91Ó°ÊÓ

To calculate the probabilities for different \(x\) values, we will use the binomial probability distribution formula: $$ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} $$ Here, \(n\) represents the number of trials (tosses), \(x\) represents the number of successful trials (number of heads), and \(p\) represents the probability of getting a head in each toss. Since the coin is balanced, \(p = 0.5\). For \(x=0,1,2,3\), we will calculate the probabilities as follows: $$P(X=0) = \binom{3}{0} (0.5)^0 (1-0.5)^{3-0} = 1 \times 1 \times 0.5^3 = 0.125$$ $$P(X=1) = \binom{3}{1} (0.5)^1 (1-0.5)^{3-1} = 3 \times 0.5 \times 0.5^2 = 0.375$$ $$P(X=2) = \binom{3}{2} (0.5)^2 (1-0.5)^{3-2} = 3 \times 0.5^2 \times 0.5 = 0.375$$ $$P(X=3) = \binom{3}{3} (0.5)^3 (1-0.5)^{3-3} = 1 \times 0.5^3 \times 1 = 0.125$$

Now, we will construct the probability distribution for number of heads as follows: | \(x\) (Number of heads) | Probability | |-----------------------|--------------| | 0 | 0.125 | | 1 | 0.375 | | 2 | 0.375 | | 3 | 0.125 |

We can calculate the mean and standard deviation using the given formulas: $$ \mu = n \cdot p = 3 \times 0.5 = 1.5 $$ $$ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{3 \times 0.5 \times 0.5} = \sqrt{0.75} \approx 0.866 $$ So the mean, \(\mu\), is 1.5 and the standard deviation, \(\sigma\), is approximately 0.866.

Using the probability distribution, we will find the fraction of population measurements lying within one and two standard deviations of the mean: - Within \(1 \sigma\) of the mean (\(1.5 - 0.866 \le x \le 1.5 + 0.866\)): \(P(X=1) = 0.375\) (only \(x=1\) lies in this range) - Within \(2 \sigma\)s of the mean (\(1.5 - 2 \times 0.866 \le x \le 1.5 + 2 \times 0.866\)): \(P(X=0) + P(X=1) + P(X=2) = 0.125 + 0.375 + 0.375 = 0.875\) (only \(x=3\) is outside this range) Now, let's compare the results with Tchebysheff's Theorem and the Empirical Rule: - Tchebysheff's Theorem states that at least \(1 - \frac{1}{k^2}\) of the data will be within \(k\) standard deviations of the mean. In our case: - For \(1\) standard deviation (\(k=1\)): \(1-\frac{1}{1^2}=0 \le 0.375\) - For \(2\) standard deviations (\(k=2\)): \(1-\frac{1}{2^2}=0.75 \le 0.875\) Our results comply with Tchebysheff's Theorem. - The Empirical Rule states that for a normal distribution, approximately \(68\%\) of the data lie within one standard deviation of the mean, and approximately \(95\%\) within two standard deviations. However, we can't compare our results with the Empirical Rule directly because we are dealing with a discrete distribution, and the Empirical Rule is applicable to continuous normal distributions only.