a six-faced die was thrown 28 times. the table shows the number that each possible score occurred.
score 1 2 3 4 5 6
frequency 8 6 6 2 4 2
(a) After the 27th throw the median score was 2. what was the least possible score on the 28th throw?
(b) The die was then thrown twice more. The mean score of all 30 throws was exactly 3. What were the scores on the two throws?
after 28, there were 14 below 3 and 14 above 2. So, what do you think? Check the last two throws.
30*3 = 90
The sum of 28 throws was 78, so the last two had to add up to 12.
(a) Well, well, well, it seems like we have a dicey situation here. If the median score after the 27th throw was 2, that means there were 13 scores that were less than or equal to 2. Looking at the table, we see that there were 8 occurrences of 1 and 6 occurrences of 2. Since the 8 occurrences of 1 were included in the 13 scores less than or equal to 2, it means that all 8 of those 1s must have been thrown before the 27th throw. Therefore, the least possible score on the 28th throw must be 2. Funny how things work out sometimes, isn't it?
(b) Ah, the mean score is playing tricks on us, is it? If the mean score of all 30 throws was exactly 3, we can use a little mathematical humor to solve this one. We can start solving this mathematical riddle by multiplying the mean of 3 by the total number of throws, which is 30. This gives us a sum of 90. Now, let's subtract the scores we already know, which are the 27 scores we have from earlier. We have a total of 8 occurrences of 1, 6 occurrences of 2, 6 occurrences of 3, 2 occurrences of 4, 4 occurrences of 5, and 2 occurrences of 6. So, the sum of those scores is 8 * 1 + 6 * 2 + 6 * 3 + 2 * 4 + 4 * 5 + 2 * 6 = 50. Now, subtracting this sum from the total sum of 90 gives us 40. Since we still have two throws left, we need to divide this remaining score of 40 by 2 to distribute it evenly between those two throws. Therefore, we can conclude that the scores on the two throws must be 20 each. Voila! The dice reveal their secrets with a touch of mathematical humor.
To find the answer, we need to analyze the given information step-by-step.
(a) After the 27th throw, the median score was 2. The median is the middle value when the data is arranged in ascending order. In this case, the scores are already given in ascending order. Let's calculate the cumulative frequencies:
score 1 2 3 4 5 6
cumulative frequency 8 14 20 22 26 28
Since the median score is 2, there must be at least 14 scores of 1 and 14 scores of 2. We already have 8 occurrences of 1 and 6 occurrences of 2. So, there must be at least 6 more occurrences of 1 and at least 8 more occurrences of 2.
To find the least possible score on the 28th throw, we need to calculate the remaining scores. Let's subtract the cumulative frequencies from the total number of throws (28):
Remaining throws = Total throws - Cumulative frequency
Remaining throws = 28 - 28 = 0
Since there are no remaining throws after the 27th throw, the least possible score on the 28th throw is 1.
Therefore, the least possible score on the 28th throw is 1.
(b) The mean score of all 30 throws was exactly 3. To calculate the mean, we need to multiply each score by its frequency and sum them up, then divide by the total number of throws. Let's calculate the sum of the scores:
Sum of scores = (1 * 8) + (2 * 6) + (3 * 6) + (4 * 2) + (5 * 4) + (6 * 2)
Sum of scores = 8 + 12 + 18 + 8 + 20 + 12
Sum of scores = 78
Now, let's calculate the score on the 29th throw. Since there are 30 throws in total, we have:
Remaining throws = Total throws - Cumulative frequency
Remaining throws = 30 - 28 = 2
To calculate the average mean (3), we need to find the sum of the two throws (score1 + score2) and divide it by 2:
(78 + score1 + score2) / 30 = 3
Simplifying the equation:
78 + score1 + score2 = 3 * 30
score1 + score2 = 3 * 30 - 78
score1 + score2 = 60 - 78
score1 + score2 = -18
Since the sum of score1 and score2 is -18, we need to find two numbers that add up to -18. One possibility is -9 and -9:
score1 = -9
score2 = -9
Therefore, the scores on the two throws are -9 and -9.
To answer these questions, we need to analyze the given data and use some statistical concepts. Let's break down each question step by step.
(a) After the 27th throw, the median score was 2. We have the frequency distribution of the scores, so we can determine the cumulative frequency of the scores to find the median.
Looking at the frequency table, we can calculate the cumulative frequencies:
Score 1: 8
Score 2: 8 + 6 = 14
Score 3: 14 + 6 = 20
Score 4: 20 + 2 = 22
Score 5: 22 + 4 = 26
Score 6: 26 + 2 = 28
Since the total number of throws is 28, the median would be the 14th throw. From the cumulative frequencies, we see that the median corresponds to a score of 2. That means, after the 27th throw, the 28th throw must have a score of 2 or higher.
Therefore, the least possible score on the 28th throw is 2.
(b) The mean score of all 30 throws is exactly 3. To find the scores on the two additional throws, we can start by finding the sum of the scores from the given data.
Sum = (1 * 8) + (2 * 6) + (3 * 6) + (4 * 2) + (5 * 4) + (6 * 2) = 8 + 12 + 18 + 8 + 20 + 12 = 78
Since the mean is the sum divided by the total number of throws, we have the equation:
Mean = Sum / Total Throws
3 = 78 / Total Throws
Now, we can solve for the Total Throws:
Total Throws = 78 / 3 = 26
Since we already have 28 throws, we know that the two additional throws make a total of 30 throws. So, we subtract the given 28 from the total:
30 - 28 = 2
The two additional throws have a total score of 2.
Therefore, the scores on the two additional throws are both 1.
To recap:
(a) The least possible score on the 28th throw is 2.
(b) The scores on the two additional throws are both 1.