If X is the number of 6s obtained in 10 throws of a die, then the variance of X is:
25/18
The variance for the number of 6s obtained in 10 throws of a die follows a binomial distribution with n = 10 and p = 1/6. The variance is calculated as np(1 - p), which results in 25/18.
Important Questions from Probability Distribution
If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
For the distribution with unknown θ
\(f(x,\theta ) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{\theta };0 \le x \le \theta }\\ {0;elsewhere} \end{array}} \right.\)
We set the testing of hypothesis H 0 ∶ θ = 1 vs H 1 ∶ θ = 2. When the critical region X ≥ 0.4, the value of probability of type-II error is:
For the cumulative distribution function \(F(x) = \left\{ {\begin{array}{*{20}{c}} {0;x < - 1}\\ {\frac{1}{2}{{(x + 1)}^2}; - 1 \le x < 0}\\ {1 - \frac{{{{(1 - x)}^2}}}{2};0 \le x < 1}\\ {1.1 < x < \infty } \end{array}} \right.\)
the upper quartile point is
Let the joint probability density function of \( (X, Y) \) be
\[f(x, y) = \begin{cases} 6xy^2 & \text{if } 0 < x < 1, 0 < y < 1 \\ 0, & \text{otherwise} \end{cases}\]
Then \( P\left(\frac{1}{2} < X < \frac{3}{4}\right) \) is:
Let X and Y have the joint p.m.f. f(x, y) = x + y / 21, where x = 1, 2, 3 and y = 1, 2. The marginal p.m.f. of X is:
