A spinner is divided into six equal-sized sectors labeled 1 through 6. Dwayne spins this spinner 12 times. Let $X$ represent the number of 1s that are spun. What are the mean and standard deviation of $X$?
$\mu_x=1, \sigma_x=1.29$
$\mu_x=2, \sigma_x=1.29$
$\mu_x=2, \sigma_x=1.67$
$\mu_x=10, \sigma_x=2.78$
Answer (1)
Recognize the problem as a binomial distribution.
Calculate the mean using the formula: μ = n p = 12 × 6 1 = 2 .
Calculate the standard deviation using the formula: σ = n p ( 1 − p ) = 12 × 6 1 × 6 5 ≈ 1.29 .
State the mean and standard deviation: μ = 2 , σ = 1.29 .
Explanation
Understand the problem We are given a problem where a spinner with 6 equal sectors is spun 12 times. We want to find the mean and standard deviation of X , where X is the number of times the spinner lands on 1.
Identify the distribution This is a binomial distribution problem because each spin is independent, and we are counting the number of successes (spinning a 1) in a fixed number of trials. The probability of success on each trial is p = 6 1 , and the number of trials is n = 12 .
Calculate the mean The mean of a binomial distribution is given by the formula μ = n p . In this case, μ = 12 × 6 1 = 2 .
Calculate the variance The variance of a binomial distribution is given by the formula σ 2 = n p ( 1 − p ) . In this case, σ 2 = 12 × 6 1 × ( 1 − 6 1 ) = 12 × 6 1 × 6 5 = 36 60 = 3 5 .
Calculate the standard deviation The standard deviation is the square root of the variance, so σ = 3 5 = 3 5 ≈ 1.29 .
State the final answer Therefore, the mean is 2 and the standard deviation is approximately 1.29.
Examples
Consider a quality control process where you inspect 12 items from a production line. If each item has a 1/6 chance of being defective, this problem helps you calculate the average number of defective items you'd expect to find (the mean) and how much the number of defective items might vary from one inspection to another (the standard deviation). This helps in predicting and managing the quality of the production line.
