Answer:
57
Step-by-step explanation:
This is because, in order to find the range you need to set the order from lowest to highest and subtract the lowest from the highest. So in this case the lowest would be 117 and the highest would be 174. So when you subtract them you get 57.
Answer:
1) Range = 57
2) Range = 12
3) Median = 88, First Quartile = 84, Third Quartile = 92, Interquartile = 8
4) Median = 81.5, Median of lower half = 67, Median of upper half = 92,
Describe the spead of the data = ______
*We cannot see the drop-down answer choices for this part of the question, so we cannot give an answer. Sorry.
5) a. Range = 172 points, Interquartile Range = 42 points
b. Outlier = 193, Range without the Outlier = 101 points, Interquartile Range without the Outlier = 48.5 points
c. The outlier affected the __________ more.
*We cannot see the drop-down answer choices here, so we cannot answer this part of the question. Sorry.
Good Luck! I hope this helps!
Step-by-step explanation:
1) Find the range of the data.
133, 117, 152, 127, 168, 146, 174.
First, put the numbers in ascending order:
117, 127, 133, 146, 152, 168, 174
To find the range, subtract the smallest number from the largest number.
174 - 117 = 57
So, the range is 57.
2) Find the range of the data.
26, 21, 27, 33, 24, 29
First, put the numbers in ascending order.
21, 24, 26, 27, 29, 33
To find the range, subtract the smallest number from the largest number.
33 - 21 = 12
So, the range is 12.
3) Find the median, first quartie, third quartile, and interquartile range of the data. 84, 75, 90, 87, 99, 91, 85, 88, 76, 92, 94
First, put the numbers in ascending order.
75, 76, 84, 85, 87, 88, 90, 91, 92, 94, 99
The median is the value in the middle, so in this case, it is 88.
To find the first quartie, third quartile, and interquartile, we must split the numbers into quarters.
75 76 84 85 87 88 90 91 92 94 99
The first quartile is 84.
The third quartile 92.
The interquartile is the first quartile subtracted from the third quartile, so
92 - 84 = 8.
4) Use grid paper to find the median of the data. Then find the median of the lower half and the median of the upper half of the data.
82, 62, 95, 81, 89, 51, 72, 56, 97, 98, 79, 85
First, put the numbers in ascending order.
51, 56, 62, 72, 79, 81, 82, 85, 89, 95, 97, 98
Find the median which is the middle of the values.
51 56 62 72 79 81 82 85 89 95 97 98
In this case, it is between 81 and 82, so we find 81+82 / 2 = 81.5.
The median is 81.5.
The median of the lower half of the data is the value in the middle of the first 6 of the 12 numbers. 51 56 62 72 79 81
We need to find 62+72 / 2 = 134 /2 = 67, so the median of the lower half of the data is 67.
The median of the upper half of the data is the value in the middle of the last 6 of the 12 numbers. 82 85 89 95 97 98
We need to find 89+95 / 2 = 184 / 2 = 92, so the median of the upper half of the data is 92.
Describe the spread of the data.
The data are ___________ . (We cannot see the drop-down options, so cannot answer this part of the question. Sorry.)
5) The table shows the number of points scored by players on a basketball team. 21 53 74 82 84 93 103 108 116 122 193
a. Find the range and interquartile range of the data.
To find the range, subtract the smallest number from the largest number.
193 - 21 = 172
The range is 172 points.
To find the interquartile range, we must split the numbers into quarters.
21 53 74 82 84 93 103 108 116 122 193
The first quartile is 74.
The third quartile 116.
The interquartile is the first quartile subtracted from the third quartile, so
116 - 74 = 42.
The interquartile range is 42 points.
b. Use the interquartile range to identify the outlier(s) in the data set.
Find the range and the interquartile range of the data set without the outlier(s).
21 53 74 82 84 93 103 108 116 122 193
Per Wikipedia, "In statistics, an outlier is a data point that differs significantly from other observations."
Per statology.org, "One popular method is to declare an observation to be an outlier if it has a value 1.5 times greater than the interquartile (IQR) or 1.5 times less than the IQR."
From statology. org:
(Note: Q1 = first quartile, Q3 = third quartile, IQR = interquartile)
The lower limit is calculated as:
Lower limit = Q1 – 1.5 * IQR = 74 – 1.5 * 42 = 74 - 63 = 9.
The lower limit of outliers is 9 points.
And the upper limited is calculated as:
Upper limit = Q3 + 1.5 * IQR = 116 + 1.5 * 42 = 116 + 63 = 179.
The upper limit of outliers is 179 points.
The outlier is 193 points.
To find the range without the outlier, subtract the smallest number from the largest number.
21 53 74 82 84 93 103 108 116 122
122 - 21 = 101 points
The range without the outlier is 101 points.
To find the interquartile range wthout the outlier, we must split the numbers into quarters.
21 53 74 82 84 93 103 108 116 122
The first quartile is 53+74 / 2 = 127 / 2 = 63.5.
The third quartile 108+116 / 2 = 224 / 2 = 112.
The interquartile is the first quartile subtracted from the third quartile, so
112 - 63.5 = 48.5.
The interquartile range without the outlier is 48.5 points.
c. Which measure did the outlier affect more?
The outlier affected __________ more. (We cannot see the drop-down options, so cannot answer this part of the question. Sorry.)
