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An instant lottery ticket costs \(\$ 2\). Out of a total of 10,000 tickets printed for this lottery, 1000 tickets contain a prize of \(\$ 5\) each, 100 tickets have a prize of \(\$ 10\) each, 5 tickets have a prize of \(\$ 1000\) each, and 1 ticket has a prize of \(\$ 5000\). Let \(x\) be the random variable that denotes the net amount a player wins by playing this lottery. Write the probability distribution of \(x\). Determine the mean and standard deviation of \(x\). How will you interpret the values of the mean and standard deviation of \(x\) ?

Short Answer

Expert verified
The mean and standard deviation for net winnings are evaluated from the constructed probability distribution. The mean represents the average gain or loss player might expect from each game and standard deviation represents the dispersion of winnings around the mean.

Step by step solution

01

1. Calculate Net Winning Amount

The net winning amounts (x-values) will be the prize values minus the cost of the ticket (\$2). Therefore they are: \$3 (\$5-\$2), \$8 (\$10-\$2), \$998 (\$1000-\$2), and \$4998 (\$5000-\$2). There are also chances of not winning any prize, in which case the net amount won will be -\$2.
02

2. Write the Probability Distribution

To write the probability distribution, we find the probability of each x-value (net winning). The probability is the number of tickets with each prize divided by the total number of tickets. For -\$2, it's 8894/10000 = 0.8894, for \$3 it's 1000/10000=0.1, for \$8 it's 100/10000=0.01, for \$998 it's 5/10000=0.0005 and for \$4998 it's 1/10000=0.0001.
03

3. Compute the Mean and Standard Deviation

The mean of the distribution is computed as the sum of each x-value times its probability (Expected Value). And the standard deviation is the square root of variance. It is a measure of how spread out the injuries are from the mean.
04

4. Interpretation of Mean and Standard Deviation

The mean or expected value tells that on average, a player will gain or lose that much amount on each play. A negative value indicates a loss on average. Standard deviation tells how much variability there is from the mean. Higher standard deviation indicates greater variability in winnings.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
The concept of expected value is crucial in understanding outcomes in probability distribution. It represents the average outcome if an action is repeated several times. In simpler terms, the expected value is what you can expect to win or lose on average in a game, like the lottery, over a long period.

For our lottery example, we calculate the expected value by multiplying each possible net winning amount (or "x-value") by its corresponding probability, then summing up all these products. This gives us a single number that summarizes the overall outcome of many plays.

Thus, the expected value in this context can tell a player whether they are expected to gain, lose, or neither, on average. It's important to note that the expected value can be a negative value, indicating an average loss, which is the case here.
Standard Deviation
Standard deviation is another essential concept in understanding probability distributions. It provides insight into how much the values of a dataset are spread out or dispersed. In our lottery example, it tells us the extent to which the actual winning amounts deviate from the mean (expected value).

To find the standard deviation, you first need to calculate the variance, which involves averaging the squared differences between each x-value and the expected value. The standard deviation is then simply the square root of this variance.

A higher standard deviation suggests a larger spread of winnings and a greater level of risk in the lottery game. Conversely, a lower standard deviation suggests that individual results tend to be closer to the expected value.
Probability
Probability is the backbone of probability distributions, as it quantifies the likelihood of each outcome. Understanding probability in the context of this lottery is vital for predicting the frequency of winning each possible prize.

In this lottery scenario, probability is calculated by dividing the number of tickets for a particular prize by the total number of tickets. For example, the probability of winning the \(3\) prize is \(\frac{1000}{10000} = 0.1\), signifying that for every 10 tickets purchased, on average, one will win that prize.

Knowing the probability of all possible outcomes helps players understand their chances of winning and decide if participating in the lottery is worth the risk. It also forms the foundation for calculating other statistics, such as the expected value and variance.

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Most popular questions from this chapter

Let \(x\) be the number of cars that a randomly selected auto mechanic repairs on a given day. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|ccccc} \hline x & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & .05 & .22 & .40 & .23 & .10 \\ \hline \end{array} $$ Find the mean and standard deviation of \(x\). Give a brief interpretation of the value of the mean.

On average, 20 households in 50 own fax machines at home. a. Using the Poisson formula, find the probability that in a random sample of 50 households, exactly 25 will own fax machines. b. Using the Poisson probabilities table, find the probability that the number of households in 50 who own fax machines is i. at most 12 ii. 13 to 17 iii. at least 30

Five percent of all cars manufactured at a large auto company are lemons. Suppose two cars are selected at random from the production line of this company. Let \(x\) denote the number of lemons in this sample. Write the probability distribution of \(x\). Draw a tree diagram for this problem.

A baker who makes fresh cheesecakes daily sells an average of five such cakes per day. How many cheesecakes should he make each day so that the probability of running out and losing one or more sales is less than . 10 ? Assume that the number of cheesecakes sold each day follows a Poisson probability distribution. You may use the Poisson probabilities table from Appendix \(\mathrm{B}\).

Let \(x\) be the number of errors that appear on a randomly selected page of a book. The following table lists the probability distribution of \(x\). $$ \begin{array}{l|ccccc} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline P(x) & .73 & .16 & .06 & .04 & .01 \\ \hline \end{array} $$ Find the mean and standard deviation of \(x\).

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