Question

a coin is weighted so that the probability of getting heads is \frac{2}{3}. suppose you toss this coin 15 times. let x represent the number of heads. what are the mean and standard deviation of x?
\\(\mu_x = 0.67, \sigma_x = 0.88\\)
\\(\mu_x = 7.5, \sigma_x = 2.33\\)
\\(\mu_x = 7.5, \sigma_x = 3.33\\)
\\(\mu_x = 10, \sigma_x = 1.83\\)

Explanation:

Step1: Identify the distribution and formula

This is a binomial distribution problem. For a binomial distribution \(X\sim B(n,p)\), the mean \(\mu = np\) and the standard deviation \(\sigma=\sqrt{np(1 - p)}\), where \(n = 15\) (number of trials) and \(p=\frac{2}{3}\) (probability of success).

Step2: Calculate the mean

Using the formula \(\mu=np\), substitute \(n = 15\) and \(p=\frac{2}{3}\):

$$ \mu=15\times\frac{2}{3}=10 $$

Step3: Calculate the standard deviation

First, find \(1 - p=1-\frac{2}{3}=\frac{1}{3}\). Then use the formula \(\sigma=\sqrt{np(1 - p)}\):

$$ \sigma=\sqrt{15\times\frac{2}{3}\times\frac{1}{3}}=\sqrt{\frac{30}{9}}=\sqrt{\frac{10}{3}}\approx1.83 $$

Answer:

\(\mu_{x}=10,\sigma_{x}=1.83\) (the fourth option)