# When dealing with the binomial distribution, why are the possible values for the random variable always 0,1,2,3,…,n where n is the number of trials or sample size? Why can't we use negative values, or fractions, or numbers greater than n?​

Question
Binomial probability
When dealing with the binomial distribution, why are the possible values for the random variable always 0,1,2,3,…,n where n is the number of trials or sample size? Why can't we use negative values, or fractions, or numbers greater than n?​

2021-03-07
The possible values of random variable is the number of successes in n trials. The number of successes cannot be more than number of trials. Maximum will be equal to number of trials n and minimum will be 0 successes. Number of success cannot be a negative value, a fraction (or decimal) or a value greater than n.
Example: Consider flipping a coin 5 times and success is defined as getting a tail. Number of tails cannot be a negative value. It cannot be more than 5 also The possible values of number of tails (here the success of the experiment) are 0, 1, 2, 3, 4 and 5.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.

A population of values has a normal distribution with $$\displaystyle\mu={29.3}$$ and $$\displaystyle\sigma={65.1}$$. You intend to draw a random sample of size $$\displaystyle{n}={142}$$.
Find the probability that a sample of size n=142 is randomly selected with a mean between 27.7 and 35.3.
$$\displaystyle{P}{\left({27.7}{<}\overline{{{X}}}{<}{35.3}\right)}=$$?

A population of values has a normal distribution with $$\displaystyle\mu={198.8}$$ and $$\displaystyle\sigma={69.2}$$. You intend to draw a random sample of size $$\displaystyle{n}={147}$$.
Find the probability that a sample of size $$\displaystyle{n}={147}$$ is randomly selected with a mean between 184 and 205.1.
$$\displaystyle{P}{\left({184}{<}{M}{<}{205.1}\right)}=$$?

A population of values has a normal distribution with $$\displaystyle\mu={204.3}$$ and $$\displaystyle\sigma={43}$$. You intend to draw a random sample of size $$\displaystyle{n}={111}$$.
Find the probability that a single randomly selected value is less than 191.2.
$$\displaystyle{P}{\left({X}{<}{191.2}\right)}=$$?
Find the probability that a sample of size $$\displaystyle{n}={111}$$ is randomly selected with a mean less than 191.2.
$$\displaystyle{P}{\left({M}{<}{191.2}\right)}=$$?

Let $$X_{1},X_{2},...,X_{6}$$ be an i.i.d. random sample where each $$X_{i}$$ is a continuous random variable with probability density function
$$f(x)=e^{-(x-0)}, x>0$$
Find the probability density function for $$X_{6}$$.
Suppose that $$X_{1}, X_{2} and X_{3}$$ are three independent random variables with the same distribution as X.
What is the ecpected value of the sum $$X_{1}+X_{2}+X_{3}$$? The product $$X_{1}X_{2}X_{3}$$?
Suppose a discrete random variable X assumes the value $$\frac{3}{2}$$ with probability 0.5 and assumes the value $$\frac{1}{2}$$ with probability 0.5.

A population of values has a normal distribution with $$\displaystyle\mu={116.5}$$ and $$\displaystyle\sigma={63.7}$$. You intend to draw a random sample of size $$\displaystyle{n}={244}$$.
Find the probability that a sample of size $$\displaystyle{n}={244}$$ is randomly selected with a mean between 104.7 and 112.8.
$$\displaystyle{P}{\left({104.7}{<}{M}{<}{112.8}\right)}=$$?
Random variables $$X_{1},X_{2},...,X_{n}$$ are independent and identically distributed. 0 is a parameter of their distribution.
If $$X_{1}, X_{2},...,X_{n}$$ are Normally distributed with unknown mean 0 and standard deviation 1, then $$\overline{X} \sim N(\frac{0,1}{n})$$. Use this result to obtain a pivotal function of X and 0.
Random variables $$X_{1},X_{2},...,X_{n}$$ are independent and identically distributed. 0 is a parameter of their distribution.
If $$q(X,0)\sim N(0,1)$$ is a pivotal function for 0, explain how you would use this result to obtain a symmetrical 95% confidence interval for 0.