Question

A person rolls a die, tosses a coin, and draws a card from an ordinary deck. He receives \(\$ 3\) for each point up on the die, \(\$ 10\) for a head and \(\$ 0\) for a tail, and \(\$ 1\) for each spot on the card (jack \(=11\), queen \(=12\), king \(=13) .\) If we assume that the three random variables involved are independent and uniformly distributed, compute the mean and variance of the amount to be received.

Short answer

The expected mean of the amount to be received is \$22.5 and the expected variance is 66.63.

Step-by-step solution

Calculate the Expected Values

First, compute the expected values (mean) for each of the individual random variables (die, coin, and card). For a fair die, the expected value E(die) is given by \((1+2+3+4+5+6)/6 = 3.5\). But as each point up on the die gets \(\$3\), the expectation becomes \(E = 3 * 3.5 = \$10.5\). \n\nFor a fair coin, the expectation E(coin) is \((\$10 * 0.5) + (\$0 * 0.5) = \$5\). \n\nFor a card, the expected value E(card) from an ordinary deck (from Ace to King) is \((1+2+3+4+5+6+7+8+9+10+11+12+13)/13 = 7\). But each spot gets \(\$1\), so the expectation remains \(E = 1 * 7 = \$7\)

Compute the Final Expected Value

Once the expected value for each of the independent random variables has been calculated, they can be added together to get the final expected value. This comes from the properties of expected values, where the expected value of the sum of random variables is the sum of their expected values.\n\nThus, the final expected value (mean) M = E(die) + E(coin) + E(card) = \$10.5 + \$5 + \$7 = \$22.5.

Compute the Variance

Following similar steps, we can compute the variance for each random variable, then add them up, as they are independent.\n\nThe variance of the die roll V(die) is given by \((2^2 + 1^2 + 0 + 1^2 + 2^2 + 3^2)/6 - 3.5^2 = 2.916\). Considering the amount received for each point, the variance becomes V(die) = 9 * 2.916 = 26.25. \n\nThe variance of the coin toss V(coin) is \([10^2 * 0.5 + 0^2 * 0.5] - 5^2 = 25\).\n\nThe variance of the card draw V(card) is given by \((2^2 + 1^2 + 0 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2)/13 - 7^2 = 15.38\) considering the amount received for each spot.\n\nFinally, sum the individual variances to get the final variance (V): V = V(die) + V(coin) + V(card) = 26.25 + 25 + 15.38 = 66.63.

Key concepts

Expected Value

The expected value is a fundamental concept in mathematical statistics. It represents the mean or average of random variables over many iterations. To find the expected value of a random variable, we look at all possible outcomes, multiply each outcome by its probability, and sum these products.

In the given exercise, we first compute the expected value for each independent random variable: the die, the coin, and the card.

• For the die, each point is worth \(3, so the expected value is calculated as multiplying the average die roll (\(3.5\)) by \)3, resulting in \( \\(10.5\).
• The expected value for a fair coin flip is the average of winning \)10 or nothing, which is \(\\(5\).
• And lastly, for drawing a card, the expected value calculated is \)7, as that's the average value of a card spot, considering Jacks, Queens, and Kings are valued higher.

Calculating these individually and then adding them gives the final expected value:\(\$22.5\). This shows how we can predict earnings on average from these combined events.

Variance

Variance measures the dispersion of a set of values relative to the mean. In simpler terms, it tells us how spread out the values are. A higher variance indicates that the numbers are more spread out around the mean.

When dealing with independent random variables, the variance of their sum is simply the sum of their variances. In this exercise:

• The variance for the die is computed considering that each value is worth \(3: \(9 * 2.916 = 26.25\).
• The variance for the coin toss is \(\\)25\), showing the spread between getting $10 or nothing.
• The variance for a card is calculated to be \(15.38\).

By adding these variances: \(26.25 + 25 + 15.38 = 66.63\), we derive the total variance for all three independent random variables combined. This gives insight into how much variation we can expect in the total earnings.

Independent Random Variables

Independent random variables are variables whose outcomes do not influence each other. In simple terms, the result of one does not change the probability or outcome of the other.

Understanding independence is crucial in statistics, particularly when analyzing multiple random events. In this exercise:

• The die roll, coin toss, and card draw are all independent events. The result from one does not affect the others.
• This property significantly simplifies calculations, allowing us to find the expected values and variances of each event separately and simply sum them for the totals.

Independence aids in making statistical models easier to analyze, as it simplifies the computation of various statistical measures, like the expected value and variance, for combined independent events.