Question

A balanced coin is tossed three times. Let \(x\) equal the number of heads observed. a. Use the formula for the binomial probability distribution to calculate the probabilities associated with \(x=0,1,2,\) and 3 b. Construct the probability distribution. c. Find the mean and standard deviation of \(x\), using these formulas: $$ \begin{array}{l} \mu=n p \\ \sigma=\sqrt{n p q} \end{array} $$ d. Using the probability distribution in part b, find the fraction of the population measurements lying within one standard deviation of the mean. Repeat for two standard deviations. How do your results agree with Tchebysheff's Theorem and the Empirical Rule?

Short answer

Based on the binomial probability distribution for the number of heads when tossing a balanced coin three times, the mean (µ) is 1.5, and the standard deviation (σ) is approximately 0.866. The fraction of population measurements within one standard deviation of the mean is 0.375, and within two standard deviations is 0.875. These results comply with Tchebysheff's Theorem but cannot be compared directly with the Empirical Rule, as we're dealing with a discrete distribution, whereas the Empirical Rule applies to continuous normal distributions.

Step-by-step solution

Calculate the probabilities

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: $$P(X=0) = \binom{3}{0} (0.5)^0 (1-0.5)^{3-0} = 1 \times 1 \times 0.5^3 = 0.125$$ $$P(X=1) = \binom{3}{1} (0.5)^1 (1-0.5)^{3-1} = 3 \times 0.5 \times 0.5^2 = 0.375$$ $$P(X=2) = \binom{3}{2} (0.5)^2 (1-0.5)^{3-2} = 3 \times 0.5^2 \times 0.5 = 0.375$$ $$P(X=3) = \binom{3}{3} (0.5)^3 (1-0.5)^{3-3} = 1 \times 0.5^3 \times 1 = 0.125$$

Construct the probability distribution

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 |

Find the mean and standard deviation

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.

Comparing with Tchebysheff's Theorem and the Empirical Rule

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.

Key concepts

Mean and Standard Deviation

When dealing with a binomial distribution like the balanced coin toss, it's helpful to understand the concepts of mean and standard deviation. These statistical measures provide insights into the expected outcome and variability of our experiment. In a binomial distribution:

• The mean, or expected value, is found using the formula \( \mu = n \cdot p \), where \( n \) is the number of trials, and \( p \) is the probability of success in each trial. For our coin toss example, \( \mu = 3 \times 0.5 = 1.5 \). This means, on average, we can expect to get 1.5 heads in three tosses.
• The standard deviation measures the dispersion of the data from its mean and is calculated using \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \). In the coin toss example, this is \( \sigma = \sqrt{3 \times 0.5 \times 0.5} \approx 0.866 \). This value indicates, on average, how much the number of heads deviates from 1.5.

Understanding these concepts helps in predicting how often the outcomes will deviate from the mean in repeated experiments.

Tchebysheff's Theorem

Tchebysheff's Theorem is a versatile tool in statistics, applicable to any data set. Its core function is to predict the parameter of data spread without assuming a specific distribution shape. The theorem asserts that at least \( 1 - \frac{1}{k^2} \) of the data will be within \( k \) standard deviations from the mean, regardless of distribution form. Here's how it applies:

• For \( k=1 \), Tchebysheff's Theorem predicts that 0% or more of data lies within 1 standard deviation of the mean. In our exercise, only 37.5% of the possibilities are within that range, reflecting a broader potential distribution allowed by the theorem.
• For \( k=2 \), it ensures at least 75% of data is within 2 standard deviations. In this exercise, 87.5% of outcomes fall within this range, conforming to the theorem.

Tchebysheff's Theorem is best employed in non-normally distributed data sets or when distribution characteristics remain unknown.

Empirical Rule

The Empirical Rule is a fundamental concept that applies primarily to bell-shaped or normal distributions. It forecasts the percentage of data points that lie within certain distances from the mean, expressed as standard deviations:

• Approximately 68% of data falls within one standard deviation of the mean.
• About 95% of data lies within two standard deviations.
• Nearly 99.7% is within three standard deviations.

While the Empirical Rule offers great insights for symmetric, normal distributions, it isn't applicable directly to binomial distributions like our coin toss example, which is discrete rather than continuous. However, it remains a powerful tool for understanding how data behaves under a normal distribution.