f X is a random variable such that: E(X) = 6.2 and E(X2) = 62.5, then what is the standard deviation of X?
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f X is a random variable such that: E(X) = 6.2 and E(X2) = 62.5, then what is the standard deviation of X?
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- X is a discrete random variable such that the possible values are {a, a+1, a+2, ... b-1, b}. The random variable has a mean of 7 and a variance of 4. What is the Pr(X<=6|X>4)?x is a gaussian random variable with a PDF as described above, where μ is the mean, σ is the standard deviation , and Fx(X) refers to the cumulative distribution function CDF. It is known that Fx(-3) = 0.500 and Fx(3.7)=0.977, what value of Xo do we find the probability Fx(Xo) = P(X<Xo) = 0.159 ?Assume that adults have IQ scores that are normally distributed with a mean of mμ=100 and a standard deviation σ= 15. Find the probability that a randomly selected adult has an IQ less than 127. The probability that a randomly selected adult has an IQ less tha 127 is
- A random sample of size n₁ = 14 is selected from a normal population with a mean of 75 and a standard deviation of 7. A second random sample of size n₂ = 9 is taken from another normal population with mean 70 and standard deviation 12. Let X₁ and X₂ be the two sample means. Find: (a) The probability that X₁ X₂ exceeds 3. (b) The probability that 4.8 X₁ X₂ ≤ 6.0. Round your answers to two decimal places (e.g. 98.76). (a) i (b) -you should calculate the required probabilities using tables A model for normal human body temperature, X, when measured orally in F, is that it is normally distributed, X N(98.2, 0.5184). According to the model, what proportion of people have a normal body temperature of 99 F or more? (0) (i) Find the normal body temperature such that, according to the model, only 10% of people have a lower normal body temperature. (ii) Let W denote normal human body temperature, when measured orally in C. Given that W =59 (X-32) and X N(98.2, 0.5184), what is the distribution of W?X is a discrete random variable with the following PMF: P(X = -9.2) = 0.25 P(X = -2.5) = 0.25 P(X = 7.7) = 0.5 Find the standard deviation of X.
- (a) Let Z follow a standard normal distribution. i. Find the equi-tailed 95% probability interval, i.e. find a > 0 such that P(Z € (-a, a)) = 0.95 and express your result in terms of the inverse function -1 of the standard normal cdf (note that the inverse function satisfies -((x)) = x for all r E R). Finally write down the value of a to three significant digits. ii. Instead of an equi-tailed interval, consider the interval (-a, b) where a, b >0 are such that P(ZE (-a, b)) = 0.95. Express b as a function b(a) of a. It may help to express your results in terms of -1 and . Show that the derivative of the length I of the interval, I = b(a) + a, is given by dl = 1 da (a) (-(0.95+ (-a))' where o denotes the standard normal pdf. Hence obtain a candidate value of a for which the length of the interval is minimal by equating this derivative to zero. You do not need to show that this candidate value is an actual minimizer.Let X and Y be independent random variables such that Var[X] = 4.8 and Var[Y ] = 9.7. How can we prove that 2X and 3Y are independent? What is the standard deviation of 2X + 3Y?X is a random variable with the probability distribution f(X)= ;X= 0,1,2. Then the mean of X is equal to:
- Let x1, x2, ..., n represent a random sample from a distribution with mean E(x) and variance Var(x). Show that Cov(x, x₁ - x) = 0.Let X be the random variable with PMF: p(x) = k(]x – 2|+ 1), for x= -2, -1, 0, 1, 2 0, elsewhere (a) Find the value of k (b)Find the PMF of X (c) Find the mean, variance, and the standard deviation of X3 The median value y of a continuous random variable is that value such that F(y) = 0.5. Find the median value of the random variable in 90.5 = f(y) = 2 25 -Y, 0 ≤ y ≤ 5, elsewhere. 0,
