Statistics: Probability Distributions (Advanced)
Working with expected value, variance, transformations, and distribution models
Statistics: Probability Distributions (Advanced)
Working with expected value, variance, transformations, and distribution models
Statistics - Grade advanced
- 1
A discrete random variable X has probability mass function P(X = x) = c x for x = 1, 2, 3, 4. Find c, E(X), and Var(X).
First normalize the probability mass function so the probabilities sum to 1.
The constant is c = 1/10 because c(1 + 2 + 3 + 4) = 1. The expected value is E(X) = (1/10)(1^2 + 2^2 + 3^2 + 4^2) = 3. The second moment is E(X^2) = (1/10)(1^3 + 2^3 + 3^3 + 4^3) = 10, so Var(X) = 10 - 3^2 = 1. - 2
Let X be a continuous random variable with density f(x) = kx^2 for 0 < x < 2 and f(x) = 0 otherwise. Find k and P(1 < X < 2).
Use the fact that the total area under a density curve must equal 1.
The constant is k = 3/8 because the integral of kx^2 from 0 to 2 is 1. The probability P(1 < X < 2) is the integral of (3/8)x^2 from 1 to 2, which equals (3/8)(7/3) = 7/8. - 3
Suppose X has moment generating function M_X(t) = exp(4t + 9t^2 / 2). Identify the distribution of X and state its mean and variance.
Compare the given function to the moment generating function of a normal random variable.
The moment generating function matches the normal form exp(mu t + sigma squared t squared divided by 2). Therefore, X is normally distributed with mean 4 and variance 9. - 4
Let X follow a binomial distribution with n = 20 and p = 0.3. Compute E(X), Var(X), and the standard deviation of X.
For a binomial random variable, E(X) = np = 20(0.3) = 6. The variance is Var(X) = np(1 - p) = 20(0.3)(0.7) = 4.2. The standard deviation is the square root of 4.2, which is approximately 2.05. - 5
Let X follow a Poisson distribution with lambda = 5. Find P(X = 3) and P(X <= 1). Give decimal approximations to four decimal places.
Use the Poisson probability formula and add individual probabilities for the cumulative event.
For a Poisson random variable, P(X = k) = e^(-5)5^k/k!. Thus P(X = 3) = e^(-5)5^3/3! is approximately 0.1404. Also, P(X <= 1) = P(X = 0) + P(X = 1) = e^(-5)(1 + 5), which is approximately 0.0404. - 6
A random variable X has cumulative distribution function F(x) = 0 for x < 0, F(x) = x^2/16 for 0 <= x <= 4, and F(x) = 1 for x > 4. Find the density f(x), the median, and P(2 < X <= 3).
Differentiate the CDF to get the density, and use differences in the CDF to find interval probabilities.
For 0 < x < 4, the density is f(x) = F'(x) = x/8, and it is 0 outside that interval. The median m satisfies F(m) = 0.5, so m^2/16 = 0.5 and m = sqrt(8) = 2sqrt(2). The probability P(2 < X <= 3) is F(3) - F(2) = 9/16 - 4/16 = 5/16. - 7
Let X have density f(x) = e^(-x) for x >= 0. Define Y = 3X + 2. Find the density of Y.
For a linear transformation Y = aX + b with a > 0, use f_Y(y) = f_X((y - b)/a)/a.
Since Y = 3X + 2, we have X = (Y - 2)/3. The support is y >= 2. Using the change of variables formula, f_Y(y) = f_X((y - 2)/3)(1/3) = (1/3)e^(-(y - 2)/3) for y >= 2, and 0 otherwise. - 8
Let X and Y have joint density f(x, y) = 6xy for 0 < x < 1 and 0 < y < 1, and f(x, y) = 0 otherwise. Determine whether X and Y are independent.
The marginal density of X is f_X(x) = integral from 0 to 1 of 6xy dy = 3x for 0 < x < 1. The marginal density of Y is f_Y(y) = integral from 0 to 1 of 6xy dx = 3y for 0 < y < 1. Since f_X(x)f_Y(y) = 9xy, which does not equal the joint density 6xy, X and Y are not independent. - 9
Let X and Y be independent normal random variables with X ~ N(10, 4) and Y ~ N(3, 9), where the second parameter is the variance. Find the distribution of Z = X - 2Y.
For independent variables, variances add after multiplying each variance by the square of its coefficient.
A linear combination of independent normal random variables is normal. The mean is E(Z) = 10 - 2(3) = 4. The variance is Var(Z) = Var(X) + 4Var(Y) = 4 + 4(9) = 40. Therefore, Z ~ N(4, 40). - 10
A population has mean 50 and variance 100. A random sample of size n = 64 is taken. Use the central limit theorem to approximate P(48 < sample mean < 52).
Convert the sample mean bounds into z-scores using the standard error sigma divided by square root of n.
The sample mean is approximately normal with mean 50 and standard error 10/sqrt(64) = 1.25. The z-scores are (48 - 50)/1.25 = -1.6 and (52 - 50)/1.25 = 1.6. Therefore, P(48 < sample mean < 52) is approximately P(-1.6 < Z < 1.6) = 0.8904. - 11
Let X follow a gamma distribution with shape alpha = 3 and rate beta = 2. Find E(X), Var(X), and the moment generating function M_X(t).
For a gamma distribution with shape alpha and rate beta, E(X) = alpha/beta = 3/2. The variance is Var(X) = alpha/beta^2 = 3/4. The moment generating function is M_X(t) = (beta/(beta - t))^alpha = (2/(2 - t))^3 for t < 2. - 12
Let X follow a uniform distribution on the interval [a, b]. Given E(X) = 7 and Var(X) = 3, find a and b.
Use the two formulas for the mean and variance of a continuous uniform distribution.
For a uniform distribution on [a, b], the mean is (a + b)/2 = 7, so a + b = 14. The variance is (b - a)^2/12 = 3, so b - a = 6. Solving the system gives a = 4 and b = 10. - 13
Let X have density f(x) = 2x for 0 < x < 1. Find E(X | X > 0.5).
The conditional expectation is E(X | X > 0.5) = integral from 0.5 to 1 of x f(x) dx divided by P(X > 0.5). The denominator is 1 - F(0.5) = 1 - 0.25 = 0.75. The numerator is integral from 0.5 to 1 of 2x^2 dx = 2(1 - 0.125)/3 = 7/12. Thus E(X | X > 0.5) = (7/12)/(3/4) = 7/9. - 14
Let X follow a geometric distribution with success probability p = 0.2, where X counts the trial number of the first success. Find P(X > 5), E(X), and Var(X).
The event X > 5 means no success occurs in the first five trials.
For this version of the geometric distribution, P(X > 5) means the first five trials are failures, so P(X > 5) = (0.8)^5 = 0.32768. The expected value is E(X) = 1/p = 5. The variance is Var(X) = (1 - p)/p^2 = 0.8/0.04 = 20. - 15
A sequence of random variables X_n has mean mu_n = 0 and variance Var(X_n) = 1/n^2. Show that X_n converges in probability to 0.
Use Chebyshev's inequality with the target value 0.
By Chebyshev's inequality, P(|X_n - 0| >= epsilon) <= Var(X_n)/epsilon^2 = 1/(n^2 epsilon^2) for any epsilon > 0. As n goes to infinity, this upper bound goes to 0. Therefore, X_n converges in probability to 0.