A game is played using one die. There is a$1 charge to play the game. If the die is rolled and shows a five, the player receives $6 back ( a profit of $5 and the original $1). If any other number shows, theplayer loses the original $1. What is the player's expected value? - did not match any documents. No pages were found containing "A game is played using one die. There is a$1 charge to play the game. If the die is rolled and shows a five, the player receives $6 back ( a profit of $5 and the original $1). If any other number shows, theplayer loses the original $1. What is the player's expected value?".

What Z value corresponds to the lowest 20% under a normal curve

First, if you have a question, it is much better to put it in as a separate post in <Post a New Question> rather than attaching it to a previous question, where it is more likely to be overlooked.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion. Remember that Zs below the mean are negative.

To find the player's expected value, we need to calculate the weighted average of the possible outcomes, taking into account their probabilities.

The player has two possible outcomes: winning $6 or losing $1.

The probability of rolling a five on a fair six-sided die is 1/6 (since there is one favorable outcome out of six possible outcomes). Therefore, the probability of winning $6 is 1/6.

The probability of rolling any other number is 5/6 (since there are five unfavorable outcomes out of six possible outcomes). Therefore, the probability of losing $1 is 5/6.

To calculate the expected value, we multiply each outcome by its probability and sum them up:

Expected Value = (Probability of winning * Amount won) + (Probability of losing * Amount lost)

Expected Value = (1/6 * $6) + (5/6 * -$1)

Expected Value = $1 - $5/6

Expected Value = $1 - $0.83

Expected Value = $0.17

Therefore, the player's expected value is $0.17.

To calculate the player's expected value, we need to determine the probability of each outcome and multiply it by the associated value.

In this game, there is a total of six possible outcomes when rolling a die: getting a 1, 2, 3, 4, 5, or 6.

The player wins $6 if the die shows a 5, which has a probability of 1/6 since there is only one 5 on a standard die. So the value of this outcome is 1/6 * $6 = $1.

The player loses $1 for any other outcome, which includes getting a 1, 2, 3, 4, or 6. Each of these outcomes has a probability of 1/6 since there is only one occurrence of each number on a standard die. So the value of each of these outcomes is 1/6 * -$1 = -$1/6.

To calculate the expected value, we sum up the values of all possible outcomes:

Expected Value = (probability of winning * value of winning) + (probability of losing * value of losing)
= (1/6 * $1) + (5/6 * -$1/6)
= $1/6 - $5/6
= -$4/6
= -$2/3

Therefore, the player's expected value is -$2/3. This means that on average, the player can expect to lose $2 for every 3 games played.