4. In a certain game of chance, your chances of winning are 0.2. If you play the game five times and outcomes are independent, the probability that you win at least once is
5. A set of data is found to have a sample standard deviation of 25. Suppose 9 were added to each of the numbers in the data. The standard deviation of the resulting data
6. A consumer group surveyed the prices for a certain item in five different stores, and reported the average price as $15. We visited four of the five stores, and found the prices to be $10,$15, $15, and $25. Assuming that the consumer group is correct, what is the price of the item at the store that we did not visit?
7. Suppose that A and B are two independent events with P(A) = 0.2 and P(B) = 0.4. P(A or B) is:
8. Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume X is normal with a mean of $360 and standard deviation $40.15. The value of P(X > $400) is?
9. An event A will occur with probability 0.4. An event B will occur with probability 0.6. The probability that both A and B will occur is 0.24. We may conclude
4. To find the probability of winning at least once in five independent attempts with a probability of winning 0.2 in each attempt, we can find the probability of losing all five times and then subtract it from 1.
The probability of losing in one attempt is 1 - 0.2 = 0.8.
Since the outcomes are independent, the probability of losing all five times is 0.8^5 = 0.32768.
Therefore, the probability of winning at least once is 1 - 0.32768 = 0.67232 (approximately).
5. When a constant value is added to each data point, it does not affect the standard deviation. Therefore, adding 9 to each number in the data will not change the sample standard deviation of 25. The standard deviation of the resulting data will remain at 25.
6. The average price reported by the consumer group is $15. This means that the sum of the prices in all five stores is 5 * $15 = $75.
We have visited four of the five stores and found the prices to be $10, $15, $15, and $25. The sum of these four prices is $10 + $15 + $15 + $25 = $65.
To find the price of the item at the store we did not visit, we subtract the sum of the prices we found from the total sum of prices reported by the consumer group:
$75 - $65 = $10.
Therefore, the price of the item at the store we did not visit is $10.
7. The probability of the union of two independent events A and B, denoted as P(A or B), can be calculated by adding their individual probabilities and then subtracting the probability of their intersection.
P(A or B) = P(A) + P(B) - P(A and B).
Given that P(A) = 0.2, P(B) = 0.4, and P(A and B) = 0 (since the events are independent), we can substitute these values into the formula:
P(A or B) = 0.2 + 0.4 - 0 = 0.6.
Therefore, the probability that either event A or event B (or both) will occur is 0.6.
8. To find the probability that the profit (random variable X) is greater than $400, we need to calculate the area under the normal distribution curve to the right of $400.
Using the normal distribution table or a statistical calculator, we can find the z-score corresponding to $400 based on the mean and standard deviation:
z = (X - mean) / standard deviation
z = ($400 - $360) / $40.15
z ≈ 0.9963
Now we can use the z-score to find the probability:
P(X > $400) = 1 - P(X ≤ $400)
P(X > $400) = 1 - P(z ≤ 0.9963)
Using the normal distribution table or a statistical calculator, we can find the probability corresponding to 0.9963:
P(z ≤ 0.9963) ≈ 0.8389
Therefore, P(X > $400) = 1 - 0.8389 = 0.1611 (approximately).
9. Since event A and event B are independent, the probability of both events A and B occurring is equal to the product of their individual probabilities:
P(A and B) = P(A) * P(B)
Given that P(A) = 0.4, P(B) = 0.6, and P(A and B) = 0.24, we can substitute these values into the formula:
0.24 = 0.4 * 0.6
Since the equation holds true, we can conclude that the events A and B are indeed independent, and the product of their probabilities is equal to the probability of both events occurring.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However, here is a start.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
Either-or probabilities are found by adding the individual probabilities.