IHave no idea how to solve any os these.

[dediganss]

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

 

[HOI]

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!

 

[HOI]

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.

 

[HOI]

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".

 

[HOI]

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.)