MHB IHave no idea how to solve any os these.

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The discussion centers on solving various probability and statistics problems, including confidence intervals, sample means, and Bernoulli distributions. A participant expresses confusion over standard homework problems, questioning their understanding of normal, uniform, and Bernoulli distributions. Detailed explanations are provided for calculating expected values and probabilities, particularly focusing on a uniform distribution and Bernoulli trials. The calculations reveal specific probabilities and expected values, emphasizing the importance of understanding the underlying statistical concepts. Overall, the conversation highlights the challenges faced by students in grasping these foundational topics in probability and statistics.
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  1. It is intended to collect samples from a Normal population with a standard deviation of 9. For a confidence level of 80%, determine the amplitude of the confidence interval for the population average in the case of a sample of size 81. Pick one:a. 1,28
b. 1,92

c. 1,44

d. 2,30

2) A sample of 16 observations independent of a Normal (2, 4) is collected. If Xb is the sample mean, determine the probability P [Xb> 1]. Pick one:a. 95,45%

b. 50,00%

c. 97,73%

d. 84,13%

3) A random variable has a uniform distribution in the set {-2, 2, 3}. For a random sample of size 2, the sample mean is Xb = (X1 + X2) / 2. Determine hope E [Xb]. Pick one:

a. 4/3

b. -2/3

c. -1/3

d. 1

4) A sample of 36 observations from a Normal (mu, 9) was collected and provided a sample mean of 8. Build a 95% Confidence Interval for the population mean. Pick one:

a. (7,28 ; 8,72)

b. (7,1775 ; 8,8225

c. (7,02 ; 8,98)

d. (7,36 ; 8,64)

5) A Bernoulli random variable has a probability of success p = 0.50. Considering random samples of size 3, the sample mean is given by Xb = (X1 + X2 + X3) / 3. Determine the probability P [Xb! = 2/3]. Pick one:

a. 7/8

b. 5/8

c. 3/8

d. 3/4
 
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I find this very troubling. Where did you get these problems? They look like pretty standard homework for an introductory Probability and Statistics class. Are you taking such a class? But then you say "I have no idea how to solve any of these". Why don't you? If you are taking a class in "probability and statistics ", you should have learned what a "normal distribution", "uniform distribution", and "Bernoulli distribution" are!
 
I will take a look at the easiest of these,
3) A random variable has a uniform distribution in the set {-2, 2, 3}. For a random sample of size 2, the sample mean is Xb = (X1 + X2) / 2. Determine hope E [Xb].
Saying this is a "uniform distribution" means that each possible outcome has the same probability. And those probabilities must sum to 1. So P(-2)= 1/3, P(2)=1/3, P(3)=1/3.
Since are 3 possible outcomes a "random sample of size 2" (without replacement?) has 3*2= 6 possible outcomes: (-2, 2), (-2, 3), (2, -2), (2, 3), (3, -2), and (3, 2). And since we have the uniform distribution, each pair has probability 1/6 of happening.

But it is not the pairs we are concerned with. It is Xb, the average of the sample pairs. Those are (-2+ 2)/2= 0, (-2+ 3)/2= 1/2, (2+ -2)/2= 0, (2+ 3)/2= 5/2, (3+ -2)/2= 1/2, and (3+ 2)/2= 5/2. Each number, 0, 1/2, and 5/2 occurs twice so each has probability 1/6+ 1/6= 1/3. The expected value is 0(1/3)+ (1/2)(1/3)+ (5/2)(1/3)= 1/6+ 5/6= 1.
 
Interestingly, if you assume "with replacement" you get the same answer!
With replacement, there are 3*3= 9 possible pairs, (-2, -2), (-2, 2) ,(-2, 3), (2, -2), (2, 2), (2, 3), (3, -2), (3, 2), and (3, 3) (those are the previous 6 pairs with the three new pairs, (-2, -2), (2, 2), and (3, 3)).

The average of each is
(-2+ -2)/2= -2
(-2+ 2)/2= 0
(-2+ 3)/2= 1/2
(2+ -2)/2= 0
(2+ 2)/2= 2
(2+ 3)/2= 5/2
(3+ -2)/2= 1/2
(3+ 2)/2= 5/2
(3+ 3)/2= 3

Again we have 0, 1/2, and 5/2 appearing twice each so each with probability 2/9 but now we also have -2, 2, and 3, each appearing once each so with probability 1/9 (and 2/9+ 2/9+ 2/9+ 1/9+ 1/9+ 1/9= 9/9= 1).

So the expected value is (0)(2/9)+ (1/2)(2/9)+ (5/2)(2/9)+ -2(1/9)+ 2(1/9)+ 3(1/9)=
4/18+ 2/18+ 10/18- 4/18+ 4/18+ 6/18= 22/18= 11/9.

But that is NOT one of the possible answer so apparently this intended "sampling without replacement".
 
5) A Bernoulli random variable has a probability of success p = 0.50. Considering random samples of size 3, the sample mean is given by Xb = (X1 + X2 + X3) / 3. Determine the probability P [Xb! = 2/3].

The other "easy one" (i.e. not "normal distribution"!).
I take it that Xb! = 2/3 should have been Xb != 2/3- that Xb is NOT equal to 2/3.

A Bernoulli distribution (also called a "binomial distribution" has only two possible outcomes: 1 ("success") and 0 ("failure"). Here success has probability p= 1/2= 0.50 and failure has probability q= 1- p= 1- 1/2= 1/2= 0.5. In N trials, the probability of a "successes" and N- a "failures" is $\begin{pmatrix}N \\ a \end{pmatrix}p^aq^{N-a}$ where $\begin{pmatrix}N \\ a \end{pmatrix}$ is the "binomial coefficient" $\frac{N!}{a! (N-a)!}$.

Here, since there are 3 random variables, each with 2 possible outcomes there are $2^3= 8$ total outcomes:
(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1)
(1, 0, 0), (1, 0, 1), (1, 1, 0), (1, 1, 1)
(0, 0, 0) has probability $q^3= 1/8= 0.125$. It has average (0+ 0+ 0)/3= 0.
(0, 0, 1), (0, 1, 0), and (1, 0, 0) each have probability $pq^2$ which is also 1/8 since p and q are both 1/2. Each of these has average (1+ 0+ 0)/3= 1/3.

(0, 1, 1), (1, 0, 1), and (1, 1, 0) each have probability $p^2q$ which is also 1/8.
Each of those has average (1+ 1+ 0)/3= 2/3.

And (1, 1, 1) has probability $p^3= 1/8$.
It has average (1+ 1+ 1)/3= 1.

The question asked for the probability that the average of the sample was 2/3. That is precisely the three possible outcomes (0, 1, 1), (1, 0, 1), and (1, 1, 0) each or which had probability 1/8. Therefore the probability that "(X1+ X2+ X3)/3 is 2/3" is 3/8.

(More generally the probability that (X1+ X2+ X3)/3 is 0 is 1/8, the probability it is 1/3 is 3/8, the probability it is 2/3 is 3/8, and the probability it is 1 is 1/8. Those are the only possible outcomes and 1/8+ 3/8+ 3/8+ 1/8= 8/8= 1.)
 
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