let x denote the duration of a randomly selected pregnancy (the time.pdf

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let x denote the duration of a randomly selected pregnancy (the time between conception and birth). Accepted value for the mean value and standard deviation of x are 266 days and 16 days, respectively. Suppose that a normal distribution is an appropriate model for the probability distribution of x. What is the probability that the duration of pregnancy is between 250 and 300 days? Solution The probability is P(250.

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Let X be an r.v. defined on a sample space S into the real line R . .pdf

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let X be a raondom variable that represents the number of infants in a group of 2000 who die before reaching their first birthdays. In the US, the probability that a child dies during his or her first year of life is .0085. What is the probability that at most 5 infants out of 2000 die in their first year of life? Solution (a) What is the mean number of infants who would die in a group of this size? Since this is a binomial distribution the mean = n*p = 17 (b) What is the probability that at most five infants out of 2000 die in their first year of life? The conditions are sufficient to approximate this random variable with either a poisson or normal distribution. Poisson: P(X<=5) = P(X=0) +.

let X be a raondom variable that represents the number of infants in.pdf

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Let X be a random variable with cumulative distribution funtion0.pdf

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let X be a random variable having Poisson density parameter Calculate E[1/(X+1)] Solution As we know that, X~poisson(); E[1/(x+1)] =[1/(x+1)](^x)[e^(-)]/x! from x=0 to x=infinity =(1/)[^(x+1)][e^(-)]/(x+1)! from x=0 to x=infinity =(1/){[^(x+1)][e^(-)]/(x+1)! from x=-1 to x=infinity - e^(-)} As we know, [^(x+1)][e^(-)]/(x+1)! from x=-1 to x=infinity will become 1 because it is the sum of all supports of x, therefore, =(1/)[1-e^(-)] =1/ - e^(-).

let X be a random variable having Poisson density parameter Calcula.pdf

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Let X be a random number between 0 and 1 produced by the idealized random number generator described in Figure 10.4 (below). Find the following probabilities: 14. P(X ? 0.4) 15. P(X < 0.4) 16. P(0.3 ? X ? 0.5) Statistics and Probability Find the following probabilities:14.P(X ? 0.4) Solution a) P(X = 0.4 ) ˜ 4 b) P(x < 0.4) ˜ 4 c) P(0.2 = X = 0.6) ˜ 4 The table is wrong. You have committed a grave error building it! The sum of all the heights must be equal to 1!.

Let X be a random number between 0 and 1 produced by the idealized r.pdf

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Let X be a binomial random variable with parameters m and 13. Find .pdf

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Let X be a discrete random variable that takes on two values -1 and.pdf

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Let X and Y have the joint normal distribution described in equation.pdf

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Let X and Y be uniformly distributed in the interval [0,1]. Compute .pdf

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Let V be a vector space containing polynomials in the form f(x)=ao+a.pdf

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Let V be the vector space of ordered pairs of complex numbers over the real field [RR] . Show that V is of dimension 4. Solution Let S={(1,0),(0,1),(i,0),(0,i)} be the set . We wish to prove S is basis of V over the real field R.Consequently we can prove that dim(V)=4, because number of elements in set S is 4. Vectors in S are linearly independent. If a(1,0)+b(0,1)+c(i,0)+d(0,i)=0 a+ic=0 => a=c=0 b+id=0 => b=d=0 a=b=c=d=0 only solution. Thus vectors in S are linearly independent. Let (x+iy,w+iz) in V then (x+iy,w+iz)=x(1,0)+y(i,0)+w(0,1)+z(0,i) (x+iy,w+iz) is an arbitrary element of V. Thus every element of V can be expressed as linear combination of vectors in S. Thus S is basis for V. Number of elements in S are 4 ,so dimension of V be 4. Hence proved..

Let V be the vector space of ordered pairs of complex numbers over t.pdf

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Let X (t) be a birth and death process with possible states 0, 1, ...pdf

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