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How does a discrete probability distribution differ from a continuous probability distribution?
A discrete probability distribution is defined over a set value (such as a value of 1 or 2 or 3, etc). A continuous probability distribution is defined over an infinite number of points (such as all values between 1 and 3, inclusive).
The probability density function for a uniform distribution ranging between 2 and 6 is?
4
What are the requirements for probability distribution?
(1) That the probabilities lie between 0 and 1. (2) The sum of all probabilities of the distribution sum up to 1.
What is the probability that exactly 2 have some kind of defect?
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
What would -2 standard deviation below the mean be?
It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.It would mean that the result was 2 standard deviations above the mean. Depending on the distribution of the variable, it may be possible to attach a probability to this, or more extreme, observations.
How does a discrete probability distribution differ from a continuous probability distribution?
A discrete probability distribution is defined over a set value (such as a value of 1 or 2 or 3, etc). A continuous probability distribution is defined over an infinite number of points (such as all values between 1 and 3, inclusive).
What is the probability that a tutor will see 0 1 2 3 or 4 students?
To determine the probability that a tutor will see 0, 1, 2, 3, or 4 students, you would typically need to know the distribution of students seen by the tutor, such as a Poisson or binomial distribution. Once you have that information, you can calculate the cumulative probability for these values. For instance, if you have a Poisson distribution with a certain average (λ), you would use the formula ( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} ) for each k from 0 to 4 and sum those probabilities to get the total probability.
What is F Probability Distribution?
Given U_i~χ_(ν_i)^2, (U_1/ν_1)/(U_2/ν_2 ) follows which distribution? F_(ν_1,ν_2 ) F Probability Distribution with ν degree of freedom Given T=Z/√(U/ν), Z~N(0,1) and U~χ_ν^2, T^2 follows an F-Distribution F_(1,ν) F Probability Distribution with one degree of freedom in the numerator and ν in the denominator
For lamda equals 5.5 the probability of x equals 2 is?
The answer depends on what the distribution is!
How do you calculate probability distribution?
calulate each of the foollowing each of the following distribution x p(X)2 0 .2 1 .8
What is Chi-square Probability Distribution?
Given Z~N(0,1), Z^2 follows χ_1^2 Chi-square Probability Distribution with one degree of freedom Given Z_i~N(0,1), ∑_(i=1)^ν▒Z_i^2 follows χ_ν^2 Chi-square Probability Distribution with ν degree of freedom Given E_ij=n×p_ij=(r_i×c_j)/n, U=∑_(∀i,j)▒(O_ij-E_ij )^2/E_ij follows χ_((r-1)(c-1))^2 Chi-square Probability Distribution with ν=(r-1)(c-1) degree of freedom Given E_i=n×p_i, U=∑_(i=1)^m▒(O_i-E_j )^2/E_i follows χ_(m-1)^2 Chi-square Probability Distribution with ν=m-1 degree of freedom
The probability density function for a uniform distribution ranging between 2 and 6 is?
4
Probability of 2 out of 12 trials being positive if the probability of each trial being positive is 5 percent?
This is a binomial probability distribution; n=12, r=2 & P=.05. Read directly from the table probability of 2 is .099 (plugging this data into my calculator gives 0.09879).
What is the probability of getting a sum of 2?
That would depend on the conditions of the experiment.
What are the requirements for probability distribution?
(1) That the probabilities lie between 0 and 1. (2) The sum of all probabilities of the distribution sum up to 1.
What is the probability that exactly 2 have some kind of defect?
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
What is the area under a curve with mu equals 15 and sigma equals 2?
If the question is to do with a probability distribution curve, the answer is ONE - whatever the values of mu and sigma. The area under the curve of any probability distribution curve is 1.
