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.
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
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
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