Question

The mgf of a random variable \(X\) is \(\left(\frac{2}{3}+\frac{1}{3} e^{t}\right)^{9} .\) Show that $$ P(\mu-2 \sigma

Short answer

The probability that \(X\) falls between the mean (- 2 times the standard deviation) and the mean (+ 2 times the standard deviation) is given by the formula \(\sum_{x=1}^{5} \binom{9}{x} (1/3)^{x}(2/3)^{9-x}\).

Step-by-step solution

Identify the Distribution

The given moment generating function (mgf) is \(\left(\frac{2}{3}+\frac{1}{3} e^{t}\right)^{9}\). This matches the mgf of a binomial distribution, with \(n = 9\) and probability \(p = 1/3\). Thus, the random variable \(X\) follows a Binomial Distribution \(B(9, 1/3)\).

Calculate Mean and Variance

For a binomial distribution, the mean (\(\mu\)) and variance (\(\sigma^2\)) are given by \(\mu = np\) and \(\sigma^2 = np(1-p)\). Substituting \(n = 9\) and \(p = 1/3\) we get, \(\mu = 9 * (1/3) = 3\) and \(\sigma^2 = 9 * (1/3) * (2/3) = 2\). Therefore, \(\sigma = \sqrt{2}\).

Determine X

The range of \(X\) is defined by \(\mu-2\sigma < X < \mu+2\sigma\). Substituting \(\mu = 3\) and \(\sigma = \sqrt{2}\), the inequality becomes: \(3 - 2\sqrt{2} < X < 3 + 2\sqrt{2}\). This simplifies to approximately \(1.17 < X < 4.83\). Thus, for integer \(X\), the possible values of \(X\) are in the range \(1 \le x \le 5\) .

Calculate Probability

We need to calculate \(P(1 \le X \le 5)\) for the binomial distribution, which is \(\sum_{x=1}^{5} \binom{9}{x} (1/3)^{x}(2/3)^{9-x}\).

Conclusion

The probability that \(X\) falls between the mean less 2 times the standard deviation and the mean plus 2 times the standard deviation is given by \(\sum_{x=1}^{5} \binom{9}{x} (1/3)^{x}(2/3)^{9-x}\). This can be calculated directly for an exact numerical result.

Key concepts

Moment Generating Function

The moment generating function (mgf) plays a critical role in probability and statistics as it uniquely characterizes the distribution of a random variable. It not only helps identify the type of distribution, but also allows us to compute the moments (mean, variance, etc.) of the distribution.

An mgf is defined as the expected value of the exponential function of the random variable, typically denoted as: \[ M_X(t) = E(e^{tX}) \] where \(X\) is the random variable and \(t\) is a real number.

The given mgf in the exercise, \( \left(\frac{2}{3}+\frac{1}{3}e^{t}\right)^{9} \), reveals that we are dealing with a binomial distribution by matching it to the standard form of an mgf for a binomially distributed variable. Recognizing the mgf is the first step in solving probability problems involving binomial distributions.

Probability

Probability quantifies the likelihood of an event occurring and ranges from 0 (impossible event) to 1 (certain event). In a binomial distribution, which models the number of successes in a fixed number of independent trials, the probability of exactly \(k\) successes is given by:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \(n\) is the number of trials, \(p\) is the probability of success on each trial, \(k\) is the number of successes, and \( \binom{n}{k} \) is the binomial coefficient representing the number of ways to choose \(k\) successes from \(n\) trials.

For our particular exercise, the range of probable outcomes and their corresponding probabilities can be represented and computed based on the binomial probability formula.

Mean and Variance of Binomial Distribution

In a binomial distribution with parameters \(n\) and \(p\), where \(n\) represents the number of trials and \(p\) the probability of success, the mean (average number of successes) and variance (measure of the spread of the distribution) are crucial concepts.

The mean is denoted by \(\mu\) and is calculated as \(\mu = np\). The variance, symbolized by \(\sigma^2\), is computed as \(\sigma^2 = np(1-p)\).

Knowing both the mean and variance is vital for understanding the distribution's behavior. For example, they help determine the likely range of outcomes and allow for the calculation of probabilities for intervals, such as in the exercise where we look for \(P(\mu - 2\sigma < X < \mu + 2\sigma)\).

Standard Deviation

The standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of a binomial distribution, it is the square root of the variance and provides insights into the spread of the distribution around the mean.

The standard deviation, \(\sigma\), is therefore determined by taking the square root of the variance, \(\sigma = \sqrt{np(1-p)}\). A lower standard deviation indicates that the values tend to be close to the mean, while a higher standard deviation points to values being spread out over a wider range.

In our exercise, we use the standard deviation to define a range around the mean, which is helpful in probabilistic forecasting and in setting confidence intervals for predictions.