How Do You Find 'a' for P(-a ≤ X ≤ a) = 0.95?

In summary, the conversation discusses finding a value for 'a' such that P(-a <= X <= a) = 0.95, given an r.v. X with a pdf of f(x) = k(1-x2), where -1<x<1, and a c.d.f of F(x) = 3/4 * (x - x3/3 + 2/3). The individual suggests using the definition of probability in a continuous distribution to write an integral expression involving 'a', the PDF, and 0.95. After some discussion, it is concluded that P(-a < X < a) = P(X < a) - P(X < -a) = F(a) - F
  • #1
MiamiThrice
5
0
Hi all,

I was having some troubles with a practise question and thought I'd ask here.

Given an r.v. X has a pdf of f(x) = k(1-x2), where -1<x<1, I found k to be 3/4.

And I found the c.d.f F(x) = 3/4 * (x - x3/3 + 2/3)

Now I have to find a value a such that P(-a <= X <= a) = 0.95.

I thought it would be 2F(a) - 1 =0.95 and just calculate a but it didn't work.

Any help?
 
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  • #2
MiamiThrice said:
Hi all,

I was having some troubles with a practise question and thought I'd ask here.

Given an r.v. X has a pdf of f(x) = k(1-x2), where -1<x<1, I found k to be 3/4.

And I found the c.d.f F(x) = 3/4 * (x - x3/3 + 2/3)

Now I have to find a value a such that P(-a <= X <= a) = 0.95.

I thought it would be 2F(c) - 1 =0.95 and just calculate c but it didn't work.

Any help?

Hey MiamiThrice and welcome to the forums.

By your post you have figured out your pdf and cdf which is good.

So now you are asked to find a given 'a' for your relation.

Think about the limits of your integration. What is the definition of probability in a continuous distribution? What is the definition for P(X < a) and how can you define P(-a < X < a) in terms of PDF/CDF?

In saying the above, can you now write an integral expression involving 'a', the PDF, and 0.95?
 
  • #3
I think this is what you mean:

P(x<a) = F(a).

I used that to get F(-a) = 1- F(a).

So for P(-a<x<a) I thought it would be F(a) - (1 - F(a)) = 2F(a) - 1 = 0.95 ?

Was I doing it correct?
 
  • #4
MiamiThrice said:
I think this is what you mean:

P(x<a) = F(a).

I used that to get F(-a) = 1- F(a).

So for P(-a<x<a) I thought it would be F(a) - (1 - F(a)) = 2F(a) - 1 = 0.95 ?

Was I doing it correct?

Almost correct.

Your first part was correct with P(X < a) = F(a). But P(X < -a) = F(-a). From this we have P(-a < X < a) = P(X < a) - P(X < -a) = F(a) - F(-a) = 0.95. Do you know where to go from here?
 
  • #5


Hi there,

It looks like you are working with continuous random variables and trying to find a specific value for a given probability. Your approach of using the c.d.f. to solve for a is correct, but it seems like there may be a mistake in your calculation. It's difficult to provide specific help without seeing your work, but here are some general steps you can follow to solve for a:

1. Start with the c.d.f. you found: F(x) = 3/4 * (x - x^3/3 + 2/3)

2. Plug in the limits of integration (-a and a) and set the resulting equation equal to 0.95. This will give you an equation in terms of a.

3. Use algebra to solve for a. You may need to use a calculator or numerical methods to solve for a.

Make sure to check your work and make sure it makes sense. It's always a good idea to double check your calculations to make sure you didn't make any mistakes. If you are still having trouble, I suggest reaching out to your professor or a tutor for additional guidance.

Hope this helps!
 

FAQ: How Do You Find 'a' for P(-a ≤ X ≤ a) = 0.95?

What is a continuous random variable?

A continuous random variable is a type of random variable that can take on an infinite number of values within a given range. Unlike discrete random variables, which can only take on specific values, continuous random variables can take on any value within a range with a certain probability.

What is the difference between a continuous random variable and a discrete random variable?

The main difference between continuous and discrete random variables is that continuous random variables can take on an infinite number of values within a range, while discrete random variables can only take on specific values. Continuous random variables are typically associated with measurements and can take on decimal values, while discrete random variables are associated with counting and can only take on whole numbers.

What is the probability density function of a continuous random variable?

The probability density function (PDF) of a continuous random variable is a mathematical function that describes the probability of a random variable taking on a certain value within a given range. It is often denoted by f(x) and represents the height of the curve at a particular value of x. The area under the curve of the PDF between two points represents the probability of the random variable taking on a value between those two points.

How do you calculate the expected value of a continuous random variable?

The expected value of a continuous random variable is calculated by taking the integral of x*f(x) over the entire range of possible values. In other words, it is the weighted average of all possible values of the random variable, where the weights are determined by the probability of each value occurring. The expected value is often denoted by E(x) or μ (mu).

What is the central limit theorem and how does it relate to continuous random variables?

The central limit theorem states that as the sample size increases, the sampling distribution of the sample mean will approach a normal distribution, regardless of the underlying distribution of the population. This means that even if the population is not normally distributed, the sample mean will be approximately normally distributed if the sample size is large enough. This is important in statistics, as many real-life phenomena can be modeled using continuous random variables, and the central limit theorem allows us to make assumptions about the sample mean and its distribution.

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