Statistics Questions

Question

The probability density of a random variable X is given in the figure below.

From this density, the probability that X is between 0.62 and 1.52 is:

Answer 1

Hi Hannah,

The distribution shown in the figure is a continuous uniform distribution over the interval [0,2]. This is usually denoted as U(0,2). The uniform distribution means that every value within the interval [a,b] has an equal probability of being drawn or selected; additionally, the probability of drawing a value outside the interval is 0 (Here that is x < 0 or x > 2).

Solution 1 (CDF):

To solve this problem, we can use the cumulative distribution function for a uniform distribution. The CDF tells us F(x) = P(X ≤ x). In other words the probability that we randomly draw a value (X) less than or equal to a point of interest (x).

1. Uniform CDF: F(x) = P(X ≤ x) = (x-a)/(b-a)
2. (Note: Since all values are equally likely, P(X ≤ x) is just the percentage of values less than x over all values)
3. (You can google the CDF for a uniform distribution to get this along with its derivation)

We can use the CDF to find the probability that X is between two values x₁ and x₂. If we subtract F(x₁) = P(X ≤ x₁) from F(x₂) = P(X ≤ x₂), we get F(x₂) - F(x₁) = P(x₁ ≤ X ≤ x₂) as long as x₂ ≥ x₁ (the order would be reverse if not).

So for a Uniform distribution U(a,b) where x₂ ≥ x₁:

1. P(x₁ ≤ X ≤ x₂) = F(x₂) - F(x₁) = (x₂ - a)/(b-a) - (x₁-a)/(b-a) = (x₂-a-(x₂-a))/(b-a) = (x₂-x₁)/(b-a)
2. P(x₁ ≤ X ≤ x₂) = (x₂-x₁)/(b-a)
3. This means that the probability of selecting a value that falls between x1 and x2 can be determined by calculating the ratio of the length of the interval [x1, x2] to the length of the entire interval [a, b] in the uniform distribution. In other words, the probability of drawing a value within the interval [x1, x2] is directly proportional to the size of that interval relative to the size of the entire range of possible values.

For x₁ = 0.62 and x₂ = 1.52 on a uniform distribution U(0,2) we get:

1. P(0.62 ≤ X ≤ 1.52) = F(1.52) - F(0.62) = (1.52 - 0.62)/(2-0) = 0.45

Solution 2 (PDF and Integration):

You could also use the probability density function and integration to solve this.

For a continuous uniform distribution U(0,2), the probability density function (PDF) is given by:

1. f(x) = 1/(b-a) = 1/(2-0) = 1/2 (for 0 ≤ x ≤ 2)

Using the formula for the probability of an event within an interval, we can find the probability that X is between 0.62 and 1.52:

1. P(0.62 ≤ X ≤ 1.52) = ∫(0.62 to 1.52) f(x) dx
2. P(0.62 ≤ X ≤ 1.52) = ∫(0.62 to 1.52) 1/(b-a) dx
3. P(0.62 ≤ X ≤ 1.52) = ∫(0.62 to 1.52) 1/2 dx (substituting the PDF for U(0,2))
4. P(0.62 ≤ X ≤ 1.52) = (1/2) ∫(0.62 to 1.52) dx
5. P(0.62 ≤ X ≤ 1.52) = (1/2) [x] from 0.62 to 1.52
6. P(0.62 ≤ X ≤ 1.52) = (1/2) (1.52 - 0.62)
7. P(0.62 ≤ X ≤ 1.52) = 0.45

Therefore, the probability that X is between 0.62 and 1.52 for a uniform distribution U(0,2) is 0.45.

I hope this helps!