Continuos probability densities with calculus
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Continuos probability densities with calculus

[From: ] [author: ] [Date: 13-03-22] [Hit: ]
Assume that our random variable x always has a smallest value m and a largest value M.(a) Let a be a fixed number between m and M. Explain why p(x=a) = 0.(b) For a discrete probability distribution the random variable x can only take on the values x1,x2,.......
Certain positive integrable functions are continuous probability densities. We use them to give probabilities to event ranges:

p(a≤x≤b)= integral from a to b of f(x) dx

In words, we say: "The probability that the variable x takes on a value between a and b is given by the integral." Let f be such a distribution. Assume that our random variable x always has a smallest value m and a largest value M.
(a) Let a be a fixed number between m and M. Explain why p(x=a) = 0. (Include your explanation in
(b) For a discrete probability distribution the random variable x can only take on the values x1,x2,...,xn with probabilities p(x1),p(x2), ,p(xn). The mean is defined as: µ=∑(i=1 to n) of (xi)(p(xi). Show that the mean of a continuous density on [m,M] should be µ=the integral from m to M of x(fx)dx
c) The standard deviation of a discrete random variable is defined to be
√(∑ (from i=1 to n) of (xi-µ)^2(p(xi)). where μ is the mean of the distribution. Derive a formula for the standard deviation of a continuous density on [m,M].


Let f be the function defined by
0, x<0
x, 0≤x≤1
0, 1 .5, 2≤x≤3
0, x>3

It can be shown that f satisfies the conditions of being a continuous density (you don't have to show this). let x be the associated random variable.

d)What is p(1.5 ≤ x ≤ 2.5)?
e) what is the mean of x
f) wht is the standard deviation of x

i think i got the a, b, and c, but not positive and im just ocnfused with d,e, and f. any help would be appreciated.

-
jo

d)What is p(1.5 ≤ x ≤ 2.5)?

Between 1.5 and 2, f(x) equals 0. From 2 to 2.5, the area of the rectangle is 0.5 x (2.5 - 2) = 0.25

p(1.5 ≤ x ≤ 2.5) = 0.25


e) what is the mean of x

Integral from x = 0 to x = 1: (x)(x) dx = 1/3 PLUS ...

Integral from x = 2 to x = 3: (x)(0.5) dx = 5/4

E(x) = 5/4 + 1/3 = 19/12



f) wht is the standard deviation of x

Same integral limits above, but instead use (x - 19/12)^2 f(x)

This will give you the variance, so simply take the square root to find the std dev.


Hope that helps
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