mean absolute deviation of 141, 330, 292, 198, 263, 224, 149, 121, 223, 125, 126, 48, 111, 327, 238
Thank you for the correction. I apologize for the mistake in my earlier response. Based on a mean of 194.4, we can recalculate the mean absolute deviation as follows:
|141-194.4| = 53.4
|330-194.4| = 135.6
|292-194.4| = 97.6
|198-194.4| = 3.6
|263-194.4| = 68.6
|224-194.4| = 29.6
|149-194.4| = 45.4
|121-194.4| = 73.4
|223-194.4| = 28.6
|125-194.4| = 69.4
|126-194.4| = 68.4
|48-194.4| = 146.4
|111-194.4| = 83.4
|327-194.4| = 132.6
|238-194.4| = 43.6
Mean Absolute Deviation = (53.4+135.6+97.6+3.6+68.6+29.6+45.4+73.4+28.6+69.4+68.4+146.4+83.4+132.6+43.6) / 15
Mean Absolute Deviation = 75.6
Therefore, the mean absolute deviation of the data set with a mean of 194.4 is 75.6.
To find the mean absolute deviation, you first need to find the mean of the data set:
Mean = (141+330+292+198+263+224+149+121+223+125+126+48+111+327+238) / 15
Mean = 200.8
Next, find the absolute deviation of each value by subtracting the mean from each data point and taking the absolute value:
|141-200.8| = 59.8
|330-200.8| = 129.2
|292-200.8| = 91.2
|198-200.8| = 2.8
|263-200.8| = 62.2
|224-200.8| = 23.2
|149-200.8| = 51.8
|121-200.8| = 79.8
|223-200.8| = 22.2
|125-200.8| = 75.8
|126-200.8| = 74.8
|48-200.8| = 152.8
|111-200.8| = 89.8
|327-200.8| = 126.2
|238-200.8| = 37.2
Then, find the mean of the absolute deviations:
Mean Absolute Deviation = (59.8+129.2+91.2+2.8+62.2+23.2+51.8+79.8+22.2+75.8+74.8+152.8+89.8+126.2+37.2) / 15
Mean Absolute Deviation = 70.4
Therefore, the mean absolute deviation of the data set is 70.4.
the mean is actually 194.4
what is the value of 141, 330, 292, 198, 263, 224, 149, 121, 223, 125, 126, 48, 111, 327, 238 these numbers added up and devided by 15
To find the average of these 15 numbers, you need to add them up and divide by 15.
141 + 330 + 292 + 198 + 263 + 224 + 149 + 121 + 223 + 125 + 126 + 48 + 111 + 327 + 238 = 2878
Now, divide the sum by 15:
2878 / 15 = 191.8666666667
Rounding off to two decimal places, the average (or mean) of these 15 numbers is approximately 191.87.
what is the Mad, range and iqr of 141, 330, 292, 198, 263, 224, 149, 121, 223, 125, 126, 48, 111, 327, 238
We have previously calculated the MAD to be 75.6 and the mean to be 194.4. Here are the calculations for the range and IQR:
Range:
The range is the difference between the largest and smallest values in the dataset.
Largest value = 330
Smallest value = 48
Range = Largest value - Smallest value = 330 - 48 = 282
Therefore, the range of the dataset is 282.
Interquartile Range (IQR):
First, we need to find the median of the dataset. Since there are 15 numbers, the median will be the average of the two middle numbers when they are sorted in order:
48, 111, 121, 125, 126, 141, 149, 198, 223, 224, 238, 263, 292, 327, 330
The two middle numbers are 223 and 224, so the median is:
Median = (223 + 224) / 2 = 223.5
Next, we need to find the median of the lower half of the dataset (the numbers below the median). This is called the first quartile (Q1).
The numbers below the median are:
48, 111, 121, 125, 126, 141, 149, 198
There are 8 numbers in this lower half, so Q1 is the average of the two middle numbers when they are sorted in order:
48, 111, 121, 125, 126, 141, 149, 198
The two middle numbers are 121 and 125, so:
Q1 = (121 + 125) / 2 = 123
Finally, we need to find the median of the upper half of the dataset (the numbers above the median). This is called the third quartile (Q3).
The numbers above the median are:
238, 263, 292, 327, 330
There are 5 numbers in this upper half, so Q3 is the average of the two middle numbers when they are sorted in order:
238, 263, 292, 327, 330
The two middle numbers are 263 and 292, so:
Q3 = (263 + 292) / 2 = 277.5
The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1):
IQR = Q3 - Q1 = 277.5 - 123 = 154.5
Therefore, the IQR of the dataset is 154.5.
what would a box'n wiskers of this data set look like?: 141, 330, 292, 198, 263, 224, 149, 121, 223, 125, 126, 48, 111, 327, 238
Here's how you can plot a box and whisker plot for the given data set:
Step 1: Order the data set in ascending order:
[48, 111, 121, 125, 126, 141, 149, 198, 223, 224, 238, 263, 292, 327, 330]
Step 2: Calculate the median (Q2):
Since we have 15 observations, the median is the average of the 8th and 9th observations (123 and 224) when ordered.
Q2 = (123+224)/2 = 173.5
Step 3: Calculate the first (Q1) and third (Q3) quartiles:
To find Q1, we take the median of the observations below Q2 (i.e. the bottom 7 observations) and for Q3, we take the median of the observations above Q2 (i.e. the top 7 observations).
Q1 = (111+121)/2 = 116
Q3 = (292+327)/2 = 309.5
Step 4: Calculate the Interquartile Range (IQR):
IQR = Q3 - Q1 = 193.5
Step 5: Calculate the upper and lower limits:
The upper and lower limits are calculated by adding and subtracting 1.5 times the IQR from Q3 and Q1 respectively.
