For all problems on this page, suppose you have data
X_1,\ldots ,X_ n \overset {\text {i.i.d.}}{\sim } \mathcal{N}(0,1)
that is a random sample of identically and independently distributed standard normal random variables.
Useful facts:
The following facts might be useful: For a standard normal random variable X_1, we have:
\mathbb E[X_1] = 0, \quad \mathbb E[X^2_1] = 1, \quad \mathbb E[X^4_1] = 3.
Suppose X_1 is an observation for Bob, X_5 is an observation for Alice, X_7 is an observation for Charlie.
Using the following facts about Bob:
P( -2 < X_1 < 2 ) \approx 0.95 \quad \text {and} \quad P(-\sqrt[5]{2} \le X_1 \le \sqrt[5]{2}) \approx 0.75,
compute the probability
P( -2 < X_5 < 2, \; -2 < (X_7)^5 < 2)
of an event involving Alice and Charlie?
(Enter the probability P( -2 < X_5 < 2, \; -2 < (X_7)^5 < 2) or if the probability is not determined uniquely, then enter -1.)
(Enter a numerical answer accurate to at least 3 decimal places.)
P( -2 < X_5 < 2, \; -2 < (X_7)^5 < 2)=
are you sure
Apologies for the mistake in the previous response. Let's recalculate the probability using the correct information.
Given:
P(-2 < X_1 < 2) ≈ 0.95
P(-√[5]{2} ≤ X_1 ≤ √[5]{2}) ≈ 0.75
We know that X_5 and X_7 are also standard normal random variables, so we can use the same probabilities for them.
P(-2 < X_5 < 2) ≈ P(-2 < X_1 < 2) ≈ 0.95
P(-2 < (X_7)^5 < 2) ≈ P(-√[5]{2} ≤ X_1 ≤ √[5]{2}) ≈ 0.75
To find the probability of the intersection of these two events, we multiply the probabilities:
P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) ≈ P(-2 < X_5 < 2) * P(-2 < (X_7)^5 < 2) ≈ 0.95 * 0.75
≈ 0.7125
Therefore, the correct probability P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) is approximately 0.7125.
To compute the probability P( -2 < X_5 < 2, -2 < (X_7)^5 < 2) involving Alice and Charlie, we can use the assumption that the X_i's are identically and independently distributed standard normal random variables.
First, let's focus on the event -2 < X_5 < 2. Since X_5 is the observation for Alice, we can use the fact that P( -2 < X_1 < 2 ) ≈ 0.95 to estimate the probability of this event. So, the probability of -2 < X_5 < 2 is approximately 0.95.
Next, let's consider the event -2 < (X_7)^5 < 2. Here, (X_7)^5 represents the fifth power of the observation for Charlie. We can use the fact that P(-√[5]{2} ≤ X_1 ≤ √[5]{2}) ≈ 0.75 to estimate the probability of this event. However, since we are dealing with (X_7)^5, we need to consider the fifth roots of -√[5]{2} and √[5]{2}.
The fifth root of -√[5]{2} is approximately -1.1487, and the fifth root of √[5]{2} is approximately 1.1487. Therefore, we can rewrite the event -2 < (X_7)^5 < 2 as -1.1487 < X_7 < 1.1487. Using the fact that P(-√[5]{2} ≤ X_1 ≤ √[5]{2}) ≈ 0.75, we can estimate the probability of this event to also be approximately 0.75.
Now, to compute the probability of the event involving both Alice and Charlie, we can multiply the probabilities of the individual events. So, P( -2 < X_5 < 2, -2 < (X_7)^5 < 2) ≈ 0.95 * 0.75 ≈ 0.7125.
Therefore, the probability P( -2 < X_5 < 2, -2 < (X_7)^5 < 2) is approximately 0.7125.
To compute the probability of the event involving Alice and Charlie, we can use the fact that the X variables are i.i.d. standard normal random variables.
First, we know that the probability of -2 < X₁ < 2 is approximately 0.95. This means that P(-2 < X₅ < 2) is also approximately 0.95 since X₅ is another observation from the same distribution as X₁.
Similarly, we know that the probability of -√(2⁵) ≤ X₁ ≤ √(2⁵) is approximately 0.75. This implies that P(-√(2⁵) ≤ (X₇)⁵ ≤ √(2⁵)) is also approximately 0.75 since X₇ is from the same distribution as X₁.
Now, to compute the probability of the event involving Alice and Charlie, we can use the fact that X₅ and (X₇)⁵ are independent random variables.
P( -2 < X₅ < 2, -2 < (X₇)⁵ < 2) = P(-2 < X₅ < 2) * P(-2 < (X₇)⁵ < 2)
Using the approximations mentioned earlier, we have:
P( -2 < X₅ < 2, -2 < (X₇)⁵ < 2) ≈ 0.95 * 0.75
Calculating this, we get:
P( -2 < X₅ < 2, -2 < (X₇)⁵ < 2) ≈ 0.7125
Therefore, the probability P( -2 < X₅ < 2, -2 < (X₇)⁵ < 2) is approximately 0.7125.
Since the variables X_1, X_5, and X_7 are identically and independently distributed, we can apply the same probabilities to X_5 and X_7 as we did for X_1.
Based on the information given, we have:
P(-2 < X_5 < 2) ≈ 0.95
P(-√[5]{2} ≤ X_5 ≤ √[5]{2}) ≈ 0.75
P(-2 < (X_7)^5 < 2) ≈ 0.95
P(-√[5]{2} ≤ (X_7)^5 ≤ √[5]{2}) ≈ 0.75
To calculate the probability of the given event, we need to find the intersection of these two events. By applying the rule of independent events, we can multiply the probabilities:
P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) ≈ P(-2 < X_5 < 2) * P(-2 < (X_7)^5 < 2)
≈ 0.95 * 0.95
≈ 0.9025
Therefore, the probability P(-2 < X_5 < 2, -2 < (X_7)^5 < 2) is approximately 0.9025.