Question

In Exercises \(\mathrm{P} .112\) to \(\mathrm{P} .115,\) calculate the mean and standard deviation of the binomial random variable. A binomial random variable with } n=30 \text { and }\\\ &p=0.5 \end{aligned} $$

Short answer

So, the mean of the binomial random variable is 15 and the standard deviation is 3.873.

Step-by-step solution

Identify the Variables

From the problem statement, we can see that the number of trials, denoted as n, is 30, and the probability of success, denoted as p, is 0.5.

Find the Mean

The mean, also referred to as the expected value, of a binomial random variable is calculated by taking the product of the number of trials, n, and the probability of success, p. So, the mean will be \( \mu = n * p = 30 * 0.5 = 15 \).

Find the Standard Deviation

The standard deviation of a binomial random variable is calculated using the formula \(\sigma = \sqrt{n * p * (1-p)}\). Substituting the given values, we get the standard deviation as \( \sigma = \sqrt{30 * 0.5 * 0.5} = 3.873 \).

Key concepts

Mean of Binomial Distribution

The mean of a binomial distribution provides insight into the expected number of successes in a fixed number of independent experiments or trials. It is symbolized by the Greek letter \( \mu \) and calculated using the formula \( \mu = n \times p \) where \( n \) represents the number of trials, and \( p \) represents the probability of success on an individual trial.

For instance, if we conduct 30 independent coin tosses (our trials) and define a success as tossing a head (with a probability of 0.5), we would expect, on average, 15 successful outcomes. That's because when \( n = 30 \) and \( p = 0.5 \) are plugged into the mean formula, the result is \( 15 \) heads. This number, \( \mu = 15 \), serves as an intuitive measure of what we might anticipate if we repeat this series of trials under the same conditions.

Standard Deviation of Binomial Distribution

The standard deviation of a binomial distribution gives us a measure of how much the number of successful outcomes can vary from the mean, providing a sense of dispersion or spread. It is denoted by \( \sigma \) and estimated using the formula \( \sigma = \sqrt{n \times p \times (1-p)}\).

Applying the formula to our example with 30 coin flips (\( n = 30 \) trials) and a success probability of \( p = 0.5 \) for each trial, the standard deviation is calculated as \( \sigma = \sqrt{30 \times 0.5 \times (1-0.5)} = 3.873 \). This means that although we expect 15 heads on average, the actual number in any given set of 30 trials is likely to fluctuate, and this variation is quantified by the standard deviation.

Probability of Success

The probability of success, denoted by \( p \), in the context of a binomial distribution, is the likelihood of a single success during a trial. It is a key factor that determines the characteristics of the distribution, influencing both the mean and the standard deviation.

The value of \( p \) ranges from 0 to 1, where 0 signifies an impossible event and 1 indicates a certainty. For example, flipping a fair coin gives us \( p = 0.5 \) for getting either heads or tails, considering that the coin has equal chances of landing on either side. The probability of success must be consistent across all trials, which is an essential assumption of the binomial distribution.

Number of Trials

In a binomial distribution, the number of trials, represented by \( n \), is the total count of independent experiments or processes we are examining where each has two possible outcomes: success or failure. The value of \( n \) must be fixed beforehand, and each trial is presumed to be identical in terms of the probability of success.

The number of trials is integral to calculating the mean and standard deviation, determining the extent of the binomial distribution. If we were to increase \( n \) in our coin flip scenario, while keeping \( p \) constant, we would expect a higher mean number of successes (due to more opportunities for success) and, concurrently, a larger standard deviation (as the outcomes vary more).