If the mean and variance of a binomial distribution are 5 and 4, respectively, then the value of n is:
25
Understanding Binomial Distribution Mean and Variance
A binomial distribution is a discrete probability distribution that describes the probability of obtaining a certain number of successes in a fixed number of independent Bernoulli trials. Two key parameters define a binomial distribution: the number of trials ($n$) and the probability of success in a single trial ($p$).
For a binomial distribution with parameters $n$ and $p$, the mean ($\mu$) and variance ($\sigma^2$) are given by specific formulas:
- Mean ($\mu$) = $np$
- Variance ($\sigma^2$) = $np(1-p)$
In this problem, we are given the mean and variance of a binomial distribution and need to find the value of $n$, the number of trials.
Solving for Binomial Parameters
We are given the following information:
- Mean, $np = 5$
- Variance, $np(1-p) = 4$
We have a system of two equations with two unknowns, $n$ and $p$. We can use these equations to solve for $p$ first, and then use the value of $p$ to find $n$.
Let's use the given equations:
- $\qquad np = 5$
- $\qquad np(1-p) = 4$
Substitute the value of $np$ from equation (1) into equation (2):
$\qquad 5(1-p) = 4$
Now, we can solve for $p$:
$\qquad 1-p = \frac{4}{5}$
$\qquad p = 1 - \frac{4}{5}$
$\qquad p = \frac{5-4}{5}$
$\qquad p = \frac{1}{5}$
The probability of success, $p$, is $\frac{1}{5}$. Since $p$ represents a probability, its value must be between 0 and 1 ($0 \le p \le 1$). Our calculated value $p=1/5$ satisfies this condition.
Now that we have the value of $p$, we can substitute it back into equation (1) to find the value of $n$:
$\qquad np = 5$
$\qquad n \left(\frac{1}{5}\right) = 5$
To solve for $n$, multiply both sides by 5:
$\qquad n = 5 \times 5$
$\qquad n = 25$
The value of $n$, the number of trials, is 25. Since $n$ represents the number of trials, it must be a positive integer. Our calculated value $n=25$ is a positive integer.
Verifying the Binomial Parameters
With $n=25$ and $p=1/5$, let's check if the mean and variance match the given values:
- Mean = $np = 25 \times \frac{1}{5} = \frac{25}{5} = 5$. This matches the given mean.
- Variance = $np(1-p) = 25 \times \frac{1}{5} \times (1 - \frac{1}{5}) = 5 \times \frac{4}{5} = \frac{20}{5} = 4$. This matches the given variance.
The values $n=25$ and $p=1/5$ are consistent with the given mean and variance of the binomial distribution.
| Parameter | Formula | Given Value | Calculated Value |
|---|---|---|---|
| Mean ($\mu$) | $np$ | 5 | $25 \times \frac{1}{5} = 5$ |
| Variance ($\sigma^2$) | $np(1-p)$ | 4 | $25 \times \frac{1}{5} \times (1 - \frac{1}{5}) = 5 \times \frac{4}{5} = 4$ |
The value of $n$ is 25.
Revision Table: Binomial Distribution Formulas
| Concept | Formula/Description |
|---|---|
| Probability Mass Function (PMF) | $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ for $k = 0, 1, \dots, n$ |
| Mean ($\mu$) | $E(X) = np$ |
| Variance ($\sigma^2$) | $Var(X) = np(1-p)$ |
| Standard Deviation ($\sigma$) | $\sigma = \sqrt{np(1-p)}$ |
Additional Information: Properties of Binomial Distribution
A binomial distribution arises from a sequence of Bernoulli trials. A Bernoulli trial is a single experiment with only two possible outcomes: success or failure. Key properties of a binomial distribution include:
- There are a fixed number of trials, denoted by $n$.
- Each trial is independent of the others.
- For each trial, there are only two possible outcomes: "success" and "failure".
- The probability of success, denoted by $p$, is constant for every trial. The probability of failure is $q = 1-p$.
- The random variable $X$ represents the number of successes in $n$ trials.
Understanding the mean and variance formulas is crucial for working with binomial distributions, as they allow us to characterize the distribution's central tendency and spread based on the parameters $n$ and $p$. In this problem, we used the inverse approach, starting from the mean and variance to find the parameters.
Important Questions from Probability Distribution
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:
For a random variable X following Poisson distribution with parameter 5, the variance of X is:
