Calculating the Mean and Standard Deviation of a Binomial Random Variable
Suppose you and a friend are playing a game that involves flipping a fair coin 3 times. Let [tex]$X=$[/tex] the number of times that the coin shows heads. You have previously shown that all conditions have been met and that this scenario describes a binomial setting.
Determine the value of [tex]$n$[/tex] and [tex]$p$[/tex] and calculate the mean and standard deviation of [tex]$X$[/tex]. Round the standard deviation to three decimal places.
[tex]$n=$[/tex]
[tex]$p=$[/tex]
[tex]$\mu_x=$[/tex]
[tex]$\sigma_x=$[/tex]
Answer (1)
Identify the number of trials: n = 3 .
Identify the probability of success: p = 0.5 .
Calculate the mean: μ x = n ∗ p = 3 ∗ 0.5 = 1.5 .
Calculate the standard deviation: σ x = n ∗ p ∗ ( 1 − p ) = 3 ∗ 0.5 ∗ 0.5 ≈ 0.866 .
Explanation
Understand the problem and provided data We are given a binomial random variable X , which represents the number of heads in 3 flips of a fair coin. Our goal is to determine the values of n and p , and then calculate the mean ( μ x ) and standard deviation ( σ x ) of X .
Identify n The number of trials, which is the number of coin flips, is n = 3 .
Identify p Since the coin is fair, the probability of success (getting heads) on each trial is p = 0.5 .
Calculate the mean The mean of a binomial random variable is calculated using the formula μ x = n ∗ p . Substituting the values of n and p , we get: μ x = 3 ∗ 0.5 = 1.5
Calculate the standard deviation The standard deviation of a binomial random variable is calculated using the formula σ x = n ∗ p ∗ ( 1 − p ) . Substituting the values of n and p , we get: σ x = 3 ∗ 0.5 ∗ ( 1 − 0.5 ) = 3 ∗ 0.5 ∗ 0.5 = 0.75 ≈ 0.866 We round the standard deviation to three decimal places as requested.
State the final answer Therefore, we have n = 3 , p = 0.5 , μ x = 1.5 , and σ x ≈ 0.866 .
Examples
Consider a quality control process where you inspect 3 items from a production line. If each item has a 50% chance of being defective, this scenario follows a binomial distribution. Calculating the mean and standard deviation helps you predict the average number of defective items you might find and the variability around that average, which is crucial for managing quality and making informed decisions.
