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.
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)=
3 answers
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.
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.