idon't know
It is the probability distribution function that is relevant for the experiment.
The answer depends on the level at which the student is expected to be. A 15-year old should know the probability of getting heads on the toss of a coin but even a mathematics graduate - who did not specialise in probability - would be expected to be able to prove the mathematical relationship between the Normal distribution and the F-distribution. If asked, most student would not even know what the second part of the sentence meant.
When you know for sure that the data you are trying to describe has a well-known theoretical probability distribution. For example, you 'know' from past experience that the heights of a certain age group in a school is normally distributed.
You need to know the standard deviation or standard error to answer the question.
This depends on what information you have. If you know the success probability and the total number of observations, you can use the given formulas. Most of the time, this is the case. If you have data or experience which allow you to estimate the parameters, it may sometimes happen that you work like this. This mostly happens when n is very large and p very small which results in an approximation with the Poisson distribution.
idon't know
It is the probability distribution function that is relevant for the experiment.
Yes, except that if you know that the distribution is uniform there is little point in using the empirical rule.
The answer depends on the level at which the student is expected to be. A 15-year old should know the probability of getting heads on the toss of a coin but even a mathematics graduate - who did not specialise in probability - would be expected to be able to prove the mathematical relationship between the Normal distribution and the F-distribution. If asked, most student would not even know what the second part of the sentence meant.
The mean and standard deviation do not, by themselves, provide enough information to calculate probability. You also need to know the distribution of the variable in question.
I will give first the non-mathematical definition as given by Triola in Elementary Statistics: A random variable is a variable typicaly represented by x that has a a single numerical value, determined by chance for each outcome of a procedure. A probability distribution is a graph, table or formula that gives the probabability for each value of the random variable. A mathematical definition given by DeGroot in "Probability and Statistics" A real valued function that is defined in space S is called a random variable. For each random variable X and each set A of real numbers, we could calculate the probabilities. The collection of all of these probabilities is the distribution of X. Triola gets accross the idea of a collection as a table, graph or formula. Further to the definition is the types of distributions- discrete or continuous. Some well know distribution are the normal distribution, exponential, binomial, uniform, triangular and Poisson.
When you know for sure that the data you are trying to describe has a well-known theoretical probability distribution. For example, you 'know' from past experience that the heights of a certain age group in a school is normally distributed.
Don't know what "this" is, but all symmetric distributions are not normal. There are many distributions, discrete and continuous that are not normal. The uniform or binomial distributions are examples of discrete symmetric distibutions that are not normal. The uniform and the beta distribution with equal parameters are examples of a continuous distribution that is not normal. The uniform distribution can be discrete or continuous.
You need to know the standard deviation or standard error to answer the question.
No, it's not a binomial. (x + 5) is an example of a binomial. A binomial has more than one term. Just look at the word binomial and you will know why. Bi means two.2xy is a monomial. Mono means one.
The successfule outcomes that fit your problem are a 3,6,9 and 12. There are 36 combinations, of which 3 can be the outcome of 2 events (1,2) and (2,1), 6 can be the outcome of 5 events, (1,5), (2,4), (3,3), (4,2), (5,1), 9 can be the outcome of 4 events (3,6),(4,5),(5,4), (6,3) and 12 is the outcome of 1 event (6,6). So out of the 36 combinations, we have 2+5+4+1 or 12 events, so 12/36 = 0.33. Now, if you throw two dice 100 times, and on the average this experiment should have 33.33 successes. Of course, some times you might have 41 successes, some times 35, but the long term average of 100 throws is 33.33 successes. This I would call the expected number of occurrences, not experimental probability. -- A bit extra to my answer: In experiments that involve chance, the results are never known. I might throw the dice 100 times and calculate a 40% of the time have success or a 25% of the time have success. These estimates are called proportions and are, I think, your "experimental probabilities" or sample estimates of success probability of your population. Of course, as given we know the success probability of the population (0.33). The binomial distribution can provide the probability of tossing two dice, "n" times and obtaining "x" successes, where the probability is 0.33. For example, I can state that 80% of the time, the number of successes will be between 27 and 39, when the dice are thrown 100 times using the binomial distribution. This is calculated by calculating the probability of 39 or fewer successes occur minus the probability of 27 or fewer sucesses occur. I have to use the cumulative distribution function (CDF). In Excel, I calculated: +binom(33-a1,100,0.33,TRUE)-binom(33-a1,100,0.33,TRUE) and varied a1 (whole numbers). When I tried a1=6, I obtained the 80%. At a1=8 (25 to 41 successes) I have the 91% confidence interval. The TRUE parameter means that I am using the CDF of the binomial distribution.