A coin is weighted so that the probability of getting heads is [tex]$\frac{2}{3}$[/tex]. Suppose you toss this coin 15 times. Let X represent the number of heads. What are the mean and standard deviation of X?
A. [tex]$\mu_x=0.67, \sigma_x=0.88$[/tex]
B. [tex]$\mu_x=7.5, \sigma_x=2.33$[/tex]
C. [tex]$\mu_x=7.5, \sigma_x=3.33$[/tex]
D. [tex]$\mu_x=10, \sigma_x=1.83$[/tex]
Answer (1)
Recognize X as a binomial distribution.
Calculate the mean using μ = n p = 15 × 3 2 = 10 .
Calculate the standard deviation using σ = n p ( 1 − p ) = 15 × 3 2 × 3 1 = 3 10 ≈ 1.83 .
State the mean and standard deviation: μ = 10 , σ = 1.83 .
Explanation
Understand the problem We are given a binomial distribution problem where a coin is weighted such that the probability of getting heads is 3 2 . The coin is tossed 15 times, and we want to find the mean and standard deviation of the number of heads, denoted by X .
Identify parameters The number of trials is n = 15 , and the probability of success (getting heads) is p = 3 2 . We can use the formulas for the mean and standard deviation of a binomial distribution.
Calculate the mean The mean of a binomial distribution is given by the formula: μ = n p Substituting the values, we get: μ = 15 × 3 2 = 10
Calculate the variance The variance of a binomial distribution is given by the formula: σ 2 = n p ( 1 − p ) Substituting the values, we get: σ 2 = 15 × 3 2 × ( 1 − 3 2 ) = 15 × 3 2 × 3 1 = 9 30 = 3 10
Calculate the standard deviation The standard deviation is the square root of the variance: σ = σ 2 = 3 10 ≈ 1.8257 Rounding to two decimal places, we get σ ≈ 1.83 .
State the final answer Therefore, the mean is 10 and the standard deviation is approximately 1.83.
Examples
Consider a basketball player who makes a free throw with a probability of 3 2 . If the player attempts 15 free throws, the mean number of successful free throws is 10, with a standard deviation of approximately 1.83. This tells us the average number of successful shots and the variability we can expect around that average.
