on a 50-point history test the median was 34 and Q1 was 28. one student got a perfect paper and qualified as an outlier under the 1.5 X IQR criterion. what is the largest value Q3 could be?
1.5*IQR=(49-28)
IQR=21/1.5=14
So
Q3=Q1+IQR=42
However, the median and Q1 should be recalculated after the outlier is removed.
Why did the student's perfect paper qualify as an outlier? Because it was "outstanding"! Now, let's calculate the value of Q3. We know that Q1 is 28, so the lower quartile is just a-t1ny bit below average. As for the median, well, it's 34 – just like my waist size after eating one too many cupcakes! Since the perfect paper is an outlier, we need to find the IQR (Interquartile Range) to determine Q3. So, let's subtract Q1 from the median: 34 - 28 = 6. Doubling that, we get 12. Now, let's add that to the median to find Q3: 34 + 12 = 46! Voila, the largest value Q3 could be is 46, just like the number of times I've told jokes today. *wink*
To find the largest value Q3 can be, we need to calculate the Interquartile Range (IQR) and then apply the 1.5 x IQR criterion to determine the outlier threshold.
Given that Q1 is 28, we know that Q1 represents the 25th percentile, which means 25% of the scores fall below Q1. The median is 34, indicating that 50% of the scores fall below it.
To find Q3, which represents the 75th percentile, we can use the fact that the median is the midpoint between Q1 and Q3. Therefore, the value of Q3 should be the same distance above the median as Q1 is below the median.
Let's calculate the IQR:
IQR = Q3 - Q1
Since the distance between the median and Q1 is the same as the distance between the median and Q3, we can rewrite the IQR equation as:
IQR = (Median - Q1) + (Q3 - Median)
Or simplifying further:
IQR = 2(Q3 - Median)
Now, we know that one student got a perfect paper and qualified as an outlier under the 1.5 x IQR criterion. Therefore, their score is 1.5 times the IQR above Q3. We can write this as:
Perfect score = Q3 + (1.5 x IQR)
Since we want to find the largest value Q3 can be, we need to maximize the perfect score while maintaining it as an outlier. This implies that the perfect score should be the largest possible outlier score.
To maximize the perfect score, we need to maximize the IQR. Since the perfect score is 1.5 times the IQR above Q3, our goal is to ensure that the IQR is as large as possible.
To maximize the IQR, we need to maximize Q3 while maintaining the distances between Q1, the median, and Q3 symmetrical. In other words, we can assume that Q1 is the same distance from the median as Q3, but in the opposite direction.
Since the distance from Q1 to the median is 6 (34 - 28) and we want the distances to be symmetrical, the distance from the median to Q3 should also be 6.
Now we can rewrite the IQR equation using these distances:
IQR = 2(Q3 - Median) = 2(6) = 12
And the perfect score threshold:
Perfect score = Q3 + (1.5 x IQR) = Q3 + (1.5 x 12) = Q3 + 18
Since the perfect score is the largest possible outlier score, we can set it equal to 50 (the maximum score on the test) and solve for Q3:
50 = Q3 + 18
Subtracting 18 from both sides:
Q3 = 50 - 18
Q3 = 32
Therefore, the largest value Q3 could be is 32.
To find the largest value for Q3, we need to understand the terms used in the question. Q1 represents the first quartile, which is the median of the lower half of the data. The median is the middle value of the data when arranged in order. Q3 represents the third quartile, which is the median of the upper half of the data.
The interquartile range (IQR) is the range between the first and third quartiles. In this case, we can calculate the IQR by subtracting Q1 from Q3. The outlier criterion mentioned is the 1.5 times the IQR. If a data point is more than 1.5 times the IQR away from either Q1 or Q3, it is considered an outlier.
Given that the median is 34 and Q1 is 28, we know that there are an equal number of data points below and above the median.
Since one student got a perfect paper and qualified as an outlier, we need to calculate the IQR without considering this outlier.
Let's assume the perfect paper score is X. Since there are an equal number of data points below and above the median, the score of this perfect paper would be equal to the median, which is 34.
Therefore, to calculate Q1 and Q3 without considering this outlier, we arrange the scores in order:
28 (Q1)
X (34 - median)
Q3
100 (perfect paper score)
Since Q1 is already given as 28 and the median is 34, we can see that X (the score of the perfect paper) is also 34.
Now we can substitute the value of X into the arrangement:
28
34
Q3
100
To find the largest value Q3 could be, we need to ensure that the outlier criterion of 1.5 times the IQR is not violated. In this case, the IQR is the range from Q1 to Q3.
The current IQR is Q3 - Q1 = Q3 - 28.
To avoid a violation, Q3 should not be more than 1.5 times the current IQR away from Q1.
Thus, Q3 - 28 ≤ 1.5(Q3 - 28).
Simplifying the inequality gives:
Q3 - 28 ≤ 1.5Q3 - 42.
Reordering the terms gives:
Q3 - 1.5Q3 ≤ -14.
Combining like terms gives:
-0.5Q3 ≤ -14.
Dividing both sides by -0.5 (remembering that when we divide or multiply both sides of an inequality by a negative number, we have to reverse the inequality sign) gives:
Q3 ≥ 28.
Therefore, the largest value Q3 could be is 28.