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cdf, expectation, and variance of a random continuous variable

marcadams267

New member
Aug 26, 2019
8
Given the probability density function f(x) = b[1-(4x/10-6/10)^2] for 1.5 < x <4. and f(x) = 0 elsewhere.

1. What is the value of b such that f(x) becomes a valid density function

2. What is the cumulative distribution function F(x) of f(x)

3. What is the Expectation of X, E[X]

4. What is the Variance of X, Var[X]

5. What is the probability that X is within one standard deviation from the mean


So far, I've gotten b by integrating the function from 1.5 to 4 and setting it equal to 1, thus getting b = 3/5.
Plugging b into the function, I also integrated it from 1.5 to x, to get a cdf of (-4x^3/125 + 18x^2/125 + 48x/125 - 99/125)
To get the E[X], I integrated f(x) from 1.5 to 4 to get 2.4375
To solve for the variance i used the equation E[X^2] - E[X]^2 and got 0.3711

However, help would be appreciated for number 5 as I am not even sure where to start on that one.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,684
Given the probability density function f(x) = b[1-(4x/10-6/10)^2] for 1.5 < x <4. and f(x) = 0 elsewhere.

1. What is the value of b such that f(x) becomes a valid density function

2. What is the cumulative distribution function F(x) of f(x)

3. What is the Expectation of X, E[X]

4. What is the Variance of X, Var[X]

5. What is the probability that X is within one standard deviation from the mean


So far, I've gotten b by integrating the function from 1.5 to 4 and setting it equal to 1, thus getting b = 3/5.
Plugging b into the function, I also integrated it from 1.5 to x, to get a cdf of (-4x^3/125 + 18x^2/125 + 48x/125 - 99/125)
To get the E[X], I integrated f(x) from 1.5 to 4 to get 2.4375
To solve for the variance i used the equation E[X^2] - E[X]^2 and got 0.3711

However, help would be appreciated for number 5 as I am not even sure where to start on that one.
Hi marcadams267, welcome to MHB!

The probability that $X$ is between two arbitrary values $a_1 \le a_2$ is:
$$P(a_1<X<a_2)=\int_{a_1}^{a_2}f(x)\,dx$$
The standard deviation $\sigma[X]$ is the square root of the variance:
$$\sigma[X]=\sqrt{\operatorname{Var}[X]}$$
So in this case:
$$P(\text{X is within one standard deviation from the mean}) \\= P(E[X]-\sigma[X]<X<E[X]+\sigma[X]) \\= \int_{E[X]-\sigma[X]}^{E[X]+\sigma[X]} f(x)\,dx$$