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(0¡ÜX¡Ü0.4)
b.P(0.4¡ÜX¡Ü1)
c.P(0.3¡ÜX0.5)
d.P(0.3(<X<0.5)
e.P(0.226¡ÜX¡Ü0.713)
f.What important fact about continuous random variable does comparing your answers to c. and d. illustrate?
How do you catch a squirrel? Climb a tree and act like a nut! Now, let's tackle those probabilities:
a. P(0 ≤ X ≤ 0.4) = 0.4 - 0 = 0.4
b. P(0.4 ≤ X ≤ 1) = 1 - 0.4 = 0.6
c. P(0.3 ≤ X ≤ 0.5) = 0.5 - 0.3 = 0.2
d. P(0.3 < X < 0.5) = P(0.3 ≤ X ≤ 0.5) = 0.2
e. P(0.226 ≤ X ≤ 0.713) = 0.713 - 0.226 ≈ 0.487
Now, comparing c. and d., we can see that the probability of X being at a specific point (0.3 or 0.5) is zero for a continuous random variable. This highlights the fact that the probability for a continuous random variable to take on a specific value is always zero. It's like the chances of me getting a date on Valentine's Day – statistically negligible!
To find the probabilities for a continuous random variable, we need to use the probability density function (PDF) and integrate it over the given range.
The PDF for a continuous uniform distribution between a and b is given by:
f(x) = 1 / (b - a) for a <= x <= b
and f(x) = 0 otherwise.
a. P(0 <= X <= 0.4)
Since X is uniformly distributed between 0 and 1, the range we are interested in is from 0 to 0.4. Therefore, the probability is:
P(0 <= X <= 0.4) = integral from 0 to 0.4 of 1 / (1 - 0) dx
= integral from 0 to 0.4 of 1 dx
= [x] from 0 to 0.4
= 0.4 - 0
= 0.4
b. P(0.4 <= X <= 1)
Similarly, the range we are interested in is from 0.4 to 1. Therefore, the probability is:
P(0.4 <= X <= 1) = integral from 0.4 to 1 of 1 / (1 - 0) dx
= integral from 0.4 to 1 of 1 dx
= [x] from 0.4 to 1
= 1 - 0.4
= 0.6
c. P(0.3 <= X <= 0.5)
This range is partially between 0 and 1, but the endpoint is not at either end. Therefore, we need to calculate the probabilities for each side of the range separately and subtract them. The probability for the left side is:
P(0 <= X <= 0.3) = integral from 0 to 0.3 of 1 / (1 - 0) dx
= integral from 0 to 0.3 of 1 dx
= [x] from 0 to 0.3
= 0.3 - 0
= 0.3
Similarly, the probability for the right side is:
P(0.5 <= X <= 1) = integral from 0.5 to 1 of 1 / (1 - 0) dx
= integral from 0.5 to 1 of 1 dx
= [x] from 0.5 to 1
= 1 - 0.5
= 0.5
Therefore, the probability of interest is:
P(0.3 <= X <= 0.5) = P(0 <= X <= 0.5) - P(0 <= X <= 0.3)
= 0.5 - 0.3
= 0.2
d. P(0.3 < X < 0.5)
Since the range excludes the endpoints, we can calculate it as:
P(0.3 < X < 0.5) = P(0 <= X <= 0.5) - P(0 <= X <= 0.3)
= 0.5 - 0.3
= 0.2
e. P(0.226 <= X <= 0.713)
Using the same approach, we can calculate it as:
P(0.226 <= X <= 0.713) = integral from 0.226 to 0.713 of 1 / (1 - 0) dx
= integral from 0.226 to 0.713 of 1 dx
= [x] from 0.226 to 0.713
= 0.713 - 0.226
= 0.487
f. Comparing answers c. and d., we can observe that the probability of the range [0.3, 0.5] is the same when we include or exclude the endpoints. This is an important fact about continuous random variables - the probability of getting an exact value is always zero. In other words, the probability of a continuous random variable falling within a specific point is always zero.
To find the probabilities for the given continuous random variable, you need to calculate the area under the probability density function (pdf) curve within the specified intervals.
For a continuous random variable with a uniform distribution between 0 and 1, the pdf is a constant 1 over the interval [0,1]. The total area under the curve is equal to 1.
a. P(0 ≤ X ≤ 0.4):
To calculate this probability, you need to find the area under the pdf curve from 0 to 0.4. Since the pdf is a constant of 1, the area is simply the length of the interval [0, 0.4]. So the probability is 0.4.
b. P(0.4 ≤ X ≤ 1):
Similarly, the length of the interval [0.4, 1] is 0.6. Therefore, the probability is 0.6.
c. P(0.3 ≤ X < 0.5):
This interval includes the probability between 0.3 and 0.5, but it doesn't include 0.5 itself. The length of the interval [0.3, 0.5) is 0.2. Therefore, the probability is 0.2.
d. P(0.3 < X < 0.5):
This interval includes the probability between 0.3 and 0.5, but excludes both endpoints. The length of the interval (0.3, 0.5) is also 0.2. Therefore, the probability is 0.2.
e. P(0.226 ≤ X ≤ 0.713):
Calculate the length of the interval [0.226, 0.713]. The length is 0.713 - 0.226 = 0.487. Therefore, the probability is 0.487.
f. Comparing the answers for c. and d., you can observe that P(0.3 ≤ X < 0.5) is equal to P(0.3 < X < 0.5). This illustrates an important fact about continuous random variables: the probability for a single point on the continuous distribution is always zero. As a result, it doesn't matter whether we use a strict inequality or not when defining intervals. The probability of getting an exact value is always zero.