Step 1: Concept Explanation:
This problem requires finding Pearson's coefficient of skewness, \(\beta_1\), for a Bernoulli distribution. Skewness measures the asymmetry of a probability distribution and is defined using central moments.
Step 2: Formula and Approach:
The coefficient of skewness \(\beta_1\) is given by:
\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3} \]
where \(\mu_2\) is the second central moment (variance) and \(\mu_3\) is the third central moment. For a Bernoulli(\(p\)) distribution, \(X=1\) with probability \(p\) and \(X=0\) with probability \(1-p\). The \(k\)-th central moment is \(\mu_k = E[(X-\mu)^k]\), with \(\mu\) being the mean.
Step 3: Detailed Solution:
For a Bernoulli(\(p\)) distribution:
The mean is \(\mu = E[X] = 1 . p + 0 . (1-p) = p\).
The second central moment (variance) is \(\mu_2 = E[(X-p)^2] = \sigma^2 = p(1-p)\).
Now, calculate the third central moment, \(\mu_3\):
\[ \mu_3 = E[(X-\mu)^3] = E[(X-p)^3] \]
We calculate the expected value by summing the possible values of \((X-p)^3\) weighted by their probabilities:
If \(X=1\), the value is \((1-p)^3\) with probability \(p\).
If \(X=0\), the value is \((0-p)^3 = -p^3\) with probability \(1-p\).
Thus,
\[ \mu_3 = p(1-p)^3 + (1-p)(-p^3) \]
Factor out \(p(1-p)\):
\[ \mu_3 = p(1-p) \left[ (1-p)^2 - p^2 \right] \]
Apply the difference of squares formula, \(a^2 - b^2 = (a-b)(a+b)\):
\[ \mu_3 = p(1-p) \left[ ((1-p)-p)((1-p)+p) \right] \]
\[ \mu_3 = p(1-p) [ (1-2p)(1) ] = p(1-p)(1-2p) \]
Now, find \(\beta_1\):
\[ \beta_1 = \frac{\mu_3^2}{\mu_2^3} = \frac{[p(1-p)(1-2p)]^2}{[p(1-p)]^3} \]
\[ \beta_1 = \frac{p^2(1-p)^2(1-2p)^2}{p^3(1-p)^3} \]
Simplify by canceling common terms:
\[ \beta_1 = \frac{(1-2p)^2}{p(1-p)} \]
Step 4: Final Result:
The coefficient of skewness \(\beta_1\) for a Bernoulli distribution is \(\frac{(1-2p)^2}{p(1-p)}\).