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 (*).