3х – 8 +7х + 10 А 10x + 2 В 4х – 80 с 21х - 80 D — 5х + 17х
1 answer:
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Answer:
a)
b)
c)
With a frequency of 4
d)
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:
Replacing we got:
Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:
Part c
The mode is the most repeated value in the sample and for this case is:
With a frequency of 4
Part d
The midrange for this case is defined as:
Part e
For this case we can calculate the deviation given by:
And replacing we got:
And we can find the limits without any outliers using two deviations from the mean and we got:
And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case