Step 1: Understanding Binomial Distribution.
A binomial distribution models the number of successes (in this case, the number of type P worms eaten) in a fixed number of independent trials (the total number of worms eaten), with two possible outcomes (P or Q worm).
Step 2: Explanation of the other options.
The Normal distribution is used for continuous data and is not suitable for discrete outcomes like counting the number of type P worms.
Log-normal distribution applies to data that is log-transformed and does not describe counts of events.
Uniform distribution assumes equal probability for all outcomes, which doesn't apply here since the bird is choosing between two types of worms.
The probability distribution of a random variable X is given by
| X | 0 | 1 | 2 |
|---|---|---|---|
| P(X) | \(1 - 7a^2\) | \(\tfrac{1}{2}a + \tfrac{1}{4}\) | \(a^2\) |
If \(a > 0\), then \(P(0 < X \leq 2)\) is equal to
The mean of the density function is \(f(x) = \lambda e^{-\lambda x}, x > 0\) is ____ .
\[ f_\theta(x) = f(x - \theta), \quad -\infty<x<\infty, \]
where \( -\infty<\theta<\infty \) and \( f(-x) = f(x) \) for \( -\infty<x<\infty \). For testing\[ H_0: \theta = 1.2 \quad \text{against} \quad H_1: \theta \neq 1.2, \]
let \( T^+ \) denote the Wilcoxon Signed-rank test statistic. If \( \eta \) denotes the probability of the event \( \{T^+<50\} \) under \( H_0 \), then \( 32\eta \) equals\[ \underline{\hspace{2cm}} \]
(round off to 2 decimal places).An ornamental shrub species was brought from Japan in the early 1800s to India, where it was planted frequently in gardens and parks. The species persisted for many decades without spreading, and then began to spread invasively fifty years ago. Which one or more of the following processes could have led to it becoming invasive?
Which one or more of the following is/are greenhouse gas(es)?