Continuous Random Variable, I Let X be a random number between 0 and 1 produced by the idealized uniform random number generator described. Find the following probabilities:

a.P(0less than or equal to X less than or equal to 0.4)

b.P(0.4 less than or equal to X less than or equal to 1)

c.P(0.3 less than or equal to X 0.5)

d.P(0.3(less than X less than 0.5)

e.P(0.226 less than or equal to X less than or equal to 0.713)

f. What important fact about continuous random variable does comparing your answers to c. and d. illustrate?

To find the probabilities for the given cases, we can use the formula for the probability density function (PDF) of a continuous random variable.

The PDF for a continuous uniform distribution on the interval [a, b] is given by:

f(x) = 1 / (b - a), for a ≤ x ≤ b

a. P(0 ≤ X ≤ 0.4):

Since X is uniformly distributed between 0 and 1, we have:
a = 0
b = 1

Using the formula for the PDF, we can calculate the probability:
P(0 ≤ X ≤ 0.4) = ∫[0, 0.4] (1 / (1 - 0)) dx
= ∫[0, 0.4] 1 dx
= [x] [0, 0.4]
= 0.4 - 0
= 0.4

Therefore, P(0 ≤ X ≤ 0.4) is 0.4.

b. P(0.4 ≤ X ≤ 1):

Using the same approach as in part a:
P(0.4 ≤ X ≤ 1) = ∫[0.4, 1] (1 / (1 - 0)) dx
= ∫[0.4, 1] 1 dx
= [x] [0.4, 1]
= 1 - 0.4
= 0.6

Therefore, P(0.4 ≤ X ≤ 1) is 0.6.

c. P(0.3 ≤ X ≤ 0.5):

Using the given range, we have:
P(0.3 ≤ X ≤ 0.5) = ∫[0.3, 0.5] (1 / (1 - 0)) dx
= ∫[0.3, 0.5] 1 dx
= [x] [0.3, 0.5]
= 0.5 - 0.3
= 0.2

Therefore, P(0.3 ≤ X ≤ 0.5) is 0.2.

d. P(0.3 < X < 0.5):

Notice that in this case, we have strict inequalities. Since X is continuous and there is no probability associated with a single point, we need to consider the probability of X being within a range but not including the endpoints.

P(0.3 < X < 0.5) = ∫[0.3, 0.5] (1 / (1 - 0)) dx
= ∫[0.3, 0.5] 1 dx
= [x] [0.3, 0.5]
= 0.5 - 0.3
= 0.2

Therefore, P(0.3 < X < 0.5) is also 0.2.

e. P(0.226 ≤ X ≤ 0.713):

Using the given range, we have:
P(0.226 ≤ X ≤ 0.713) = ∫[0.226, 0.713] (1 / (1 - 0)) dx
= ∫[0.226, 0.713] 1 dx
= [x] [0.226, 0.713]
= 0.713 - 0.226
= 0.487

Therefore, P(0.226 ≤ X ≤ 0.713) is 0.487.

f. Comparing answers c. and d., we can observe that the probabilities are the same. This illustrates an important fact about continuous random variables, which is that the probability of an individual point occurring in a continuous distribution is always zero. In other words, P(X = a) = 0 for any specific value a. Instead, we consider the probability of ranges or intervals for continuous random variables.