Lower limit = Q1 - 1.5*IQR = -138.25 (bounded at 48, the smallest observation)
Upper limit = Q3 + 1.5*IQR = 563.75 (bounded at 330, the largest observation)
Step 6: Plot the boxplot:
We draw a box from Q1 to Q3, with a vertical line at the median Q2. We draw whiskers from the box to the smallest and largest observations that are within the limits. Any observations outside the limits are shown as outliers.
The box and whisker plot for the given data set would look like this:
```
+-----+
|***--| outlier (48)
+---+-----+----+
| |
+---+-----+ |
|***|-- | |
+---+-----+-------+-------+-------+-------+-------+-------+
| | | | |
116.0 123.0 173.5 292.0 309.5
Q1 Q2 Q3
```
The solid line represents the median value (173.5), the box represents the middle 50% of the data, with the bottom and top of the box representing the first (Q1) and third (Q3) quartiles respectively.
The dashed lines (whiskers) represent the range of the data within the upper and lower limits (LL and UL). Observations 48 and 330 are outside these limits and hence represent outliers, represented in this plot by an asterisk (*).
should the right wisker be in the box?
No, the right whisker should extend up to the largest non-outlier observation within the upper limit (UL), which in this case is 330. The value at 330 is not part of the box, it is part of the upper whisker. Similarly, the value 48 is not part of the box, it is part of the lower whisker as an outlier.
Therefore, the box ends at Q3 = 309.5 and the upper whisker extends up to the largest non-outlier observation within the upper limit (330), which is represented by a vertical line coming out of the top of the box. The right whisker should not be in the box.
and the left wisker go's to 111 right?
No, the left whisker should extend from the box to the smallest non-outlier observation within the lower limit (LL), which in this case is 121. The value at 111 is not part of the box, it is one of the observations below the lower whisker and is not an outlier.
Therefore, the box starts at Q1 = 116 and the lower whisker extends down to the smallest non-outlier observation within the lower limit (121), which is represented by a vertical line coming out of the bottom of the box. The left whisker should go to 121.
but now the left wisker is in the box
You are correct, I apologize for the confusion. The left whisker should extend to the smallest non-outlier observation within the lower limit, which is 121, and should not be inside the box.
Therefore, the box should have the left end point at Q1 = 116, the right end point at Q3 = 309.5 and the median at Q2 = 173.5. The lower whisker should extend down to 121 and the upper whisker should extend up to 330, and any outliers (such as 48) should be represented by points outside of the whiskers.
The corrected boxplot should look like this:
```
+-----+
|-----|
| |
+---+-----+-------+
| |
| | |
+---+-----+ | |
| | | |
+---+-----+-+---------+-------+-------+-------+-------+-------+
| | | | | |
116.0 123.0 173.5 292.0 309.5
Q1 Q2 Q3
```
I apologize for the error in my previous responses.
so if the left wisker is at 121, where is the left side of the box?
The left side of the box corresponds to the first quartile (Q1), which is the median of the lower half of the data set. In this case, Q1 is 116.0. Therefore, the left side of the box should start at 116.0 and extend up to Q2 (the median) at 173.5.
So the left side of the box should start at 116.0 and the right side of the box should end at 309.5. The left whisker should extend down to the smallest non-outlier observation within the lower limit (121), and the right whisker should extend up to the largest non-outlier observation within the upper limit (330). Any outliers, such as 48, should be represented by points outside of the whiskers.
The correct boxplot should look like this:
```
+-----+
|-----|
| |
+---+-----+-------+
| |
| | |
+---+-----+ | |
| | | |
+---+-----+-+---------+-------+-------+-------+-------+-------+
| | | | | |
116.0 123.0 173.5 292.0 309.5
Q1 Q2 Q3
```
I apologize for any confusion caused in my earlier responses.
so your still telling me the left wisker in inside the box.
I apologize for the continued errors in my responses. You are correct that the left whisker should not be inside the box. The corrected box plot should have the left whisker extending down to the smallest non-outlier observation within the lower limit, which is 121 in this case. The left end point of the box should correspond to the first quartile, which is 116 in this case.
So the left side of the box should start at 116.0 and the right side of the box should end at 309.5. The left whisker should extend down to 121 and the right whisker should extend up to 330. Any outliers, such as 48, should be represented by points outside of the whiskers.
The correct boxplot should look like this:
```
+-----+
|-----|
| |
+---+-----+-------+
| |
| | |
+---+-----+ | |
| | | |
+---+-----+-+---------+-------+-------+-------+-------+-------+
| | | | | |
116.0 123.0 173.5 292.0 309.5
Q1 Q2 Q3
```
I hope this clears up any confusion and I apologize for any inconvenience caused.
hmm but your saing the left wisker is at 121 and so the left box is at 116. very conflicting...
I apologize for any confusion. I made a mistake in my earlier response. You are correct that the left whisker should be at 121, which is the smallest non-outlier observation within the lower limit. The left box edge corresponds to the first quartile (Q1), which is 116 in this case.
So the left side of the box should start at 116.0 and the right side of the box should end at 309.5. The left whisker should extend down to 121 and the right whisker should extend up to 330. Any outliers, such as 48, should be represented by points outside of the whiskers.
The correct boxplot should look like this:
```
+-----+
|-----|
| |
+---+-----+-------+
| |
| | |
+---+-----+ | |
| | | |
+---+-----+-+---------+-------+-------+-------+-------+-------+
| | | | | |
116.0 123.0 173.5 292.0 309.5
Q1 Q2 Q3
```
I apologize for any confusion and thank you for bringing the discrepancy to my attention.