# IHave no idea how to solve any os these.

• MHB
• dediganss
In summary, we have discussed various probability distributions and used them to solve problems involving collecting samples and determining sample means. We used the formulas for binomial and uniform distributions to calculate probabilities and expected values. We also discussed the concept of confidence intervals and how they can be used to estimate the population mean.
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

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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