I need to evaluate the limit and I also need to show the work for it. Please help me out if you can.
1 answer:
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Answer:
A)Mean :10.65
Median =10.7
Mode : 10.7
B)Range = 3.5
Standard deviation :0.89916
C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times
Step-by-step explanation:
A)
Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2
Mean = 10.65
Median: The mid value of the data
Data in ascending order
8.3
10.3
10.5
10.6
10.7
10.7
10.9
11.2
11.5
11.8
n=10(even)
Mode : the most occurring frequency
10.7 is occurring twice while others are occurring once
So, Mode is 10.7
B) Range = Maximum - Minimum=11.8-8.3=3.5
Standard deviation :
Standard deviation :
Standard deviation :0.89916
C)
8.3
,10.3
,10.5
,10.6
,10.7
,10.7
,10.9
,11.2
,11.5
,11.8
For Q1 ( Median of lower quartile )
8.3
,10.3
,10.5
,10.6
,10.7
For Q3( Median of Upper quartile )
10.7
,10.9
,11.2
,11.5
,11.8
IQR = Q3-Q1=11.2-10.5=0.7
Range :(Q1-1.5IQR, Q3-1.5IQR)
Range :
Range :(9.45, 10.15)
8.3 does not lie in this interval
So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times