2)/2sigma^2).

Putting x = mu in (here mu = 0 and sigma = 2), you find that the peak value is 1/sqrt(2*pi)*sigma = 1/(2*sqrt(2 pi)).

Answer 2

The equation for y = f(x) = [1/sqrt(2pi sigma^2)] e^-(x - mu)/2sigma^2 becomes f(x = mu) = 1/sqrt(2 pi sigma^2) = 1/[sigma sqrt(2pi)] when x = mu, where your peak is. Thus, the height when sigma = 2 is 1/(2sqrt(2pi)) for your N(0,4) normal curve.

Practically, a normal curve runs six sigma (perhaps you've heard of six sigma project manaement) from tail to tail. Thus, your tail to tail spread, for all practical purposes is del x = 6 sigma = 12 units or plus/minus 6 units around x = mu = 0. The area under the curve outside three sigmas on either side is so small, it can be ignored in most problems.

Answer 3

while a ingredient is on the x-axis the value of y is 0. for example if the line intersects the x-axis at -3 then the ingredient is (-3,0). consequently interior the equation: 5x-7y=one hundred and five, to locate the x intercept you're making y equivalent to 0. 5x-7(0)=one hundred and five 5x=one hundred and five x=21 additionally the comparable subject with the y through fact while a ingredient is on the y axis, x is 0. consequently interior the equation: 5x-7y=one hundred and five, to locate the y intercept you're making x equivalent to 0. 5(0)-7y=one hundred and five -7y=one hundred and five y=-15