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?

<<the idealized uniform random number generator described.>>

is not described

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To find the probabilities, we need to use the formula for the probability density function (pdf) of a continuous random variable, which is equal to the area under the probability distribution curve.

a. P(0 ≤ X ≤ 0.4):
Since X is a continuous random variable, the probability of getting an exact value is zero. Therefore, P(0 ≤ X ≤ 0.4) is equal to the area under the curve from 0 to 0.4.
P(0 ≤ X ≤ 0.4) = 0.4 - 0 = 0.4

b. P(0.4 ≤ X ≤ 1):
Similarly, the probability here is equal to the area under the curve from 0.4 to 1.
P(0.4 ≤ X ≤ 1) = 1 - 0.4 = 0.6

c. P(0.3 ≤ X ≤ 0.5):
Again, this probability is equal to the area under the curve from 0.3 to 0.5.
P(0.3 ≤ X ≤ 0.5) = 0.5 - 0.3 = 0.2

d. P(0.3 < X < 0.5):
Comparing this to the previous case in part c, we find that the probability is still 0.2 since the inequality is strict.

e. P(0.226 ≤ X ≤ 0.713):
Similarly, this probability is equal to the area under the curve from 0.226 to 0.713.
P(0.226 ≤ X ≤ 0.713) = 0.713 - 0.226 = 0.487

f. The important fact about continuous random variables illustrated by comparing parts c and d is that the probability for an exact value of a continuous random variable is zero. In both cases, the probability of getting an exact value is zero, so the probability is calculated as the area under the curve between the specified values.

To find the probabilities for the given values, we can use the fact that X is uniformly distributed between 0 and 1. The probability density function (pdf) for a continuous random variable is a constant 1 over the interval of possible values.

a. P(0 ≤ X ≤ 0.4)
Since the pdf is a constant, the probability is equal to the length of the interval.
P(0 ≤ X ≤ 0.4) = 0.4 - 0 = 0.4

b. P(0.4 ≤ X ≤ 1)
Similar to the previous case, the probability is equal to the length of the interval.
P(0.4 ≤ X ≤ 1) = 1 - 0.4 = 0.6

c. P(0.3 ≤ X ≤ 0.5)
Again, we calculate the length of the interval.
P(0.3 ≤ X ≤ 0.5) = 0.5 - 0.3 = 0.2

d. P(0.3 < X < 0.5)
As this is an open interval, we need to use strict inequalities.
P(0.3 < X < 0.5) = P(0.3 ≤ X ≤ 0.5) = 0.2

e. P(0.226 ≤ X ≤ 0.713)
This probability is calculated as the length of the interval.
P(0.226 ≤ X ≤ 0.713) = 0.713 - 0.226 ≈ 0.487

f. The important fact about continuous random variables illustrated by comparing c. and d. is that the probability of X taking any specific value in a continuous interval is always zero. In c., we use a closed interval [0.3, 0.5], and the probability is a positive value (0.2). However, in d., when we change to an open interval (0.3 < X < 0.5), the probability remains the same (0.2). This highlights that the probability lies in the interval and not at specific values.